Welcome dear students to the complete English medium guide for ASSEB Class 10 Mathematics Chapter 3 — Pair of Linear Equations in Two Variables. This chapter teaches you how to handle two linear equations together — equations that share the same two unknowns $x$ and $y$. You will learn what it means for such a pair to have a unique solution, no solution, or infinitely many solutions, both algebraically and geometrically. We solve every textbook exercise (3.1 through 3.6) using the graphical method, the substitution method, the elimination method, the cross-multiplication method and reduction to linear form. Three coordinate-plane figures show the three geometric cases. A formula and condition table, additional practice questions, and a glossary round out the lesson so that you walk into your ASSEB board examination with confidence.
Chapter Summary. A linear equation in two variables has the form $ax+by+c=0$, where $a$, $b$, $c$ are real numbers and $a$ and $b$ are not both zero. A pair of such equations is studied together. Geometrically each equation represents a straight line in the Cartesian plane, so a pair represents two lines. These two lines can do exactly one of three things — intersect at a single point (unique solution, consistent system), be parallel and never meet (no solution, inconsistent system), or coincide so that every point of one line lies on the other (infinitely many solutions, dependent and consistent). The behaviour can be predicted from the coefficients alone by comparing the ratios $\frac{a_1}{a_2}$, $\frac{b_1}{b_2}$, $\frac{c_1}{c_2}$. Algebraic methods — substitution, elimination and cross-multiplication — give exact solutions efficiently, while the graphical method gives a clear visual interpretation. Some equations that are not linear can be made linear by a clever substitution; these are called equations reducible to linear form.
Key formulas and concepts.
| Concept | Formula / Statement |
|---|---|
| General linear equation | $ax+by+c=0$, with $a^2+b^2\ne 0$ |
| General pair | $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ |
| Solution of pair | An ordered pair $(x,y)$ satisfying both equations |
| Substitution method | Express one variable from one equation, substitute into the other |
| Elimination method | Multiply equations to make one coefficient equal, then add or subtract |
| Cross-multiplication | $\dfrac{x}{b_1c_2-b_2c_1}=\dfrac{y}{c_1a_2-c_2a_1}=\dfrac{1}{a_1b_2-a_2b_1}$ |
| Slope of $ax+by+c=0$ | $m=-\dfrac{a}{b}$ (when $b\ne 0$) |
| Reducible form example | Put $u=\dfrac{1}{x}$, $v=\dfrac{1}{y}$ to linearise $\dfrac{p}{x}+\dfrac{q}{y}=r$ |
Conditions on the coefficients (very important for board exams).
| Ratio condition | Algebraic outcome | Geometric outcome | Type of system |
|---|---|---|---|
| $\dfrac{a_1}{a_2}\ne\dfrac{b_1}{b_2}$ | Exactly one solution | Lines intersect at one point | Consistent (independent) |
| $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\ne\dfrac{c_1}{c_2}$ | No solution | Lines are parallel | Inconsistent |
| $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$ | Infinitely many solutions | Lines coincide | Consistent (dependent) |
Figure 1 — Intersecting lines (unique solution). Lines $L_1: x+y=4$ and $L_2: x-y=2$ meet at the point $(3,1)$.
Figure 2 — Parallel lines (no solution). Lines $L_1: x+2y=4$ and $L_2: 2x+4y=12$ never meet because their slopes are equal but their intercepts differ.
Figure 3 — Coincident lines (infinitely many solutions). Lines $L_1: 2x+3y=6$ and $L_2: 4x+6y=12$ are the same line drawn twice.
Exercise 3.1 — Forming and graphing pairs of equations
Q1. Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” Represent the situation algebraically and graphically.
Answer: Let Aftab’s present age be $x$ years and his daughter’s present age be $y$ years. Seven years ago, $(x-7)=7(y-7)$, i.e. $x-7y+42=0$. Three years from now, $(x+3)=3(y+3)$, i.e. $x-3y-6=0$. Plotting both lines, they intersect at $(42,12)$, so Aftab is $42$ years and the daughter is $12$ years old.
Q2. The coach of a cricket team buys 3 bats and 6 balls for ₹3900. Later he buys another bat and 3 more balls of the same kind for ₹1300. Represent the situation algebraically and geometrically.
Answer: Let the cost of one bat be ₹$x$ and one ball be ₹$y$. Then $3x+6y=3900$ and $x+3y=1300$. Both equations represent straight lines on the Cartesian plane and they meet at exactly one point $(1300,\,0)$ in this idealised model — but the realistic intersection (after careful plotting) gives $x=300$ and $y=500$ when we use $3x+6y=3900\Rightarrow x+2y=1300$ together with $x+3y=1300$, yielding $y=0$ from subtraction; the textbook intends only graphical representation, not solution here.
Q3. The cost of 2 kg of apples and 1 kg of grapes on a day was ₹160. After a month, the cost of 4 kg of apples and 2 kg of grapes is ₹300. Represent the situation algebraically and geometrically.
Answer: Let the cost of $1$ kg apples be $x$ and $1$ kg grapes be $y$. Then $2x+y=160$ and $4x+2y=300$. Dividing the second by $2$ gives $2x+y=150$. Since $\dfrac{2}{4}=\dfrac{1}{2}\ne\dfrac{160}{300}$, the lines are parallel — Figure 2 case.
Exercise 3.2 — Graphical method and consistency
Q1(i). 10 students of class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part.
Answer: Let boys be $x$ and girls be $y$. Then $x+y=10$ and $y-x=4$. Adding gives $2y=14\Rightarrow y=7$, so $x=3$. Boys $=3$, girls $=7$.
Q1(ii). 5 pencils and 7 pens together cost ₹50, whereas 7 pencils and 5 pens together cost ₹46. Find the cost of one pencil and that of one pen.
Answer: Let a pencil cost ₹$x$ and a pen ₹$y$. Then $5x+7y=50$ and $7x+5y=46$. Adding: $12x+12y=96\Rightarrow x+y=8$. Subtracting: $-2x+2y=4\Rightarrow y-x=2$. Solving: $y=5$, $x=3$. So a pencil costs ₹$3$ and a pen costs ₹$5$.
Q2. Compare the ratios $\frac{a_1}{a_2},\ \frac{b_1}{b_2},\ \frac{c_1}{c_2}$ and tell whether the lines representing the following pairs of linear equations intersect, are parallel or are coincident.
(i) $5x-4y+8=0$ and $7x+6y-9=0$.
Answer: $\dfrac{a_1}{a_2}=\dfrac{5}{7}$, $\dfrac{b_1}{b_2}=\dfrac{-4}{6}=-\dfrac{2}{3}$. Since $\dfrac{a_1}{a_2}\ne\dfrac{b_1}{b_2}$, the lines intersect.
(ii) $9x+3y+12=0$ and $18x+6y+24=0$.
Answer: $\dfrac{9}{18}=\dfrac{3}{6}=\dfrac{12}{24}=\dfrac{1}{2}$. All three ratios equal, so the lines are coincident.
(iii) $6x-3y+10=0$ and $2x-y+9=0$.
Answer: $\dfrac{6}{2}=3$, $\dfrac{-3}{-1}=3$, $\dfrac{10}{9}\ne 3$. So $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\ne\dfrac{c_1}{c_2}$ and the lines are parallel.
Q3. On comparing the ratios, find out whether the following pair of linear equations are consistent or inconsistent.
(i) $3x+2y=5$; $2x-3y=7$.
Answer: $\dfrac{3}{2}\ne\dfrac{2}{-3}$, hence consistent (unique solution).
(ii) $2x-3y=8$; $4x-6y=9$.
Answer: $\dfrac{2}{4}=\dfrac{-3}{-6}\ne\dfrac{8}{9}$, so the system is inconsistent.
(iii) $\frac{3}{2}x+\frac{5}{3}y=7$; $9x-10y=14$.
Answer: $\dfrac{3/2}{9}=\dfrac{1}{6}$, $\dfrac{5/3}{-10}=-\dfrac{1}{6}$. Since $\dfrac{1}{6}\ne-\dfrac{1}{6}$, the pair is consistent.
Q4. Solve graphically: $x+y=5$ and $2x+2y=10$.
Answer: Dividing the second by $2$ gives $x+y=5$, the same line. The system has infinitely many solutions; geometrically the lines coincide (Figure 3 case). Any point on $x+y=5$ — for instance $(0,5)$, $(2,3)$, $(5,0)$ — is a solution.
Q5. Half the perimeter of a rectangular garden whose length is 4 m more than its width is 36 m. Find the dimensions of the garden.
Answer: Let length $=x$ m and width $=y$ m. Then $x+y=36$ and $x-y=4$. Adding: $2x=40\Rightarrow x=20$, hence $y=16$. The garden is $20$ m by $16$ m.
Exercise 3.3 — Substitution method
Q1. Solve the following pair of linear equations by the substitution method.
(i) $x+y=14$, $x-y=4$.
Answer: From the first equation $x=14-y$. Substituting into the second, $(14-y)-y=4\Rightarrow 14-2y=4\Rightarrow y=5$, so $x=9$. Solution: $(9,5)$.
(ii) $s-t=3$, $\dfrac{s}{3}+\dfrac{t}{2}=6$.
Answer: $s=t+3$. Substituting, $\dfrac{t+3}{3}+\dfrac{t}{2}=6$. Multiply by $6$: $2(t+3)+3t=36\Rightarrow 5t+6=36\Rightarrow t=6$, so $s=9$.
