NCERT Solutions Class 10 Maths Chapter 3: Pair of Linear Equations in Two Variables

Q: Solve the following pair of linear equations by the substitution method: x + y = 14; x – y = 4

From x – y = 4, we get x = 4 + y. Substituting this in x + y = 14, we get (4 + y) + y = 14, so 2y = 10 and y = 5. Putting y = 5 in x = 4 + y, we get x = 9. The solution is x = 9, y = 5.

Q: The difference between two numbers is 26 and one number is three times the other. Find them.

Let the two numbers be x and y (x > y). According to the problem, x – y = 26 and x = 3y. Substituting x = 3y in the first equation, we get 3y – y = 26, which means 2y = 26, so y = 13. Then x = 3(13) = 39. The numbers are 39 and 13.

Q: Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.

Solving the equations, we get x = -2 and y = 5. Substituting these in y = mx + 3, we get 5 = m(-2) + 3, which means 2 = -2m. Therefore, m = -1.

Detailed solutions for Exercise 3.2 (Algebraic Solution by Substitution Method).

Ex 3.1 Ex 3.2 Ex 3.3

#### **Q1. Solve the following pair of linear equations by the substitution method.** **(i)** $x + y = 14; \quad x – y = 4$ **(ii)** $s – t = 3; \quad \frac{s}{3} + \frac{t}{2} = 6$ **(iii)** $3x – y = 3; \quad 9x – 3y = 9$ **(iv)** $0.2x + 0.3y = 1.3; \quad 0.4x + 0.5y = 2.3$ **(v)** $\sqrt{2}x + \sqrt{3}y = 0; \quad \sqrt{3}x – \sqrt{8}y = 0$ **(vi)** $\frac{3x}{2} – \frac{5y}{3} = -2; \quad \frac{x}{3} + \frac{y}{2} = \frac{13}{6}$ #### **A1. Solution:** **(i) $x + y = 14 \dots(1)$ and $x – y = 4 \dots(2)$** From (2): $x = y + 4$. Substitute in (1): $(y + 4) + y = 14$ $2y = 10 \Rightarrow y = 5$ Put $y=5$ in $x = y + 4$: $x = 5 + 4 = 9$ **Answer: $x = 9, y = 5$** **(ii) $s – t = 3 \dots(1)$ and $\frac{s}{3} + \frac{t}{2} = 6 \dots(2)$** From (1): $s = t + 3$. Substitute in (2): $\frac{t+3}{3} + \frac{t}{2} = 6$ Multiply by 6 to clear fractions: $2(t + 3) + 3t = 36$ $2t + 6 + 3t = 36$ $5t = 30 \Rightarrow t = 6$ Put $t=6$ in $s = t + 3$: $s = 6 + 3 = 9$ **Answer: $s = 9, t = 6$** **(iii) $3x – y = 3 \dots(1)$ and $9x – 3y = 9 \dots(2)$** From (1): $y = 3x - 3$. Substitute in (2): $9x - 3(3x - 3) = 9$ $9x - 9x + 9 = 9$ $9 = 9$ This statement is always true. **Answer: Infinite solutions.** (The lines are coincident). **(iv) $0.2x + 0.3y = 1.3 \dots(1)$ and $0.4x + 0.5y = 2.3 \dots(2)$** Multiply both equations by 10 to remove decimals: 1) $2x + 3y = 13$ 2) $4x + 5y = 23$ From (1): $2x = 13 - 3y \Rightarrow x = \frac{13-3y}{2}$. Substitute in (2): $4(\frac{13-3y}{2}) + 5y = 23$ $2(13 - 3y) + 5y = 23$ $26 - 6y + 5y = 23$ $-y = -3 \Rightarrow y = 3$ Put $y=3$ in $x$: $x = \frac{13 - 3(3)}{2} = \frac{4}{2} = 2$ **Answer: $x = 2, y = 3$** **(v) $\sqrt{2}x + \sqrt{3}y = 0 \dots(1)$ and $\sqrt{3}x – \sqrt{8}y = 0 \dots(2)$** From (1): $x = -\frac{\sqrt{3}}{\sqrt{2}}y$. Substitute in (2): $\sqrt{3}(-\frac{\sqrt{3}}{\sqrt{2}}y) - \sqrt{8}y = 0$ $-\frac{3}{\sqrt{2}}y - 2\sqrt{2}y = 0$ (Since $\sqrt{8}=2\sqrt{2}$) $y(-\frac{3}{\sqrt{2}} - 2\sqrt{2}) = 0$ For this to hold, $y$ must be 0. $y = 0$. Substitute $y=0$ in (1): $x = 0$. **Answer: $x = 0, y = 0$** **(vi) $\frac{3x}{2} – \frac{5y}{3} = -2 \dots(1)$ and $\frac{x}{3} + \frac{y}{2} = \frac{13}{6} \dots(2)$** Simplify (1) by multiplying by 6: $9x - 10y = -12$ Simplify (2) by multiplying by 6: $2x + 3y = 13$ From simplified (2): $2x = 13 - 3y \Rightarrow x = \frac{13-3y}{2}$. Substitute in simplified (1): $9(\frac{13-3y}{2}) - 10y = -12$ Multiply by 2: $9(13 - 3y) - 20y = -24$ $117 - 27y - 20y = -24$ $-47y = -141$ $y = 3$ Put $y=3$ in $x$: $x = \frac{13 - 3(3)}{2} = \frac{4}{2} = 2$ **Answer: $x = 2, y = 3$** #### **Q2. Solve $2x + 3y = 11$ and $2x – 4y = – 24$ and hence find the value of ‘m’ for which $y = mx + 3$.