NCERT Math Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables solutions (English Medium) || CBSE Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 Math Solution

Explore the NCERT Solutions for Class 10 Maths Chapter 3: Pair of Linear Equations in Two Variables (English Medium), with detailed explanations for Exercise 3.3. This resource is designed to simplify complex Pair of Linear Equations in Two Variables concepts, helping Class 10 Maths students grasp the subject effortlessly. For any queries regarding NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3, feel free to leave a comment, and we’ll respond as soon as possible. Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables solutions Ex 3.3 NCERT Solutions for Class 10 Maths Chapter 3: Pair of Linear Equations in Two Variables (English Medium)

NCERT Math Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables solutions (English Medium) || CBSE Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 Math Solution
Exercise-3.3

NCERT Math Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables solutions (English Medium) || CBSE Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 Math Solution

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(i) Solve the following pair of linear equations by the substitution method.

x + y = 14

$x+y=14$

x - y = 4

$x-y=4$

Answer

x + y = 14 \dots (i)

$x+y=14 \ldots(i)$

x - y = 4 \dots (i i)

$x-y=4 \ldots(ii)$
From equation (i)
take

x

$x$ on one side and when we take

y

$y$ to the other side its sign changes and we get,

x = 14 - y \dots (i i i)

$x=14-y \ldots(iii)$
Putting value of

x

$x$ in equation (ii) we get,

(14 - y) - y = 4

$(14-y)-y=4$

14 - 2 y = 4

$14-2 y=4$

2 y = 10

$2 y=10$

y = \frac{10}{2} = 5

$y=\frac{10}{2}=5$
Putting value of

y

$y$ in equation (iii) we get,

x = 14 - 5 = 9

$x=14-5=9$
Hence,

x = 9

$x=9$ and

y = 5

$y=5$

(ii) Solve the following pair of linear equations by the substitution method.

s - t = 3

$\mathrm{s}-\mathrm{t}=3$

\frac{s}{3} + \frac{t}{2} = 6

$\frac{s}{3}+\frac{t}{2}=6$

Answer

s - t = 3 \dots (i)

$\mathrm{s}-\mathrm{t}=3 \ldots(i)$
and,

\frac{s}{3} + \frac{t}{2} = 6 \dots (i i)

$\frac{s}{3}+\frac{t}{2}=6 \ldots(ii)$
From equation (i) we get,
taking

t

$t$ to the other side, the sign of

t

$t$ changes to positive

s = t + 3 \dots (i i i)

$\mathrm{s}=\mathrm{t}+3 \ldots(iii)$
Putting value of

x

$x$ from (iii) to (ii)

\Rightarrow \frac{t + 3}{3} + \frac{t}{2} = 6

$\Rightarrow \frac{t+3}{3}+\frac{t}{2}=6$

\Rightarrow 2 t + 6 + 3 t = 36

$\Rightarrow 2 \mathrm{t}+6+3 \mathrm{t}=36$

\Rightarrow 5 t = 30

$\Rightarrow 5 \mathrm{t}=30$

\Rightarrow t = \frac{30}{5} = 6

$\Rightarrow \mathrm{t}=\frac{30}{5}=6$
Putting value of

t

$t$ in equation (iii), we get,

s = 6 + 3 = 9

$s=6+3=9$
Hence,

s = 9, t = 6

$s=9, t=6$

(iii) Solve the following pair of linear equations by the substitution method.

3 x - y = 3

$3 x-y=3$

9 x - 3 y = 9

$9 x-3 y=9$

Answer

3 x - y = 3 \dots (i)

$3 x-y=3 \ldots(i)$

9 x - 3 y = 9 \dots (i i)

$9 x-3 y=9 \ldots(ii)$
Comparing with general pair of equations i.e.

a_{1} x + b_{1} y + c_{1} = 0

$a_{1} x+b_{1} y+c_{1}=0$

a_{2} x + b_{2} y + c_{2} = 0

$a_{2} x+b_{2} y+c_{2}=0$ , we have

a_{1} = 3, b_{1} = - 1, c_{1} = - 3

$a_{1}=3, b_{1}=-1, c_{1}=-3$

a_{2} = 9, b_{2} = - 2

$\mathrm{a}_{2}=9, \mathrm{~b}_{2}=-2$ and

c_{2} = - 9

$\mathrm{c}_{2}=-9$
Here,

\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} = 3

$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}=3$
and In this case, the system of linear equation is consistent and has infinite solutions.

