NCERT Solutions Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3

NCERT Solutions Class 10 Maths
Chapter – 3 (Pair of Linear Equations in Two Variables)

The NCERT Solutions in English Language for Class 10 Mathematics Chapter – 3 Pair of Linear Equations in Two Variables Exercise 3.3 has been provided here to help the students in solving the questions from this exercise.

Chapter : 3 Pair of Linear Equations in Two Variables

Exercise – 3.3

1. Solve the following pair of linear equations by the substitution method.
(i) x + y = 14
x – y = 4

(ii) s – t = 3
$\mathbf{\frac{s}{3}}$ + $\mathbf{\frac{t}{2}}$ = 6

(iii) 3x – y = 3
9x – 3y = 9

(iv) 0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3

(v) √2x + √3y = 0
√3x – √8y = 0

(vi) $\mathbf{\frac{3x}{2}}$ – $\mathbf{\frac{5y}{3}}$ = -2

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

Solutions –

(i) x + y = 14 ————— (i)
x – y = 4 ————— (ii)
From the (i) equation, we get,
x = 14 – y
Now, substitute the value of x to the second equation to get,
(14 – y) – y = 4
14 – 2y = 4
2y = 10
y = 10/2
y = 5
By the value of y, we can now find the exact value of x.
∵ x = 14 – y
∴ x = 14 – 5
x = 9
Hence, x = 9 and y = 5

(ii) s – t = 3 ————— (i)
$\mathbf{\frac{s}{3}}$ + $\mathbf{\frac{t}{2}}$ = 6 ————— (ii)
From the (i) equation, we get,
s = 3 + t
Now, substitute the value of s to the second equation to get,
$\frac{3+t}{3} + \frac{t}{2}$ = 6

$\frac{2(3+t) + 3(t))}{6}$ = 6
(6 + 2t + 3t) = 6 × 6
(6 + 5t) = 36
5t = 30
t = 6
Now, substitute the value of t to the equation (1)
s – t = 3
s – 6 = 3
s = 3 + 6
s = 9
Therefore, s = 9 and t = 6

(iii) 3x – y = 3 ————— (i)
9x – 3y = 9 ————— (ii)
From the 1^st equation, we get,
x = $\frac{(3+y)}{3}$
Now, substitute the value of x to the second equation to get,
$9\frac{(3+y)}{3} -$ 3y = 9
3(3 + y) – 3y = 9
9 + 3y – 3y = 9
9 = 9
Therefore, y has infinite values, and since x = $\frac{(3+y)}{3}$ , x also has infinite values.

(iv) 0.2x + 0.3y = 1.3 ————— (i)
0.4x + 0.5y = 2.3 ————— (ii)
From the (i) equation, we get,
x = $\frac{(1.3-0.3y)}{0.2}$
Now, substitute the value of x to the second equation to get,
0.4 $\frac{(1.3-0.3y)}{0.2}$ + 0.5y = 2.3
2(1.3 – 0.3y) + 0.5y = 2.3
2.6 – 0.6y + 0.5y = 2.3
2.6 – 0.1 y = 2.3
0.1 y = 0.3
y = 3
Now, substitute the value of y in equation (i), and we get,
0.2x + 0.3y = 1.3
0.2x + 0.3 × 3 = 1.3
0.2x = 1.3 – 0.9
0.2x = 0.4
x = 0.4/0.2
x = 2
Therefore, x = 2 and y = 3

(v) √2x + √3y = 0 ————— (i)
√3x – √8y = 0 ————— (ii)
From the (i) equation, we get,
x = – $\frac{\sqrt{3}}{\sqrt{2}}y$ ————— (iii)
Putting the value of x in the given (ii) equation to get,
√3 $\left (- \frac{\sqrt{3}}{\sqrt{2}} \right )$ y – √8y = 0
$\left (- \frac{3}{\sqrt{2}} \right )$ y – 2√2 y = 0
y $\left ( -\frac{3}{\sqrt{2}} - 2\sqrt{2}\right )$ = 0
y = 0
Now, substitute the value of y in equation (iii), and we get,
x = 0
Therefore, x = 0 and y = 0

(vi) $\mathbf{\frac{3x}{2}}$ – $\mathbf{\frac{5y}{3}}$ = -2 ————— (i)

$\mathbf{\frac{x}{3}}$ + $\mathbf{\frac{y}{2}}$ = $\mathbf{\frac{13}{6}}$ ————— (ii)
From (i) equation, we get,
$\mathbf{\frac{3x}{2}}$ = -2 + $\mathbf{\frac{5y}{3}}$
x = $\frac{2(-6+5y)}{9}$
x = (-12 + 10y)/9 ………………………(iii)
Putting the value of x in the (ii) equation, we get,
$\frac{\frac{(-12+10y)}{9}}{3} + \frac{y}{2}$ = 13/6
$\frac{y}{2}$ = $\frac{13}{6} - \frac{(-12 + 10y)}{27}$