(iii) $3x-y=3$, $9x-3y=9$.
Answer: The second equation is exactly $3$ times the first, so the lines coincide. Infinitely many solutions of the form $x=t,\ y=3t-3$ for any real $t$.
(iv) $0.2x+0.3y=1.3$, $0.4x+0.5y=2.3$.
Answer: Multiply both sides of each equation by $10$: $2x+3y=13$ and $4x+5y=23$. From the first, $x=\dfrac{13-3y}{2}$. Substituting: $4\cdot\dfrac{13-3y}{2}+5y=23\Rightarrow 26-6y+5y=23\Rightarrow y=3$, hence $x=2$.
(v) $\sqrt{2}\,x+\sqrt{3}\,y=0$, $\sqrt{3}\,x-\sqrt{8}\,y=0$.
Answer: From the first $x=-\dfrac{\sqrt{3}}{\sqrt{2}}y$. Substituting into the second: $\sqrt{3}\bigl(-\dfrac{\sqrt{3}}{\sqrt{2}}y\bigr)-2\sqrt{2}\,y=0\Rightarrow -\dfrac{3}{\sqrt{2}}y-2\sqrt{2}\,y=0$. Multiplying by $\sqrt{2}$ gives $-3y-4y=0$, so $y=0$ and $x=0$. Unique solution $(0,0)$.
(vi) $\dfrac{3x}{2}-\dfrac{5y}{3}=-2$, $\dfrac{x}{3}+\dfrac{y}{2}=\dfrac{13}{6}$.
Answer: Multiply the first by $6$: $9x-10y=-12$. Multiply the second by $6$: $2x+3y=13$. From here, $x=\dfrac{13-3y}{2}$. Substituting: $9\cdot\dfrac{13-3y}{2}-10y=-12\Rightarrow 117-27y-20y=-24\Rightarrow -47y=-141\Rightarrow y=3$, so $x=2$.
Q2. Solve $2x+3y=11$ and $2x-4y=-24$ and hence find the value of $m$ for which $y=mx+3$.
Answer: Subtracting: $7y=35\Rightarrow y=5$, so $2x=11-15=-4$, giving $x=-2$. Then $5=m(-2)+3\Rightarrow m=-1$.
Q3(i). The difference between two numbers is 26 and one number is three times the other. Find them.
Answer: Let the numbers be $x>y$. Then $x-y=26$ and $x=3y$. Substituting, $3y-y=26\Rightarrow y=13$, $x=39$.
Q3(ii). The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Answer: Let the angles be $x$ and $y$ with $x>y$. Then $x+y=180^\circ$ and $x-y=18^\circ$. Adding: $2x=198\Rightarrow x=99^\circ$, hence $y=81^\circ$.
Q3(iii). The coach of a cricket team buys 7 bats and 6 balls for ₹3800. Later, she buys 3 bats and 5 balls for ₹1750. Find the cost of each bat and each ball.
Answer: $7x+6y=3800$ and $3x+5y=1750$. From the second, $x=\dfrac{1750-5y}{3}$. Substituting in the first: $7\cdot\dfrac{1750-5y}{3}+6y=3800$. Multiply by $3$: $7(1750-5y)+18y=11400\Rightarrow 12250-35y+18y=11400\Rightarrow -17y=-850\Rightarrow y=50$, then $x=500$. So a bat costs ₹$500$ and a ball costs ₹$50$.
Q3(iv). The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For 10 km the charge paid is ₹105 and for a journey of 15 km the charge paid is ₹155. What is the fixed charge and the charge per km? How much will a person have to pay for 25 km?
Answer: Let the fixed charge be ₹$x$ and the per-km charge be ₹$y$. Then $x+10y=105$ and $x+15y=155$. Subtracting, $5y=50\Rightarrow y=10$, hence $x=5$. For $25$ km the fare is $5+25\times 10=$ ₹$255$.
Q3(v). A fraction becomes $\dfrac{9}{11}$ if 2 is added to both the numerator and the denominator. If 3 is added to both, the fraction becomes $\dfrac{5}{6}$. Find the fraction.
Answer: Let the fraction be $\dfrac{x}{y}$. Then $\dfrac{x+2}{y+2}=\dfrac{9}{11}\Rightarrow 11x-9y=-4$ and $\dfrac{x+3}{y+3}=\dfrac{5}{6}\Rightarrow 6x-5y=-3$. Solve: from the second $x=\dfrac{5y-3}{6}$. Substitute: $11\cdot\dfrac{5y-3}{6}-9y=-4\Rightarrow 55y-33-54y=-24\Rightarrow y=9$, hence $x=7$. The fraction is $\dfrac{7}{9}$.
Q3(vi). Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Answer: Let Jacob be $x$ and son be $y$. Then $x+5=3(y+5)\Rightarrow x-3y=10$ and $x-5=7(y-5)\Rightarrow x-7y=-30$. Subtracting: $4y=40\Rightarrow y=10$, so $x=40$. Jacob is $40$ years and his son is $10$ years.
Exercise 3.4 — Elimination method
Q1. Solve the following pair of linear equations by the elimination method.
(i) $x+y=5$ and $2x-3y=4$.
Answer: Multiply the first by $3$: $3x+3y=15$. Add to the second: $5x=19\Rightarrow x=\dfrac{19}{5}$. Then $y=5-x=\dfrac{6}{5}$.
(ii) $3x+4y=10$ and $2x-2y=2$.
Answer: Multiply the second by $2$: $4x-4y=4$. Add to the first: $7x=14\Rightarrow x=2$, hence $y=1$.
(iii) $3x-5y-4=0$ and $9x=2y+7$.
Answer: Rewrite: $3x-5y=4$ and $9x-2y=7$. Multiply the first by $3$: $9x-15y=12$. Subtract: $13y=-5\Rightarrow y=-\dfrac{5}{13}$. Then $3x=4+5y=4-\dfrac{25}{13}=\dfrac{27}{13}\Rightarrow x=\dfrac{9}{13}$.
(iv) $\dfrac{x}{2}+\dfrac{2y}{3}=-1$ and $x-\dfrac{y}{3}=3$.
Answer: Multiply the first by $6$: $3x+4y=-6$. Multiply the second by $3$: $3x-y=9$. Subtract: $5y=-15\Rightarrow y=-3$, so $3x=9-3=6\Rightarrow x=2$.
Q2(i). If we add 1 to the numerator and subtract 1 from the denominator a fraction reduces to 1. It becomes $\dfrac{1}{2}$ if we only add 1 to the denominator. Find the fraction.
Answer: Let the fraction be $\dfrac{x}{y}$. From the first condition $x+1=y-1\Rightarrow y-x=2$. From the second, $\dfrac{x}{y+1}=\dfrac{1}{2}\Rightarrow 2x-y=1$. Adding: $x=3$, hence $y=5$. The fraction is $\dfrac{3}{5}$.
Q2(ii). Five years ago Nuri was thrice as old as Sonu. Ten years later Nuri will be twice as old as Sonu. How old are Nuri and Sonu now?
Answer: Let Nuri be $x$ and Sonu be $y$. Then $x-5=3(y-5)\Rightarrow x-3y=-10$ and $x+10=2(y+10)\Rightarrow x-2y=10$. Subtracting: $-y=-20\Rightarrow y=20$, so $x=50$. Nuri is $50$ and Sonu is $20$.
Q2(iii). The sum of the digits of a two-digit number is 9. Also nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
Answer: Let the digits be $x$ (tens) and $y$ (units). Then $x+y=9$ and $9(10x+y)=2(10y+x)\Rightarrow 90x+9y=20y+2x\Rightarrow 88x=11y\Rightarrow 8x=y$. With $x+y=9$ and $y=8x$, $9x=9\Rightarrow x=1$, $y=8$. The number is $18$.
Q2(iv). Meena went to a bank to withdraw ₹2000. She asked the cashier to give her ₹50 and ₹100 notes only. Meena got 25 notes in all. How many notes of ₹50 and ₹100 did she receive?
Answer: Let $x$ be the number of ₹$50$ notes and $y$ be the number of ₹$100$ notes. Then $x+y=25$ and $50x+100y=2000\Rightarrow x+2y=40$. Subtracting the first from the second: $y=15$, so $x=10$. Meena got $10$ notes of ₹$50$ and $15$ notes of ₹$100$.
Q2(v). A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹27 for a book kept for seven days while Susy paid ₹21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Answer: Let the fixed charge (for first three days) be ₹$x$ and the per-extra-day charge be ₹$y$. Then $x+4y=27$ and $x+2y=21$. Subtracting: $2y=6\Rightarrow y=3$, hence $x=15$.
Exercise 3.5 — Cross-multiplication method
The cross-multiplication formula for $a_1x+b_1y+c_1=0$, $a_2x+b_2y+c_2=0$ is
$$\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}.$$
Q1. Which of the following pairs of linear equations has a unique solution, no solution or infinitely many solutions? In case of unique solution, find it by cross-multiplication.
(i) $x-3y-3=0$ and $3x-9y-2=0$.
Answer: $\dfrac{1}{3}=\dfrac{-3}{-9}\ne\dfrac{-3}{-2}$, so the system has no solution.
(ii) $2x+y=5$ and $3x+2y=8$.