NCERT Math Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables solutions (English Medium) || CBSE Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 Math Solution

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(iv) Solve the following pair of linear equations by the substitution method.

0.2 x + 0.3 y = 1.3

$0.2 \mathrm{x}+0.3 \mathrm{y}=1.3$

0.4 x + 0.5 y = 2.3

$0.4 x+0.5 y=2.3$

Answer

0.2 x + 0.3 y = 1.3 \dots (i)

$0.2 \mathrm{x}+0.3 \mathrm{y}=1.3 \ldots(i)$

0.4 x + 0.5 y = 2.3 \dots (i i)

$0.4 x+0.5 y=2.3 \ldots(ii)$
From equation (i) we get,

0.2 x = 1.3 - 0.3 y

$0.2 \mathrm{x}=1.3-0.3 \mathrm{y}$

x = \frac{1.3 - 0.3 y}{0.2}

$x=\frac{1.3-0.3 y}{0.2}$

x = \frac{1.3}{0.2} - \frac{0.3}{0.2} y

$x=\frac{1.3}{0.2}-\frac{0.3}{0.2} y$

x = 6.5 - 1.5 y

$x=6.5-1.5 y$
Putting value of

x

$x$ in equation (ii) we get,

(6.5 - 1.5 y) \times 0.4 + 0.5 y = 2.3

$(6.5-1.5 y) \times 0.4+0.5 y=2.3$

6.5 \times 0.4 - 1.5 y \times 0.4 + 0.5 y = 2.3

$6.5 \times 0.4-1.5 y \times 0.4+0.5 y=2.3$

2.6 - 0.6 y + 0.5 y = 2.3

$2.6-0.6 y+0.5 y=2.3$

- 0.1 y = - 0.3

$-0.1 y=-0.3$

y = \frac{- 0.3}{- 0.1} = 3

$y=\frac{-0.3}{-0.1}=3$
Putting value of y in equation (iii) we get,

x = 6.5 - 1.5 \times 3 = 6.5 - 4.5 = 2

$x=6.5-1.5 \times 3=6.5-4.5=2$
Hence,

x = 2

$x=2$ and

y = 3

$y=3$

(v) Solve the following pair of linear equations by the substitution method.

\sqrt{2} x + \sqrt{3} y = 0

$\sqrt{2} x+\sqrt{3} y=0$

\sqrt{3} x - \sqrt{8} y = 0

$\sqrt{3} x-\sqrt{8} y=0$

Answer

\sqrt{2} x + \sqrt{3} y = 0 \dots (i)

$\sqrt{2} x+\sqrt{3} y=0 \ldots(i)$

\sqrt{3} x - \sqrt{8} y = 0 \dots (i i)

$\sqrt{3} x-\sqrt{8} y=0 \ldots(ii)$
From equation (i), we get,

\sqrt{2} x = - \sqrt{3} y

$\sqrt{2} x=-\sqrt{3} y$

x = \frac{- \sqrt{3} y}{\sqrt{2}} \dots (i i i)

$x=\frac{-\sqrt{3} y}{\sqrt{2}} \ldots(iii)$
Putting value of

x

$x$ in equation (ii). we get,

\Rightarrow \sqrt{3} (\frac{- \sqrt{3}}{\sqrt{2}}) - \sqrt{8} y = 0

$\Rightarrow \sqrt{3}\left(\frac{-\sqrt{3}}{\sqrt{2}}\right)-\sqrt{8} y=0$

\Rightarrow - \frac{3 y}{\sqrt{2}} - 2 \sqrt{2} y = 0 \sqrt{8} = 2 \sqrt{2}

$\Rightarrow -\frac{3 y}{\sqrt{2}}-2 \sqrt{2} y=0 \quad \sqrt{8}=2 \sqrt{2}$

\Rightarrow y (\frac{- 3}{\sqrt{2}} - 2 \sqrt{2}) = 0

$\Rightarrow y\left(\frac{-3}{\sqrt{2}}-2 \sqrt{2}\right)=0$
so,

y = 0

$\mathrm{y}=0$
Putting value of y in equation (iii) we get.

x = 0

$x=0$
Hence,

x = 0

$x=0$ and

y = 0

$y=0$

NCERT Math Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables solutions (English Medium) || CBSE Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 Math Solution

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(vi) Solve the following pair of linear equations by the substitution method.