** #### **A2. Solution:** **Equations:** 1. $2x + 3y = 11$ 2. $2x - 4y = -24$ From (2): $2x = 4y - 24 \Rightarrow x = 2y - 12$. Substitute in (1): $2(2y - 12) + 3y = 11$ $4y - 24 + 3y = 11$ $7y = 35 \Rightarrow y = 5$ Put $y=5$ in $x = 2y - 12$: $x = 2(5) - 12 = 10 - 12 = -2$. **Solution:** $x = -2, y = 5$. **Find m:** Substitute $x = -2, y = 5$ into $y = mx + 3$: $5 = m(-2) + 3$ $5 - 3 = -2m$ $2 = -2m \Rightarrow m = -1$ **Answer: $m = -1$** #### **Q3. Form the pair of linear equations for the following problems and find their solution by substitution method.** **(i)** The difference between two numbers is 26 and one number is three times the other. Find them. **(ii)** The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them. **(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. **(iv)** The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹ 105 and for a journey of 15 km, the charge paid is ₹ 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km? **(v)** A fraction becomes $\frac{9}{11}$, if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes $\frac{5}{6}$. Find the fraction. **(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? #### **A3. Solution:** **(i) Solution:** Let the numbers be $x$ and $y$ ($x > y$). 1. Difference is 26: $x - y = 26$ 2. One is 3 times other: $x = 3y$ Substitute $x=3y$ into eq(1): $3y - y = 26$ $2y = 26 \Rightarrow y = 13$ $x = 3(13) = 39$ **Answer: The numbers are 39 and 13.** **(ii) Solution:** Let the larger angle be $x$ and smaller be $y$. 1. Supplementary: $x + y = 180^\circ$ 2. Difference: $x = y + 18$ Substitute (2) into (1): $(y + 18) + y = 180$ $2y = 162 \Rightarrow y = 81^\circ$ $x = 81 + 18 = 99^\circ$ **Answer: The angles are $99^\circ$ and $81^\circ$.** **(iii) Solution:** Let cost of 1 bat = ₹$x$ and 1 ball = ₹$y$. 1. $7x + 6y = 3800$ 2. $3x + 5y = 1750 \Rightarrow 3x = 1750 - 5y \Rightarrow x = \frac{1750 - 5y}{3}$ Substitute $x$ into (1): $7(\frac{1750 - 5y}{3}) + 6y = 3800$ Multiply by 3: $7(1750 - 5y) + 18y = 11400$ $12250 - 35y + 18y = 11400$ $-17y = -850 \Rightarrow y = 50$ Put $y=50$ in $x$: $x = \frac{1750 - 250}{3} = \frac{1500}{3} = 500$ **Answer: Cost of bat = ₹500, Cost of ball = ₹50.** **(iv) Solution:** Let Fixed charge = ₹$x$ and Charge per km = ₹$y$. 1. $x + 10y = 105 \Rightarrow x = 105 - 10y$ 2. $x + 15y = 155$ Substitute (1) into (2): $(105 - 10y) + 15y = 155$ $5y = 50 \Rightarrow y = 10$ $x = 105 - 10(10) = 5$ Fixed charge = ₹5, Charge per km = ₹10. **Cost for 25km:** $x + 25y = 5 + 25(10) = 5 + 250 = 255$ **Answer: Fixed charge = ₹5, Per km = ₹10, Cost for 25km = ₹255.** **(v) Solution:** Let the fraction be $\frac{x}{y}$. 1. $\frac{x+2}{y+2} = \frac{9}{11} \Rightarrow 11(x+2) = 9(y+2) \Rightarrow 11x - 9y = -4$ 2. $\frac{x+3}{y+3} = \frac{5}{6} \Rightarrow 6(x+3) = 5(y+3) \Rightarrow 6x - 5y = -3$ From (2): $x = \frac{5y - 3}{6}$. Substitute into (1): $11(\frac{5y - 3}{6}) - 9y = -4$ $55y - 33 - 54y = -24$ $y - 33 = -24 \Rightarrow y = 9$ Put $y=9$ in $x$: $x = \frac{5(9) - 3}{6} = \frac{42}{6} = 7$ **Answer: The fraction is $\frac{7}{9}$.** **(vi) Solution:** Let Jacob's age = $x$ and Son's age = $y$. 1. 5 years hence: $(x+5) = 3(y+5) \Rightarrow x - 3y = 10 \Rightarrow x = 10 + 3y$ 2. 5 years ago: $(x-5) = 7(y-5) \Rightarrow x - 7y = -30$ Substitute (1) into (2): $(10 + 3y) - 7y = -30$ $-4y = -40 \Rightarrow y = 10$ $x = 10 + 3(10) = 40$ **Answer: Jacob's age = 40 years, Son's age = 10 years.**