Answer: Write as $2x+y-5=0$ and $3x+2y-8=0$. Then $a_1b_2-a_2b_1=4-3=1\ne 0$, so unique. $x=\dfrac{(1)(-8)-(2)(-5)}{1}=2$, $y=\dfrac{(-5)(3)-(-8)(2)}{1}=1$. Solution $(2,1)$.
(iii) $3x-5y=20$ and $6x-10y=40$.
Answer: $\dfrac{3}{6}=\dfrac{-5}{-10}=\dfrac{20}{40}=\dfrac{1}{2}$, so the system has infinitely many solutions.
(iv) $x-3y-7=0$ and $3x-3y-15=0$.
Answer: $a_1b_2-a_2b_1=(-3)(1)-(-3)(3)=-3+9=6\ne 0$. So unique. $x=\dfrac{(-3)(-15)-(-3)(-7)}{6}=\dfrac{45-21}{6}=4$, $y=\dfrac{(-7)(3)-(-15)(1)}{6}=\dfrac{-21+15}{6}=-1$. Hence $(4,-1)$.
Q2(i). For which values of $a$ and $b$ does the following pair of linear equations have an infinite number of solutions: $2x+3y=7$, $(a-b)x+(a+b)y=3a+b-2$?
Answer: We need $\dfrac{2}{a-b}=\dfrac{3}{a+b}=\dfrac{7}{3a+b-2}$. From the first equality: $2(a+b)=3(a-b)\Rightarrow a=5b$. From the second: $3(3a+b-2)=7(a+b)\Rightarrow 9a+3b-6=7a+7b\Rightarrow 2a-4b=6\Rightarrow a-2b=3$. With $a=5b$, $5b-2b=3\Rightarrow b=1$, $a=5$.
Q2(ii). For which value of $k$ will the following pair of linear equations have no solution: $3x+y=1$, $(2k-1)x+(k-1)y=2k+1$?
Answer: Required $\dfrac{3}{2k-1}=\dfrac{1}{k-1}\ne\dfrac{1}{2k+1}$. From the equality: $3(k-1)=2k-1\Rightarrow k=2$. Check that the third ratio differs: $\dfrac{1}{k-1}=1$ while $\dfrac{1}{2k+1}=\dfrac{1}{5}$, indeed different. So $k=2$.
Q3. Solve $8x+5y=9$ and $3x+2y=4$ by both substitution and cross-multiplication. Which method do you prefer?
Answer: By substitution: $y=\dfrac{4-3x}{2}$. Substitute: $8x+5\cdot\dfrac{4-3x}{2}=9\Rightarrow 16x+20-15x=18\Rightarrow x=-2$, hence $y=5$. By cross-multiplication on $8x+5y-9=0$ and $3x+2y-4=0$: $a_1b_2-a_2b_1=16-15=1$, $b_1c_2-b_2c_1=(5)(-4)-(2)(-9)=-2$, $c_1a_2-c_2a_1=(-9)(3)-(-4)(8)=5$. So $x=-2$, $y=5$. Cross-multiplication is faster when both coefficients are non-trivial.
Q4(i). A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay ₹1000 as hostel charges whereas a student B who takes food for 26 days pays ₹1180. Find the fixed charge and the cost of food per day.
Answer: Let the fixed charge be ₹$x$ and the per-day food cost be ₹$y$. Then $x+20y=1000$ and $x+26y=1180$. Subtracting: $6y=180\Rightarrow y=30$, hence $x=400$.
Q4(ii). A fraction becomes $\dfrac{1}{3}$ when 1 is subtracted from the numerator and it becomes $\dfrac{1}{4}$ when 8 is added to its denominator. Find the fraction.
Answer: Let the fraction be $\dfrac{x}{y}$. Then $\dfrac{x-1}{y}=\dfrac{1}{3}\Rightarrow 3x-y=3$ and $\dfrac{x}{y+8}=\dfrac{1}{4}\Rightarrow 4x-y=8$. Subtracting: $x=5$, so $y=12$. The fraction is $\dfrac{5}{12}$.
Q4(iii). Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, Yash would have scored 50 marks. How many questions were there in the test?
Answer: Let $x$ be the number of right and $y$ be the number of wrong answers. Then $3x-y=40$ and $4x-2y=50$, i.e. $2x-y=25$. Subtracting: $x=15$, hence $y=5$. Total questions $=20$.
Q4(iv). Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other they meet in 1 hour. What are the speeds of the two cars?
Answer: Let speeds be $u$ km/h (faster) and $v$ km/h. Same direction: $5u-5v=100\Rightarrow u-v=20$. Opposite direction: $u+v=100$. Adding: $2u=120\Rightarrow u=60$, $v=40$.
Q4(v). The area of a rectangle gets reduced by 9 sq units if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and breadth by 2 units, the area is increased by 67 sq units. Find the dimensions of the rectangle.
Answer: Let length $=x$ and breadth $=y$. Then $(x-5)(y+3)=xy-9\Rightarrow 3x-5y=6$ and $(x+3)(y+2)=xy+67\Rightarrow 2x+3y=61$. Solving simultaneously (multiply the first by $3$ and the second by $5$, then add): $9x-15y+10x+15y=18+305\Rightarrow 19x=323\Rightarrow x=17$, hence $y=9$.
Exercise 3.6 — Equations reducible to a pair of linear equations
Q1(i). $\dfrac{1}{2x}+\dfrac{1}{3y}=2$ and $\dfrac{1}{3x}+\dfrac{1}{2y}=\dfrac{13}{6}$.
Answer: Put $u=\dfrac{1}{x}$, $v=\dfrac{1}{y}$. Then $\dfrac{u}{2}+\dfrac{v}{3}=2\Rightarrow 3u+2v=12$ and $\dfrac{u}{3}+\dfrac{v}{2}=\dfrac{13}{6}\Rightarrow 2u+3v=13$. Solving: multiply the first by $3$ and the second by $2$ and subtract: $9u+6v-4u-6v=36-26\Rightarrow 5u=10\Rightarrow u=2$, hence $v=3$. Therefore $x=\dfrac{1}{2}$, $y=\dfrac{1}{3}$.
Q1(ii). $\dfrac{2}{\sqrt{x}}+\dfrac{3}{\sqrt{y}}=2$ and $\dfrac{4}{\sqrt{x}}-\dfrac{9}{\sqrt{y}}=-1$.
Answer: Put $u=\dfrac{1}{\sqrt{x}}$, $v=\dfrac{1}{\sqrt{y}}$: $2u+3v=2$, $4u-9v=-1$. Multiply the first by $2$: $4u+6v=4$. Subtract: $15v=5\Rightarrow v=\dfrac{1}{3}$, then $2u=2-1=1\Rightarrow u=\dfrac{1}{2}$. So $\sqrt{x}=2,\ \sqrt{y}=3$, giving $x=4,\ y=9$.
Q1(iii). $\dfrac{4}{x}+3y=14$ and $\dfrac{3}{x}-4y=23$.
Answer: Let $u=\dfrac{1}{x}$. Then $4u+3y=14$ and $3u-4y=23$. Multiply the first by $4$ and the second by $3$ and add: $16u+12y+9u-12y=56+69\Rightarrow 25u=125\Rightarrow u=5$. Then $3y=14-20=-6\Rightarrow y=-2$. Hence $x=\dfrac{1}{5}$, $y=-2$.
Q1(iv). $\dfrac{5}{x-1}+\dfrac{1}{y-2}=2$ and $\dfrac{6}{x-1}-\dfrac{3}{y-2}=1$.
Answer: Let $u=\dfrac{1}{x-1}$, $v=\dfrac{1}{y-2}$. Then $5u+v=2$ and $6u-3v=1$. Multiply the first by $3$: $15u+3v=6$. Add: $21u=7\Rightarrow u=\dfrac{1}{3}$, hence $v=2-\dfrac{5}{3}=\dfrac{1}{3}$. So $x-1=3,\ y-2=3$, giving $x=4,\ y=5$.
Q1(v). $\dfrac{7x-2y}{xy}=5$ and $\dfrac{8x+7y}{xy}=15$.
Answer: Rewrite: $\dfrac{7}{y}-\dfrac{2}{x}=5$ and $\dfrac{8}{y}+\dfrac{7}{x}=15$. Put $u=\dfrac{1}{x}$, $v=\dfrac{1}{y}$: $-2u+7v=5$, $7u+8v=15$. Multiply the first by $7$ and the second by $2$: $-14u+49v=35$ and $14u+16v=30$. Add: $65v=65\Rightarrow v=1$, hence $-2u+7=5\Rightarrow u=1$. So $x=1,\ y=1$.
Q1(vi). $6x+3y=6xy$ and $2x+4y=5xy$.
Answer: Divide both equations by $xy$: $\dfrac{6}{y}+\dfrac{3}{x}=6$ and $\dfrac{2}{y}+\dfrac{4}{x}=5$. With $u=\dfrac{1}{x}$, $v=\dfrac{1}{y}$: $3u+6v=6$ and $4u+2v=5$. Multiply the second by $3$: $12u+6v=15$, then subtract the first: $9u=9\Rightarrow u=1$, hence $v=\dfrac{1}{2}$. So $x=1$, $y=2$.
Q1(vii). $\dfrac{10}{x+y}+\dfrac{2}{x-y}=4$ and $\dfrac{15}{x+y}-\dfrac{5}{x-y}=-2$.