$\frac{y}{2}$ = $\frac{13\times 9 -2(-12+10y)}{54}$

$\frac{y}{2}$ = $\frac{117 + 24 - 20y}{54}$

$\frac{y}{2}$ = $\frac{141 -20y}{54}$
27y = 141 – 20y
27y + 20y = 141
47y = 141
y = 141/47
y = 3
Now, substitute the value of y in equation (iii), and we get,
x = (-12 + 10y)/9 ………………………(iii)
x = (-12 + 10×3)/9
x = (-12 + 30)/9
x = 18/9
x = 2
Therefore, x = 2 and y = 3

2. Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.
Solution –
2x + 3y = 11 …………………………..(i)
2x – 4y = -24 ………………………… (ii)
From equation (i), we get
x = (11 – 3y)/2 ………………….(iii)
Substituting the value of x to equation (ii), we get
2(11-3y)/2 – 4y = 24
11 – 3y – 4y = -24
-7y = -35
y = 5 …………………… (iv)
Putting the value of y in equation (iii), we get
x = (11 – 3×5)/2
x = (11 – 15)/2
x = -4/2
x = -2
Hence, x = -2, y = 5
Also,
y = mx + 3
5 = -2m +3
-2m = 2
m = -1
Therefore, the value of m is -1.

3. Form the pair of linear equations for the following problems and find their solution by the substitution method.
(i) The difference between two numbers is 26, and one number is three times the other. Find them.

Solution –
Let the two numbers be x and y, respectively, such that y > x.
According to the question,
y = 3x ……………… (i)
y – x = 26 ….………….. (ii)
Substituting the value of (i) to (ii), we get
3x – x = 26
x = 13 ……………. (iii)
Substituting (iii) in (i), we get
y = 3x
y = 3 × 13
y = 39
Hence, the numbers are 13 and 39.

(ii) The larger of two supplementary angles exceeds, the smaller by 18 degrees. Find them.
Solution –
Let the larger angle be x^o and the smaller angle be y^o.
We know that the sum of two supplementary pairs of angles is always 180^o.
According to the question,
x + y = 180^o……………. (i)
x – y = 18^o……………..(ii)
From (i), we get
x = 180^o – y …………. (iii)
Substituting (iii) in (ii), we get
180^o– y – y =18^o162^o = 2y
y = 81^o ………….. (iv)
Using the value of y in (iii), we get
x = 180^o – 81^o= 99^oHence, the angles are 99^o and 81^o.

(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.
Solution –
Let the cost of a bat be x and the cost of a ball be y.
According to the question,
7x + 6y = 3800 ………………. (i)
3x + 5y = 1750 ………………. (ii)
From (i), we get
y = (3800 – 7x)/6 ……………….. (iv)
Substituting (iii) to (ii), we get,
3x + 5 $\left ( \frac{3800 - 7x}{6} \right )$ = 1750

3x + $\frac{9500}{3}$ – $\frac{35x}{6}$ = 1750

3x – $\frac{35x}{6}$ = 1750 – $\frac{9500}{3}$

(18x – 35x)/6 = (5250 – 9500)/3
-17x/6 = -4250/3
-17x = -8500
x = 500 ……………………….. (iv)
Substituting the value of x to (iii), we get
y = (3800 – 7 × 500)/6
y = 300/6
y = 50
Hence, the cost of a bat is Rs 500, and the cost of a ball is 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?
Solution –
Let the fixed charge be Rs x and the per km charge be Rs y.
According to the question,
x + 10y = 105 …………….. (i)
x + 15y = 155 …………….. (ii)
From (1), we get
x = 105 – 10y ………………. (iii)
Substituting the value of x to (ii), we get
105 – 10y + 15y = 155
5y = 50
y = 10 …………….. (iv)
Putting the value of y in (3), we get
x = 105 – 10 × 10
x = 5
Hence, the fixed charge is Rs 5 and the per km charge = Rs 10
Charge for 25 km = x + 25y
= 5 + 250
= Rs 255

(v) A fraction becomes 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 5/6. Find the fraction.
Solution –
Let the fraction be x/y.
According to the question,
(x + 2) / (y + 2) = 9/11
11x + 22 = 9y + 18
11x – 9y = -4 …………….. (i)
(x+3) /(y+3) = 5/6
6x + 18 = 5y +15
6x – 5y = -3 ………………. (ii)
From (i), we get
x = (-4 + 9y)/11 …………….. (iii)
Substituting the value of x to (ii), we get
6(-4 + 9y)/11 – 5y = -3
-24 + 54y – 55y = -33
-y = -9
y = 9 ………………… (iv)
Substituting the value of y to (iii), we get
x = (-4 + 9 × 9 )/11
x = 7
Hence, the fraction is 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?
Solutions –
Let the age of Jacob and his son be x and y, respectively.
According to the question,
(x + 5) = 3(y + 5)
x – 3y = 10 ………………….. (i)
(x – 5) = 7(y – 5)
x – 7y = -30 ………………….. (ii)
From (1), we get
x = 3y + 10 ………………….. (iii)
Substituting the value of x to (ii), we get
3y + 10 – 7y = -30
-4y = -40
y = 10 ………………….. (iv)
Substituting the value of y to (iii), we get
x = 3 × 10 + 10
x = 40

Go Back To Chapters

NCERT Solutions Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3