\frac{3 x}{2} - \frac{5 y}{2} = - 2

$\frac{3 x}{2}-\frac{5 y}{2}=-2$

\frac{x}{3} + \frac{y}{2} = \frac{13}{6}

$\frac{x}{3}+\frac{y}{2}=\frac{13}{6}$

Answer

\frac{3 x}{2} - \frac{5 y}{3} = - 2 \dots (i)

$\frac{3 x}{2}-\frac{5 y}{3}=-2 \ldots(i)$
and

\frac{x}{3} + \frac{y}{2} = \frac{13}{6} \dots (i i)

$\frac{x}{3}+\frac{y}{2}=\frac{13}{6} \ldots(ii)$
From equation (i) we get,
By taking L.C.M and solving we get,

\frac{3 \times 3 x - 2 \times 5 y}{6} = - 2

$\frac{3 \times 3 x-2 \times 5 y}{6}=-2$

9 x - 10 y = - 12

$9 \mathrm{x}-10 \mathrm{y}=-12$

x = \frac{- 12 + 10 y}{9}

$x=\frac{-12+10 y}{9}$
Putting this value of

x

$x$ in equation (ii), we get,

\Rightarrow \frac{\frac{- 12 + 10 y}{9}}{3} + \frac{y}{2} = \frac{13}{6}

$\Rightarrow \frac{\frac{-12+10 y}{9}}{3}+\frac{y}{2}=\frac{13}{6}$

$\Rightarrow \frac{-12+10 y}{27}+\frac{y}{2}=\frac{13}{6}$

$\Rightarrow 47 \mathrm{y}=117+24$

$\Rightarrow 47 \mathrm{y}=141$

$\Rightarrow \mathrm{y}=\frac{141}{47}=3$
Putting value of

$y$ in (iii)

$\Rightarrow x=\frac{-12+10 y}{9}=\frac{-12+10 \times 3}{9}=\frac{18}{9}=2$
Hence,

$x=2$ and

$y=3$

2. Solve

$2 x+3 y=11$ and

$2 x-4 y=-24$ and hence find the value of ' m ' for which

$\mathrm{y}=\mathrm{mx}+\mathrm{x}$ .

Answer

To Solve:

$2 \mathrm{x}+3 \mathrm{y}=11 \ldots(i)$

$2 x-4 y=-24 \ldots(ii)$
Solving by substitution method,
From equation (i)

$2 x=11-3 y \ldots(iii)$
putting value of

$2 x$ from equation (iii) to equation (ii)

$(11-3 y)-4 y=-24$

$\Rightarrow 11-7 y=-24$

$\Rightarrow-7 y=-35$

$y=\frac{-35}{-7}$

$y=5$
Putting value of

$y$ in equation (iii) we get,

$2 \mathrm{x}=11-3 \times 5=11-15=-4$

$\mathrm{x}=-\frac{4}{2}=-2$
Now, putting values of

$x$ and

$y$ in equation

$y=m x+3$

$5=-2 m+3$

$=-2 m=2$

$m=-\frac{2}{2}=-1$
Value of

$m$ is -1

NCERT Solutions for Class 10 Maths Chapter 3: Pair of Linear Equations in Two Variables || CBSE Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables solutions Ex 3.3 Math Solution

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3. 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.

Answer

i) Let larger number

$=\mathrm{x}$
Let smaller number

$=\mathrm{y}$
According to the question,

$=x-y=26$

$=x=26+y \ldots(i)$
And,

$=x=3 y \ldots(ii)$
Comparing values of

$x$ from both equation, we get,

$26+y=3 y$

$-2 y=-26$

$y=13$
So,

$x=3 y=3 \times 13=39$
Hence, the numbers are 13 and 39.

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Answer

Let the first angle

$=x$
Let second angle

$=y$
According to the question,

$=x+y=180$ (sum of supplementary angles is always 180)

$\mathrm{x}=180-\mathrm{y} \ldots(i)$
And,

$=x-y=18 \ldots(ii)$ Given
Putting value of

$x$ from equation (i) to (ii). we get,

$=180-y-y=18$

$=-2 y=18-180=-162$

$=y=\frac{-162}{-2}=81$

$\text { so, } x=180=y=180-81=99$
Hence the angles are

$99^{\circ}$ and

$81^{\circ}$

NCERT Math Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables solutions (English Medium) || CBSE Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 Math Solution

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(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800 . Later, she buys 3 bats and 5 balls for Rs 1750 . Find the cost of each bat and each ball.