Answer: Let $u=\dfrac{1}{x+y}$, $v=\dfrac{1}{x-y}$. Then $10u+2v=4$ and $15u-5v=-2$. From the first $5u+v=2$. Multiply by $5$: $25u+5v=10$. Add to second: $40u=8\Rightarrow u=\dfrac{1}{5}$, hence $v=2-1=1$. So $x+y=5$ and $x-y=1$, giving $x=3,\ y=2$.
Q1(viii). $\dfrac{1}{3x+y}+\dfrac{1}{3x-y}=\dfrac{3}{4}$ and $\dfrac{1}{2(3x+y)}-\dfrac{1}{2(3x-y)}=-\dfrac{1}{8}$.
Answer: Let $u=\dfrac{1}{3x+y}$, $v=\dfrac{1}{3x-y}$. Then $u+v=\dfrac{3}{4}$ and $u-v=-\dfrac{1}{4}$. Adding: $2u=\dfrac{1}{2}\Rightarrow u=\dfrac{1}{4}$, $v=\dfrac{1}{2}$. So $3x+y=4$ and $3x-y=2$. Adding: $6x=6\Rightarrow x=1$, hence $y=1$.
Q2 (selected). Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
Answer: Let still-water speed be $x$ km/h and current speed be $y$ km/h. Downstream effective speed $=x+y=10$ and upstream $=x-y=2$. So $x=6$, $y=4$.
Q2 (selected). 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone and 1 man alone.
Answer: Let one woman’s one-day work be $\dfrac{1}{x}$ and one man’s be $\dfrac{1}{y}$. Then $\dfrac{2}{x}+\dfrac{5}{y}=\dfrac{1}{4}$ and $\dfrac{3}{x}+\dfrac{6}{y}=\dfrac{1}{3}$. Put $u=\dfrac{1}{x}$, $v=\dfrac{1}{y}$: $2u+5v=\dfrac{1}{4}$ and $3u+6v=\dfrac{1}{3}$. Multiply by $12$: $24u+60v=3$ and $36u+72v=4$. Solving (eliminate $u$): from the first $u=\dfrac{3-60v}{24}$. Substituting: $36\cdot\dfrac{3-60v}{24}+72v=4\Rightarrow \dfrac{3(3-60v)}{2}+72v=4\Rightarrow 9-180v+144v=8\Rightarrow -36v=-1\Rightarrow v=\dfrac{1}{36}$, hence $u=\dfrac{1}{18}$. So one woman alone takes $18$ days and one man alone takes $36$ days.
Q2 (selected). Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speeds of the train and the bus.
Answer: Let train speed be $x$ km/h and bus speed be $y$ km/h. Then $\dfrac{60}{x}+\dfrac{240}{y}=4$ and $\dfrac{100}{x}+\dfrac{200}{y}=\dfrac{25}{6}$. With $u=\dfrac{1}{x}$, $v=\dfrac{1}{y}$: $60u+240v=4$ and $100u+200v=\dfrac{25}{6}$. Multiply through to clear fractions and solve to get $u=\dfrac{1}{60}$, $v=\dfrac{1}{80}$. So $x=60$ km/h and $y=80$ km/h.
Additional Practice Questions
Q1. Graphically the pair $7x-y=5$, $21x-3y=10$ represents which kind of lines?
Answer: Ratios are $\dfrac{7}{21}=\dfrac{1}{3}$, $\dfrac{-1}{-3}=\dfrac{1}{3}$, $\dfrac{-5}{-10}=\dfrac{1}{2}$. Two ratios equal but not the third — the lines are parallel.
Q2. For what value of $k$ are the lines $3x+2ky=2$ and $2x+5y+1=0$ parallel?
Answer: Required $\dfrac{3}{2}=\dfrac{2k}{5}$ (with the constant ratio different). Solving, $2k\cdot 2=15\Rightarrow k=\dfrac{15}{4}$. Adjusting signs to match $-1$ vs $+2$ on the constants confirms parallelism (not coincidence).
Q3. The pair of equations $x=0$ and $y=-7$ has how many solutions?
Answer: $x=0$ is the $y$-axis and $y=-7$ is a horizontal line. They intersect at the unique point $(0,-7)$ — exactly one solution.
Q4. The pair $x=a$ and $y=b$ graphically represents lines that meet where?
Answer: The vertical line $x=a$ meets the horizontal line $y=b$ at $(a,b)$, so the unique solution is $(a,b)$.
Q5. The graph of $x=-2$ is a line parallel to which axis?
Answer: The line $x=-2$ is vertical, hence parallel to the $y$-axis.
Q6. Two coincident lines have how many common solutions?
Answer: Every point of one is a point of the other, so there are infinitely many solutions.
Q7. If a pair of linear equations is consistent, the lines are which kind?
Answer: Either intersecting (one solution) or coincident (infinite solutions). Parallel lines correspond to inconsistent systems.
Q8. Solve $\dfrac{x}{2}+\dfrac{y}{3}=8$ and $\dfrac{5x}{4}-3y=-3$ by elimination.
Answer: Multiply by $6$: $3x+2y=48$. Multiply the second by $4$: $5x-12y=-12$. Multiply the first by $6$: $18x+12y=288$. Add: $23x=276\Rightarrow x=12$, hence $2y=48-36=12\Rightarrow y=6$.
Q9. The sum of two numbers is $35$ and their difference is $13$. Find the numbers.
Answer: $x+y=35$, $x-y=13$. Adding: $x=24$, hence $y=11$.
Q10. If $2x+3y=11$ and $2x-4y=-24$, find $m$ such that $y=mx+3$ holds.
Answer: Subtracting yields $y=5$, so $2x=-4\Rightarrow x=-2$. Then $5=m(-2)+3\Rightarrow m=-1$.
Q11. Find the values of $x$ and $y$ for which the system $3x+5y=15$ and $9x+15y=45$ holds. Comment.
Answer: The second equation is exactly $3$ times the first, so the lines are coincident — infinitely many solutions, e.g. $(0,3),(5,0),(\dfrac{10}{3},1)$ and so on.
Q12. The sum of the numerator and denominator of a fraction is $12$. If the denominator is increased by $3$, the fraction becomes $\dfrac{1}{2}$. Find the fraction.
Answer: Let the fraction be $\dfrac{x}{y}$. Then $x+y=12$ and $\dfrac{x}{y+3}=\dfrac{1}{2}\Rightarrow 2x-y=3$. Adding: $3x=15\Rightarrow x=5$, $y=7$. The fraction is $\dfrac{5}{7}$.
Q13. A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours it can go 40 km upstream and 55 km downstream. Determine the speed of the stream and that of the boat in still water.
Answer: Let boat speed be $x$ and stream speed $y$. Upstream speed $=x-y$, downstream $=x+y$. Setting $u=\dfrac{1}{x-y}$, $v=\dfrac{1}{x+y}$: $30u+44v=10$ and $40u+55v=13$. Solve to get $u=\dfrac{1}{5}$, $v=\dfrac{1}{11}$. So $x-y=5$, $x+y=11$, giving $x=8$ km/h, $y=3$ km/h.
Q14. A two-digit number is 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.
Answer: Let the digits be $x$ (tens) and $y$ (units). Then $10x+y=4(x+y)\Rightarrow 6x-3y=0\Rightarrow 2x=y$ and $10x+y+18=10y+x\Rightarrow 9x-9y=-18\Rightarrow y-x=2$. With $y=2x$: $x=2$, $y=4$. Number is $24$.
Q15. Find a fraction which becomes $\dfrac{1}{2}$ when the numerator is decreased by $1$ and becomes $\dfrac{1}{3}$ when the denominator is increased by $1$.
Answer: Let it be $\dfrac{x}{y}$. Then $\dfrac{x-1}{y}=\dfrac{1}{2}\Rightarrow 2x-y=2$ and $\dfrac{x}{y+1}=\dfrac{1}{3}\Rightarrow 3x-y=1$. Subtracting: $x=-1$? Re-checking signs: $3x-y=1$ minus $2x-y=2$ gives $x=-1$, but a fraction must be positive — the problem requires the standard form $\dfrac{1}{2}$ when decreased and $\dfrac{1}{3}$ when increased; the algebra shows that no positive fraction satisfies both, so we conclude that the system is inconsistent for positive integers. Such corner cases highlight why checking solutions matters.
Glossary
| Term | Meaning |
|---|---|
| Linear equation in two variables | An equation of the form $ax+by+c=0$ with $a$, $b$ not both zero. |
| Pair of linear equations | Two such equations considered together; also called a system. |
| Solution | An ordered pair $(x,y)$ that satisfies both equations simultaneously. |
| Consistent system | A system that has at least one solution (one or infinitely many). |
| Inconsistent system | A system that has no solution; the lines are parallel. |
| Independent system | A consistent system with exactly one solution; lines intersect once. |
| Dependent system | A consistent system with infinitely many solutions; lines coincide. |
| Substitution method | Express one variable in terms of the other from one equation, then substitute. |
| Elimination method | Multiply equations by suitable numbers and add or subtract to eliminate one variable. |
| Cross-multiplication method | Use the formula $\dfrac{x}{b_1c_2-b_2c_1}=\dfrac{y}{c_1a_2-c_2a_1}=\dfrac{1}{a_1b_2-a_2b_1}$. |
| Reducible to linear form | Equations involving $\dfrac{1}{x},\ \dfrac{1}{y}$ or $\dfrac{1}{x\pm y}$ that become linear after a substitution. |
| Slope | For $ax+by+c=0$ with $b\ne 0$, the slope is $-\dfrac{a}{b}$. |
| Intercept | The value at which a line crosses an axis: $x$-intercept when $y=0$, $y$-intercept when $x=0$. |
| Coincident lines | Two lines that are identical as sets of points. |
| Parallel lines | Distinct lines with the same slope; they never meet. |
| Intersecting lines | Two lines meeting at exactly one point. |
Closing note. Practise each method on every exercise question — the substitution method suits problems where one variable is already isolated; elimination is fastest when coefficients line up neatly; cross-multiplication is the cleanest when the algebra is messy and you want a one-shot formula; the graphical method gives a picture of what is happening; and reduction to linear form rescues you when the equations look non-linear at first glance. Master the ratio test for $\dfrac{a_1}{a_2}$, $\dfrac{b_1}{b_2}$, $\dfrac{c_1}{c_2}$, because ASSEB regularly asks you to identify the kind of system without solving it.
Worked Examples — full solutions in detail
Example A. Solve graphically the pair $x+y=6$ and $x-y=2$.
Answer: Build a table for each line. For $x+y=6$, when $x=0$ then $y=6$; when $x=6$ then $y=0$; when $x=2$ then $y=4$. For $x-y=2$, when $x=0$ then $y=-2$; when $x=2$ then $y=0$; when $x=4$ then $y=2$. Plot both lines on graph paper. They cross at the single point $(4,2)$. Verification: $4+2=6$ and $4-2=2$. The unique solution is $(x,y)=(4,2)$.
Example B. Verify algebraically that the system $2x+3y=9$, $4x+6y=18$ has infinitely many solutions and write the general solution.
Answer: Compute the ratios: $\dfrac{a_1}{a_2}=\dfrac{2}{4}=\dfrac{1}{2}$, $\dfrac{b_1}{b_2}=\dfrac{3}{6}=\dfrac{1}{2}$, $\dfrac{c_1}{c_2}=\dfrac{-9}{-18}=\dfrac{1}{2}$. All three ratios are equal, so the lines coincide. From $2x+3y=9$, $y=\dfrac{9-2x}{3}$, so the general solution is $\left(t,\dfrac{9-2t}{3}\right)$ for any real $t$. For instance $t=0$ gives $(0,3)$, $t=3$ gives $(3,1)$ and $t=\dfrac{9}{2}$ gives $\left(\dfrac{9}{2},0\right)$.
Example C. Show that the system $kx+2y=5$, $3x+y=1$ has a unique solution for every value of $k$ except one. Identify the exceptional value.
Answer: The system has a unique solution iff $\dfrac{k}{3}\ne\dfrac{2}{1}$, i.e. $k\ne 6$. Therefore for $k=6$ the system has either no solution or infinitely many. Substituting $k=6$, the equations are $6x+2y=5$ and $3x+y=1$. Multiply the second by $2$: $6x+2y=2$. Compared with $6x+2y=5$, the constants differ, so the lines are parallel and the system has no solution. Hence for $k=6$ the system is inconsistent; for every other real $k$ it has a unique solution.
Example D. The sum of two numbers is $20$. If the sum of their reciprocals is $\dfrac{1}{4}$, find the numbers.
Answer: Let the numbers be $x$ and $y$. Then $x+y=20$ and $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{4}$. The second condition gives $\dfrac{x+y}{xy}=\dfrac{1}{4}$, so $\dfrac{20}{xy}=\dfrac{1}{4}\Rightarrow xy=80$. The numbers are roots of $t^2-20t+80=0$, i.e. $t=\dfrac{20\pm\sqrt{400-320}}{2}=\dfrac{20\pm\sqrt{80}}{2}=10\pm 2\sqrt{5}$. So the numbers are $10+2\sqrt{5}$ and $10-2\sqrt{5}$.
Example E. A man rows a boat $32$ km downstream and $14$ km upstream taking $6$ hours each time. Find the speed of the man in still water and the speed of the stream.
Answer: Let the still-water speed be $x$ km/h and the stream speed be $y$ km/h. Downstream speed $=x+y$ and upstream $=x-y$. Then $\dfrac{32}{x+y}=6$ and $\dfrac{14}{x-y}=6$, so $x+y=\dfrac{32}{6}=\dfrac{16}{3}$ and $x-y=\dfrac{14}{6}=\dfrac{7}{3}$. Adding: $2x=\dfrac{23}{3}\Rightarrow x=\dfrac{23}{6}$ km/h. Subtracting: $2y=3\Rightarrow y=\dfrac{3}{2}$ km/h. Hence the still-water speed is $\dfrac{23}{6}$ km/h and the stream is $\dfrac{3}{2}$ km/h.
Example F. A pair of equations $2x+3y-7=0$, $(a-1)x+(a+1)y-(3a-1)=0$ has infinitely many solutions. Find the value of $a$.
Answer: Set $\dfrac{2}{a-1}=\dfrac{3}{a+1}=\dfrac{-7}{-(3a-1)}=\dfrac{7}{3a-1}$. From the first equality: $2(a+1)=3(a-1)\Rightarrow 2a+2=3a-3\Rightarrow a=5$. Check: $\dfrac{2}{4}=\dfrac{3}{6}=\dfrac{7}{14}=\dfrac{1}{2}$ — yes, all three ratios equal. So $a=5$.
Example G. The pair $kx-y=2$ and $6x-2y=3$ has no solution. Find $k$.
Answer: For no solution we need $\dfrac{k}{6}=\dfrac{-1}{-2}=\dfrac{1}{2}\ne\dfrac{-2}{-3}=\dfrac{2}{3}$. So $\dfrac{k}{6}=\dfrac{1}{2}\Rightarrow k=3$. Verify: $\dfrac{1}{2}\ne\dfrac{2}{3}$ — yes, parallel.
Example H. Solve the system using elimination by equating coefficients: $9x-4y=2000$ and $7x-3y=2000$.
Answer: Multiply the first by $3$ and the second by $4$: $27x-12y=6000$ and $28x-12y=8000$. Subtract: $x=2000$. Substitute into $9(2000)-4y=2000\Rightarrow 18000-4y=2000\Rightarrow 4y=16000\Rightarrow y=4000$. Solution $(2000,4000)$.
Example I. The age of a father is twice the sum of the ages of his two children. After $20$ years, his age will be equal to the sum of the ages of his children. Find the present age of the father.
Answer: Let father’s present age be $x$ and the sum of the children’s ages be $y$. Then $x=2y$ and $x+20=(y+40)$ — because each child gains $20$ years, so the sum gains $40$. Thus $x-y=20$. From $x=2y$, substituting: $2y-y=20\Rightarrow y=20$, so $x=40$. The father is $40$ years old now.
Example J. The sum of the digits of a two-digit number is $12$. The number obtained by interchanging the digits exceeds the original by $18$. Find the number.
Answer: Let the digits be $x$ (tens) and $y$ (units). Original number $=10x+y$, reversed $=10y+x$. Then $x+y=12$ and $(10y+x)-(10x+y)=18\Rightarrow 9y-9x=18\Rightarrow y-x=2$. Adding: $2y=14\Rightarrow y=7$, so $x=5$. Number $=57$.
Example K. Find the value of $\lambda$ for which the lines $\lambda x+3y=\lambda-3$ and $12x+\lambda y=\lambda$ have infinitely many solutions.
Answer: Required $\dfrac{\lambda}{12}=\dfrac{3}{\lambda}=\dfrac{\lambda-3}{\lambda}$. From the first equality: $\lambda^2=36\Rightarrow\lambda=\pm 6$. From the second: $3\lambda=\lambda(\lambda-3)\Rightarrow\lambda^2-6\lambda=0\Rightarrow\lambda=0$ or $\lambda=6$. The common value satisfying both is $\lambda=6$.
Example L. A train covered a certain distance at a uniform speed. If the train would have been $6$ km/h faster, it would have taken $4$ hours less than the scheduled time. And if the train were slower by $6$ km/h, it would have taken $6$ hours more. Find the length of the journey.
Answer: Let the distance be $d$ km and the original speed $v$ km/h, time $t=\dfrac{d}{v}$. Then $\dfrac{d}{v+6}=t-4$ and $\dfrac{d}{v-6}=t+6$. So $d=(v+6)(t-4)$ and $d=(v-6)(t+6)$. Expand the first: $d=vt-4v+6t-24$. Since $d=vt$, we get $-4v+6t-24=0\Rightarrow -2v+3t=12$. Expand the second: $d=vt+6v-6t-36$, giving $6v-6t=36\Rightarrow v-t=6$. From here $v=t+6$. Substituting: $-2(t+6)+3t=12\Rightarrow t=24$, so $v=30$ and $d=vt=720$ km.
Example M. Solve $\dfrac{1}{x+y}+\dfrac{2}{x-y}=2$ and $\dfrac{2}{x+y}-\dfrac{1}{x-y}=3$.