Answer

Let cost of each bat

$=$ Rs. X
Let cost of each ball

$=$ Rs.

$Y$
According to the question,

$=7 x+6 y=3800 \text { Given }$

$=6 y=3800-7 x$

$=y=\frac{3800-7 x}{6} \ldots(i)$
And

$=3 x+5 y=1750 \ldots(ii)$ Given
Putting value of

$y$ from Equation (i) to equation (ii)

$=3 x+5 \times \frac{3800-7 x}{6}=1750$

$=18 \mathrm{x}+1900-35 \mathrm{x}=1750 \times 6$

$=-17 x=10500-19000=-8500$

$=x=\frac{-8500}{-17}=500$
Putting value of

$x$ in equation (i), we get

$=y=\frac{3800-7 \times 500}{6}=\frac{300}{6}=50$
Hence,
Cost of each bat

$=$ Rs. 500 and Cost of each ball

$=$ Rs. 50

(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 Rs 105 and for a journey of 15 km , the charge paid is Rs 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 ?

Answer

Let the fixed charge for taxi

$=$ Rs. X
Let variable cost per

$\mathrm{km}=$ Rs. y
We know that,
Total cost

$=$ Fixed charge + Variable Charge
According to the question,

$=x+10 y=105 \text { given }$

$=x=105-10 y \ldots(i)$
And, For a journey of 15 km charge paid = Rs. 155

$\text { so, } x+15 y=155 \ldots(ii)$
Putting value of

$x$ from equation (i) to equation (ii). we get,

$=105-10 y+15 y=155$

$=5 y=155-105=50$

$y=\frac{50}{5}=10$
Putting value of

$y$ in equation (i). we get,

$=x=105-10 \times 10=105-100=5$
So, People have to pay for travelling a distance of 25 km

$=x+25 y=5+25 \times 10=\text { Rs. } 255$

(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.

Answer

Let numerator

$=x$
Let denominator

$=\mathrm{y}$
Fraction will be

$=\frac{x}{y}$
According to the question,
Fraction become

$\frac{9}{11}$ , if 2 is added in both. Numerator and denominator

$=\frac{x+2}{y+2}=\frac{9}{11}$

$=11 \mathrm{x}+22=9 \mathrm{y}+18 \text { (by cross multiplication) }$

$=11 \mathrm{x}=9 \mathrm{y}-4$

$=x=\frac{9 y-4}{11} \ldots(i)$
And, if 3 is added to both numerator and denominator it become

$\frac{5}{6}$

$=\frac{x+3}{y+3}=\frac{5}{6}$

$=6 x+18=5 y+15 \ldots(ii)$ By cross multiplication
Putting value of

$x$ from equation (i) to equation (ii)

$=6 \times \frac{9 y-4}{11}+18=5 y+15$

$=54 y-24=55 y-33$

$=-y=-9$

$=y=9$
Putting value of

$y$ in equation (i)

$=x=\frac{9 y-4}{11}=\frac{9 \times 9-4}{11}=\frac{77}{11}=7$
Hence, the fraction is

$\frac{7}{9}$ .

(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 present age of Jacob

$=X$ years
Let present age of his son

$=$ Y years
Five year hence,
Age of Jacob

$=X+5$
Age of son

$=Y+5$
And, age of Jacob is 3 times of his son Given

$=x+5=3(y+5)$

$=x+5=3 y+15$

$=x=3 y+10 \ldots(i)$
Five years ago,
Age of Jacob

$=\mathrm{X}-5$
Age of son

$=$ Y - 5
And, Jacob's age was 7 times of his son Given

$=x-5=7(y-5)$

$=x-5=7 y-35$

$=x-7 y=-30 \ldots(ii)$
Putting value of X from equation (i) to equation (ii)

$=3 y+10-7 y=-30$

$=-4 y=-40$

$=y=\frac{-40}{-4}=10$
Putting value of

$Y$ in (i)

$=x=3 \times 10+10=30+10=40$
Hence, present age of Jacob

$=40$ years and his son's age

$=10$ years.

NCERT Math Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables solutions (English Medium) || CBSE Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 Math Solution

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Central Board of Secondary Education Official Site
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Class 10 : CBSE Class 10 Maths Chapter 2 Polynomials Ex 2.2
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NCERT Math Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables solutions (English Medium) || CBSE Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 Math Solution

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