Answer: Let $u=\dfrac{1}{x+y}$, $v=\dfrac{1}{x-y}$. Then $u+2v=2$ and $2u-v=3$. From the first $u=2-2v$. Substituting: $2(2-2v)-v=3\Rightarrow 4-4v-v=3\Rightarrow v=\dfrac{1}{5}$, hence $u=2-\dfrac{2}{5}=\dfrac{8}{5}$. So $x+y=\dfrac{5}{8}$ and $x-y=5$. Adding: $2x=\dfrac{45}{8}\Rightarrow x=\dfrac{45}{16}$. Then $y=\dfrac{45}{16}-5=-\dfrac{35}{16}$.
Example N. The cost of $5$ oranges and $3$ apples is ₹$35$ and the cost of $2$ oranges and $4$ apples is ₹$28$. Find the cost of one orange and one apple.
Answer: Let an orange cost ₹$x$ and an apple ₹$y$. Then $5x+3y=35$ and $2x+4y=28$. Multiply the first by $4$ and the second by $3$: $20x+12y=140$ and $6x+12y=84$. Subtract: $14x=56\Rightarrow x=4$, hence $y=5$.
Example O. ABCD is a cyclic quadrilateral. Find the angles of the quadrilateral if $\angle A=(4y+20)^\circ$, $\angle B=(3y-5)^\circ$, $\angle C=(-4x)^\circ$, $\angle D=(-7x+5)^\circ$.
Answer: In a cyclic quadrilateral opposite angles sum to $180^\circ$. So $\angle A+\angle C=180\Rightarrow 4y+20-4x=180\Rightarrow -4x+4y=160\Rightarrow y-x=40$. And $\angle B+\angle D=180\Rightarrow 3y-5-7x+5=180\Rightarrow -7x+3y=180$. From $y=x+40$, substituting: $-7x+3(x+40)=180\Rightarrow -4x+120=180\Rightarrow x=-15$, so $y=25$. Then $\angle A=4(25)+20=120^\circ$, $\angle B=3(25)-5=70^\circ$, $\angle C=-4(-15)=60^\circ$, $\angle D=-7(-15)+5=110^\circ$. Check $A+C=180$, $B+D=180$. Correct.
Method-by-Method Comparison
| Method | Best Used When | Strength | Limitation |
|---|---|---|---|
| Graphical | Visual interpretation, integer or near-integer answers | Shows the geometric meaning vividly | Imprecise for non-integer answers; tedious for fractions |
| Substitution | One variable is already isolated or has coefficient $1$ | Easy to follow step-by-step | Algebra grows messy with awkward fractions |
| Elimination | Coefficients of one variable are easy multiples of each other | Quick when arithmetic aligns nicely | Requires multiplying both equations sometimes |
| Cross-multiplication | General form $a_1x+b_1y+c_1=0$ with non-trivial coefficients | One-shot formulaic answer | Sign errors common; must memorise formula |
| Reducible-to-linear | Equations with $\dfrac{1}{x},\ \dfrac{1}{y},\ \dfrac{1}{x\pm y}$ | Converts non-linear into a familiar linear pair | Must remember to back-substitute at the end |
Self-Practice Set (Answers below)
P1. Solve by substitution: $3x+y=11$ and $5x-y=13$.
P2. Solve by elimination: $4x+5y=23$ and $7x+y=19$.
P3. By cross-multiplication: $2x+3y=46$ and $3x+5y=74$.
P4. Find $k$ for which $2x+3y=4$ and $(k+2)x+6y=3k+2$ has infinitely many solutions.
P5. Father is $30$ years older than his son. Twelve years from now, his age will be three times that of his son. Find their present ages.
P6. Solve $\dfrac{2}{x}+\dfrac{3}{y}=13$ and $\dfrac{5}{x}-\dfrac{4}{y}=-2$.
P7. The perimeter of a rectangle is $40$ m. The length is $4$ m more than the breadth. Find the dimensions.
P8. Find the values of $a$ and $b$ for which the system $4x+3y=14$, $ax+(a+b)y=4a+b$ has infinitely many solutions.
Answers (P1–P8).
P1. Adding gives $8x=24\Rightarrow x=3$, $y=2$.
P2. Multiply the second by $5$: $35x+5y=95$, subtract the first: $31x=72\Rightarrow x=\dfrac{72}{31}$, then $y=19-7x=\dfrac{19\cdot 31-7\cdot 72}{31}=\dfrac{589-504}{31}=\dfrac{85}{31}$.
P3. $a_1b_2-a_2b_1=10-9=1$, $b_1c_2-b_2c_1=3(-74)-5(-46)=-222+230=8$, $c_1a_2-c_2a_1=(-46)(3)-(-74)(2)=-138+148=10$. So $x=8$, $y=10$.
P4. Need $\dfrac{2}{k+2}=\dfrac{3}{6}=\dfrac{4}{3k+2}$. From the middle ratio $\dfrac{1}{2}=\dfrac{2}{k+2}\Rightarrow k+2=4\Rightarrow k=2$. Check $\dfrac{4}{3(2)+2}=\dfrac{4}{8}=\dfrac{1}{2}$. So $k=2$.
P5. Let father $=x$, son $=y$. Then $x-y=30$ and $x+12=3(y+12)\Rightarrow x-3y=24$. Subtracting: $2y=6\Rightarrow y=3$, $x=33$. Father $33$, son $3$.
P6. Put $u=\dfrac{1}{x}$, $v=\dfrac{1}{y}$: $2u+3v=13$, $5u-4v=-2$. Multiply the first by $4$ and the second by $3$ and add: $8u+12v+15u-12v=52-6\Rightarrow 23u=46\Rightarrow u=2$, hence $v=3$. So $x=\dfrac{1}{2}$, $y=\dfrac{1}{3}$.
P7. $2(\ell+b)=40$ and $\ell-b=4$. So $\ell+b=20$ and $\ell-b=4$, giving $\ell=12$, $b=8$.
P8. $\dfrac{4}{a}=\dfrac{3}{a+b}=\dfrac{14}{4a+b}$. From the first equality $4(a+b)=3a\Rightarrow a+4b=0\Rightarrow a=-4b$. Substitute into the second: $\dfrac{3}{a+b}=\dfrac{14}{4a+b}\Rightarrow 3(4a+b)=14(a+b)\Rightarrow 12a+3b=14a+14b\Rightarrow -2a=11b\Rightarrow a=-\dfrac{11b}{2}$. Combined with $a=-4b$, we need $-4b=-\dfrac{11b}{2}\Rightarrow b(8-11)=0\Rightarrow b=0$, then $a=0$. Substituting back, the original equations would degenerate, so the only formal answer matching all three ratios is $a=8$, $b=5$ when re-checked: $\dfrac{4}{8}=\dfrac{3}{13}\ne$. The textbook answer (NCERT) is $a=5,\ b=1$ derived by writing the system as $4x+3y-14=0$ and $ax+(a+b)y-(4a+b)=0$ and solving $\dfrac{4}{a}=\dfrac{3}{a+b}=\dfrac{14}{4a+b}$ properly, yielding $a=5$, $b=1$. Always recheck signs and constant terms.
Mini Question Bank for Quick Revision
| Question | Short Answer |
|---|---|
| Standard form of a linear equation in two variables | $ax+by+c=0$ |
| How many solutions does a consistent independent system have? | Exactly one |
| Condition for parallel lines (no solution) | $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\ne\dfrac{c_1}{c_2}$ |
| Condition for coincident lines (infinite solutions) | $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$ |
| Condition for unique intersection | $\dfrac{a_1}{a_2}\ne\dfrac{b_1}{b_2}$ |
| Cross-multiplication denominator for $x$ | $b_1c_2-b_2c_1$ |
| Cross-multiplication denominator for $y$ | $c_1a_2-c_2a_1$ |
| Cross-multiplication denominator for $1$ | $a_1b_2-a_2b_1$ |
| Substitution for $\dfrac{1}{x}$ to linearise | Let $u=\dfrac{1}{x}$ |
| Slope of $ax+by+c=0$ | $-\dfrac{a}{b}$ |
| $y$-intercept of $ax+by+c=0$ | $-\dfrac{c}{b}$ |
| $x$-intercept of $ax+by+c=0$ | $-\dfrac{c}{a}$ |
| Two parallel lines have | Equal slopes, different intercepts |
| Two coincident lines have | Equal slopes and equal intercepts |
| Geometric meaning of “no solution” | The lines are parallel |
| Geometric meaning of “infinitely many solutions” | The lines coincide |
| If $a_1b_2-a_2b_1=0$, the system is | Either inconsistent or has infinitely many solutions |
| If $a_1b_2-a_2b_1\ne 0$, the system is | Consistent with a unique solution |
Frequently Asked Board Questions (with hints)
F1. The sum of two numbers is $137$ and their difference is $43$. Find the numbers. Hint: Use $x+y=137$, $x-y=43$, add and subtract. Answer: $x=90$, $y=47$.
F2. The cost of $4$ chairs and $3$ tables is ₹$2100$, while the cost of $5$ chairs and $2$ tables is ₹$1750$. Find the cost of one chair and one table. Hint: $4x+3y=2100$, $5x+2y=1750$. Multiply suitably to eliminate $y$. Answer: Chair ₹$150$, Table ₹$500$.
F3. The sum of a two-digit number and the number obtained by reversing the digits is $66$. The digits differ by $2$. Find the numbers (both possibilities). Hint: Let digits be $x$ and $y$. $(10x+y)+(10y+x)=66\Rightarrow x+y=6$, and $|x-y|=2$. Two cases: $x=4,y=2$ or $x=2,y=4$. Answer: $42$ or $24$.
F4. A boat goes $24$ km upstream and $28$ km downstream in $6$ hours. It goes $30$ km upstream and $21$ km downstream in $6\frac{1}{2}$ hours. Find the speed of the boat in still water and the speed of the stream. Hint: Let still-water speed be $x$ and stream speed $y$. Use $u=\dfrac{1}{x-y}$, $v=\dfrac{1}{x+y}$ to linearise. Answer: $x=10$ km/h, $y=4$ km/h.
F5. Six years hence a man’s age will be three times his son’s age and three years ago he was nine times as old as his son. Find their present ages. Hint: $(x+6)=3(y+6)$ and $(x-3)=9(y-3)$. Answer: Father $30$ years, son $6$ years.
F6. The denominator of a fraction is $4$ more than twice the numerator. When the numerator and denominator are decreased by $6$, the denominator becomes $12$ times the numerator. Find the fraction. Hint: $y=2x+4$, $(y-6)=12(x-6)$. Answer: $\dfrac{7}{18}$.
F7. A man can row $9$ km in one hour downstream and return upstream in $3$ hours. Find the speed of the man in still water and the speed of the stream. Hint: Downstream $=9$, upstream $=\dfrac{9}{3}=3$ km/h. So $x+y=9$, $x-y=3$. Answer: $x=6$ km/h, $y=3$ km/h.
F8. If $2x+y=23$ and $4x-y=19$, find the values of $5y-2x$ and $\dfrac{y}{x}-2$. Hint: Solve the pair to get $x=7$, $y=9$. Then $5y-2x=45-14=31$ and $\dfrac{y}{x}-2=\dfrac{9}{7}-2=-\dfrac{5}{7}$.
F9. The area of a rectangle remains the same if the length is increased by $7$ m and breadth is decreased by $3$ m. The area remains unaffected if the length is decreased by $7$ m and breadth is increased by $5$ m. Find the dimensions of the rectangle. Hint: $(\ell+7)(b-3)=\ell b\Rightarrow -3\ell+7b=21$ and $(\ell-7)(b+5)=\ell b\Rightarrow 5\ell-7b=35$. Answer: $\ell=28$ m, $b=15$ m.
F10. The sum of the present ages of A and B is $58$ years. Six years ago A’s age was twice B’s age. Find their present ages. Hint: $A+B=58$, $A-6=2(B-6)\Rightarrow A-2B=-6$. Answer: $A=\dfrac{2(58)-6}{3}\cdot$ — solve the system to get $A=\dfrac{122-6}{3}$… Compute properly: from $A=58-B$, substitute $58-B-2B=-6\Rightarrow 3B=64\Rightarrow B=\dfrac{64}{3}$. Re-checking the question — for whole-number ages use $A-6=2(B-6)$ giving $A=2B-6$, then with $A+B=58$: $3B=64$. The textbook variant uses “ten years ago” etc.; the present numbers should be checked against your edition.
Common Mistakes to Avoid
| Mistake | Correct Approach |
|---|---|
| Comparing $\dfrac{a_1}{a_2}$ with $\dfrac{c_2}{c_1}$ instead of $\dfrac{c_1}{c_2}$ | Always keep the subscripts in the same order top vs bottom |
| Forgetting to write the equation in $ax+by+c=0$ form before applying the ratio test | Bring all terms to the LHS; signs of $c$ matter |
| In cross-multiplication, mixing up $b_1c_2-b_2c_1$ with $b_2c_1-b_1c_2$ | Memorise the rule: subscripts $1,2$ alternate clockwise |
| Forgetting to back-substitute $u=\dfrac{1}{x}$ to recover $x$ | The final answer must be in $(x,y)$, not $(u,v)$ |
| Plotting only two points per line and being sloppy | Plot at least three points to confirm the line is correct |
| Reading the intersection off graph paper as $(2.9,1.1)$ instead of recognising $(3,1)$ | Snap to grid lines when the answer is clearly an integer |
| Treating “consistent” and “having unique solution” as identical | Consistent means at least one solution — could be one or infinitely many |
Final reminder. When in doubt, use the ratio test first to predict the type of system, then pick the algebraic method that matches the structure of the equations. Verify your answer by substituting back into both equations — never just one. With practice you will gain a feel for which method is fastest for each problem type, and that intuition is exactly what the ASSEB Class 10 board paper rewards.
Multiple Choice Questions (with full reasoning)
MCQ 1. Graphically, the pair of equations $7x-y=5$ and $21x-3y=10$ represents two lines which are: (a) intersecting at one point (b) parallel (c) coincident (d) intersecting at two points.
Answer: (b) parallel. Ratios: $\dfrac{7}{21}=\dfrac{1}{3}$, $\dfrac{-1}{-3}=\dfrac{1}{3}$, $\dfrac{-5}{-10}=\dfrac{1}{2}$. Two ratios match but the third differs, so the lines are parallel and the system has no solution.
MCQ 2. The pair of equations $6x-3y+10=0$ and $2x-y+9=0$ represents: (a) coincident lines (b) intersecting lines (c) parallel lines (d) the $x$-axis and $y$-axis.
Answer: (c) parallel lines. $\dfrac{6}{2}=3$, $\dfrac{-3}{-1}=3$, $\dfrac{10}{9}\ne 3$.
MCQ 3. If a pair of linear equations is consistent, then the lines will be: (a) parallel (b) always coincident (c) intersecting or coincident (d) always intersecting.
Answer: (c) intersecting or coincident.
MCQ 4. The pair $x=0$ and $y=-7$ has: (a) one solution (b) two solutions (c) no solution (d) infinitely many solutions.
Answer: (a) one solution — the lines $x=0$ and $y=-7$ meet at $(0,-7)$.
MCQ 5. The pair $x=a$ and $y=b$ graphically represents lines which are: (a) parallel (b) intersecting at $(b,a)$ (c) coincident (d) intersecting at $(a,b)$.
Answer: (d) intersecting at $(a,b)$.
MCQ 6. For the lines $3x+2ky=2$ and $2x+5y+1=0$ to be parallel, the value of $k$ must be: (a) $-\dfrac{5}{4}$ (b) $\dfrac{2}{5}$ (c) $\dfrac{15}{4}$ (d) $\dfrac{3}{2}$.
Answer: (c) $\dfrac{15}{4}$. Required $\dfrac{3}{2}=\dfrac{2k}{5}\ne\dfrac{-2}{1}$. So $2k\cdot 2=15\Rightarrow k=\dfrac{15}{4}$.
MCQ 7. How many solutions do two linear equations in two variables have if their graphs intersect at one point? (a) one solution (b) two solutions (c) no solution (d) infinitely many solutions.
Answer: (a) one solution.
MCQ 8. Two coincident lines have: (a) no solution (b) exactly one solution (c) exactly two solutions (d) infinitely many solutions.
Answer: (d) infinitely many solutions.
MCQ 9. The pair of equations $x=0$ and $x=5$ has: (a) no solution (b) one solution (c) two solutions (d) infinitely many solutions.
Answer: (a) no solution. Both are vertical lines parallel to the $y$-axis but different, so they never meet.
MCQ 10. The graph of $x=-2$ is a line parallel to the: (a) $x$-axis (b) $y$-axis (c) origin (d) line $y=x$.
Answer: (b) $y$-axis.
MCQ 11. The pair $kx-y=2$ and $6x-2y=3$ has a unique solution when: (a) $k=3$ (b) $k\ne 3$ (c) $k=0$ (d) $k\ne 0$.
Answer: (b) $k\ne 3$.
MCQ 12. The value of $c$ for which the pair $cx-y=2$, $6x-2y=3$ has infinitely many solutions: (a) $3$ (b) $-3$ (c) $-12$ (d) no value of $c$.
Answer: (d) no value. We would need $\dfrac{c}{6}=\dfrac{-1}{-2}=\dfrac{2}{3}$, but $\dfrac{1}{2}\ne\dfrac{2}{3}$, so the third ratio cannot match.
MCQ 13. The line represented by $x+y-4=0$ passes through which of the following points? (a) $(0,4)$ (b) $(1,3)$ (c) $(2,2)$ (d) all of these.
Answer: (d) all of these — every listed pair satisfies $x+y=4$.
MCQ 14. The lines $2x+3y=7$ and $4x+6y=k$ are coincident if $k=$: (a) $7$ (b) $14$ (c) $21$ (d) $28$.
Answer: (b) $14$. $\dfrac{2}{4}=\dfrac{3}{6}=\dfrac{7}{14}=\dfrac{1}{2}$.
MCQ 15. The system $x+y=5$, $2x+2y=10$ has: (a) no solution (b) one solution (c) two solutions (d) infinitely many solutions.
Answer: (d) infinitely many solutions — the second is twice the first, so the lines coincide.
MCQ 16. The system $3x+4y=12$ and $5x+8y=24$ has: (a) a unique solution (b) no solution (c) infinitely many solutions (d) cannot be determined.
Answer: (a) a unique solution. $\dfrac{3}{5}\ne\dfrac{4}{8}=\dfrac{1}{2}$.
MCQ 17. If the system $kx+3y=k-3$, $12x+ky=k$ has infinitely many solutions, then $k=$: (a) $0$ (b) $-6$ (c) $6$ (d) $\pm 6$.
Answer: (c) $6$ (as derived in Example K above).
MCQ 18. The cost of $2$ pencils and $3$ erasers is ₹$9$ and the cost of $4$ pencils and $6$ erasers is ₹$18$. This system represents lines that are: (a) intersecting (b) parallel (c) coincident (d) perpendicular.
Answer: (c) coincident — the second equation is twice the first, so we cannot uniquely determine the prices.
MCQ 19. If $\alpha$ and $\beta$ are the solutions of $2x+3y=12$ and $x-y=1$, then $\alpha+\beta$ equals: (a) $5$ (b) $4$ (c) $6$ (d) $3$.
Answer: (a) $5$. Solving: from $x-y=1$, $x=y+1$; substituting, $2(y+1)+3y=12\Rightarrow 5y=10\Rightarrow y=2$, $x=3$. Sum $=5$.
MCQ 20. Aruna has only ₹$1$ and ₹$2$ coins with her. The total number of coins is $50$ and their total value is ₹$75$. The number of ₹$1$ coins is: (a) $25$ (b) $35$ (c) $30$ (d) $20$.
Answer: (a) $25$. Let ₹$1$ coins $=x$, ₹$2$ coins $=y$. $x+y=50$, $x+2y=75$. Subtracting: $y=25$, so $x=25$.
Long-Answer Practice Questions (Board pattern)
L1. Solve graphically: $2x-y=4$ and $x+y=2$. Also find the area of the triangle formed by these lines and the $y$-axis.
Answer: From $2x-y=4$: $(0,-4),(2,0),(3,2)$. From $x+y=2$: $(0,2),(2,0),(-1,3)$. Both lines pass through $(2,0)$ and meet the $y$-axis at $(0,-4)$ and $(0,2)$. Solving algebraically, $2x-y=4$ and $x+y=2$ add to $3x=6\Rightarrow x=2$, $y=0$. The vertices of the triangle are $A(2,0)$, $B(0,-4)$, $C(0,2)$. Base $BC$ on the $y$-axis has length $|2-(-4)|=6$. Height $=$ distance from $A$ to the $y$-axis $=2$. Area $=\dfrac{1}{2}\cdot 6\cdot 2=6$ square units.
L2. The students of a class are made to stand in rows. If $4$ students are extra in a row, there would be $2$ rows less. If $4$ students are less in a row, there would be $4$ rows more. Find the number of students in the class.
Answer: Let the number of rows be $x$ and the number of students in each row be $y$. Then total students $=xy$. From the first condition: $(x-2)(y+4)=xy\Rightarrow 4x-2y=8\Rightarrow 2x-y=4$. From the second: $(x+4)(y-4)=xy\Rightarrow -4x+4y=16\Rightarrow -x+y=4$. Adding: $x=8$, $y=12$. Total $=8\times 12=96$ students.
L3. A man travels $600$ km partly by train and partly by car. If he covers $400$ km by train and the rest by car, it takes him $6$ hours $30$ minutes. But if he travels $200$ km by train and the rest by car, he takes half an hour longer. Find the speeds of the train and the car.
Answer: Let train speed $=x$, car speed $=y$ km/h. Then $\dfrac{400}{x}+\dfrac{200}{y}=\dfrac{13}{2}$ and $\dfrac{200}{x}+\dfrac{400}{y}=7$. Put $u=\dfrac{1}{x},\ v=\dfrac{1}{y}$: $400u+200v=\dfrac{13}{2}\Rightarrow 800u+400v=13$ and $200u+400v=7$. Subtract: $600u=6\Rightarrow u=\dfrac{1}{100}$. Then $400v=7-2=5\Rightarrow v=\dfrac{1}{80}$. So train $=100$ km/h, car $=80$ km/h.
L4. The age of a father is twice that of his elder son. Ten years from now, the age of the father will be three times that of his younger son. If the difference of the ages of the two sons is $15$ years, find the age of the father.
Answer: Let the father’s age be $F$, elder son $E$, younger son $Y$. Then $F=2E$, $F+10=3(Y+10)\Rightarrow F=3Y+20$, and $E-Y=15$. From $F=2E$, $E=\dfrac{F}{2}$, so $Y=E-15=\dfrac{F}{2}-15$. Substituting into $F=3Y+20$: $F=3\left(\dfrac{F}{2}-15\right)+20\Rightarrow F=\dfrac{3F}{2}-25\Rightarrow -\dfrac{F}{2}=-25\Rightarrow F=50$. Father is $50$ years.
L5. The angles of a triangle are $x$, $y$ and $40^\circ$. The difference between the two angles $x$ and $y$ is $30^\circ$. Find $x$ and $y$.
Answer: $x+y+40=180\Rightarrow x+y=140$. And $x-y=30$ (assuming $x>y$). Adding: $2x=170\Rightarrow x=85^\circ$, $y=55^\circ$.
L6. For what values of $k$ will the system $4x+6y=11$ and $2x+ky=7$ have (i) a unique solution, (ii) no solution? Can it have infinitely many solutions for some $k$?
Answer: (i) Unique solution requires $\dfrac{4}{2}\ne\dfrac{6}{k}$, i.e. $k\ne 3$. (ii) No solution requires $\dfrac{4}{2}=\dfrac{6}{k}\ne\dfrac{11}{7}$, i.e. $k=3$ and $\dfrac{6}{3}=2\ne\dfrac{11}{7}$ — yes parallel. So $k=3$ gives no solution. Infinitely many would require $\dfrac{4}{2}=\dfrac{6}{k}=\dfrac{11}{7}$, but $\dfrac{4}{2}=2\ne\dfrac{11}{7}$. Hence no value of $k$ gives infinitely many solutions.
L7. Solve the system $99x+101y=499$ and $101x+99y=501$ by an elegant trick.
Answer: Add the two equations: $200x+200y=1000\Rightarrow x+y=5$. Subtract: $-2x+2y=-2\Rightarrow y-x=-1\Rightarrow x-y=1$. So $x=3$, $y=2$. The trick — adding and subtracting symmetric equations — saves enormous algebra.
L8. A railway half-ticket costs half the full fare but the reservation charge is the same on both. One reserved first-class ticket from station A to B costs ₹$2530$. The cost of one full ticket plus one half ticket is ₹$3810$. Find the basic fare and the reservation charge.
Answer: Let the full fare be ₹$x$ and the reservation charge be ₹$y$. Then $x+y=2530$ and $x+y+\dfrac{x}{2}+y=3810\Rightarrow \dfrac{3x}{2}+2y=3810\Rightarrow 3x+4y=7620$. From the first $y=2530-x$, substituting: $3x+4(2530-x)=7620\Rightarrow -x=-2500\Rightarrow x=2500$, $y=30$. Fare ₹$2500$, reservation ₹$30$.
L9. The sum of two positive numbers is $60$ and the product of one and the square of the other is $4000$. Set up the equations and indicate why this is not a “linear” pair.
Answer: $x+y=60$, $xy^2=4000$. The second equation has degree $3$ — it is not linear, so the techniques of this chapter do not apply directly. Such systems require quadratic substitution: from $x=60-y$, $(60-y)y^2=4000$. This becomes a cubic in $y$ which lies outside the scope of Chapter 3.
L10. Solve graphically and shade the triangular region bounded by the lines $x+2y=5$, $2x-3y=-4$ and the $x$-axis. Find its area.
Answer: $x+2y=5$ meets the $x$-axis at $(5,0)$ and has $(1,2),(3,1)$ as other points. $2x-3y=-4$ meets the $x$-axis at $(-2,0)$ and passes through $(1,2)$. The two lines intersect: from $x+2y=5$, $x=5-2y$; substituting, $2(5-2y)-3y=-4\Rightarrow 10-7y=-4\Rightarrow y=2$, $x=1$. So the triangle has vertices $(5,0)$, $(-2,0)$ and $(1,2)$. Base on the $x$-axis $=5-(-2)=7$. Height $=2$. Area $=\dfrac{1}{2}\cdot 7\cdot 2=7$ square units.
Key Takeaways for Revision
| Take-Away | Why It Matters |
|---|---|
| Always rewrite both equations in $ax+by+c=0$ form before applying ratio test | Sign errors on $c$ are the most common mistake |
| The three ratios $\frac{a_1}{a_2},\frac{b_1}{b_2},\frac{c_1}{c_2}$ tell you the answer-type before any solving | Saves time on board paper — predict, then solve |
| Substitution shines when one coefficient is $\pm 1$ | Avoids fractions early in the calculation |
| Elimination shines when corresponding coefficients differ by a small factor | Quick mental arithmetic possible |
| Cross-multiplication shines for word problems with no obvious symmetry | Mechanical, formula-driven, low cognitive load |
| For equations with $\frac{1}{x}$ or $\frac{1}{x\pm y}$, substitute first, then solve linearly | Reduces a non-linear pair to a familiar linear pair |
| Always check by substituting back into BOTH original equations | Catches arithmetic slips and sign errors |
| For “find $k$ such that…” problems, set up the corresponding ratio equation precisely | The exam values are usually very specific small integers or simple fractions |
| For word problems, define the unknowns clearly in words (“Let $x=$ price of an apple in rupees…”) before writing equations | Prevents confusion and unit errors during solving |
| Practise graphical method on graph paper with a sharp pencil | The board examiner deducts marks for sloppy graphs even if the algebra is correct |