# Class10 NCRT Pair of Linear Equations in Two Variables Exercise – 3.3 pdf || UP Boad

# __Pair of Linear Equations in Two
Variables__

__Exercise – 3.3__

__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**

**Solution****:**

x + y = 14……………..(1)

x – y =
4……………..(2)

From equation (1),

y = 14 – x
……………….(3)

Substituting the value of equation (3) in (2),

x – (14 –
x) = 4

x – 14 + x
= 4

2x = 14 +
4

2x = 18

x = 9

Substituting
the value of x in equation (2),

9 – y = 4

y = 9 – 4

y = 5

Hence, x = 9, y = 5.

**Solution****:**

s - t =
3……………..(1)

s/3 + t/2
= 6 ……………..(2)

From equation (1),

s = 3 + t
……………….(3)

Substituting the value of equation (3) in (2),

5t + 6 =
36

t = 6

Substituting
the value of x in equation (3),

s = 3 + 6

s = 9

Hence, s = 9, t = 6.

**(****iii****)**** 3x - y = 3, 9x – 3y = 9**

**Solution****:**

3x - y =
3……………..(1)

9x – 3y =
9……………..(2)

From equation (1),

y = 3x - 3
……………….(3)

Substituting the value of equation (3) in (2),

9x – 3(3x
– 3) = 9

9x – 9x +
9 = 9

9 = 9

Hence, y = 3x – 3 where x can take any value, that is, there are infinitely many
solutions.

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

**Solution****:**

0.2x +
0.3y = 1.3……………..(1)

0.4x +
0.5y = 2.3……………..(2)

From equation (1),

Substituting the value of equation (3) in (2),

0.12x +
0.65 – 0.10x = 0.69

0.02x =
0.04

x = 2

Substituting
the value of x in equation (2),

0.4×2 +
0.5y = 2.3

0.8 + 0.5y
= 2.3

0.5y = 1.5

y = 3

Hence, x = 2, y = 3.

**(****v****)**** √2x + √3y = 0, √3x - √2y = 0**

**Solution****:**

√2x + √3y
= 0……………..(1)

√3x - √2y
= 0……………..(2)

From equation (1),

Substituting the value of equation (3) in (2),

3x – 2x =
0

x = 0

Substituting
the value of x in equation (2),

√3×0 - √2y
= 0

y = 0

Hence, x = 0, y = 0.

**Solution****:**

From equation (1),

Substituting the value of equation (3) in (2),

282x + 216
= 780

282x = 564

x = 2

Substituting
the value of x in equation (2),

2/3 + y/2
= 13/6

y/2 = 9/6

y = 3

Hence, x = 2, 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…………………(1)

2x – 4y =
-24………………(2)

From equation (1),

y =
(11-2x)/3 ………………(3)

Substituting the value of equation (3) in (2),

2x –
4[(11-2x)/3] = -24

6x – 44 +
8x = -72

14x = -28

x = -2

Substituting
the value of x in equation (2),

2(-2) – 4y
= -24

-4 – 4y = -24

-4y = -20

y = 5

So, x = -2, y = 5.

Substituting
the values of x and y in y = mx + 3,

5 = -2m +
3

m = -1

**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**** **

**Solution****:**

Let the first number = x

and second number = y

According to Question,

x = 3y
…………(1)

The difference of both the numbers is 26, so

x - y = 26

x = y + 26
…………………..(2)

Substituting the value of x in Equation (1),

y + 26 =
3y

3y – y =
26

2y = 26

y = 13

Substituting
the value of y in equation (1),

x = 3×13 = 39

Hence, one number = 39, and the second number = 13.

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

**Solution****:**

Let the greater angle = x

and smaller angle = y

According to Question,

x = 18 +
y …………(1)

Both angles are supplementary, so

x + y = 180 …………………..(2)

[The sum of
supplementary angles is 180
degrees.]

Substituting
the value of x in equation (2),

y + 18 + y
= 180

2y = 180 –
18

2y = 162

y = 81

Substituting
the value of y in equation (1),

x = 81 + 18

x = 99

Hence, greater angle = 99, and smaller angle = 81.

**(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.****
**

**Solution****:**

Let cost of one bat = ₹x

and cost of one ball = ₹y

according to the first condition,

7x + 6y = 3800

according to the second condition,

3x + 5y =
1750 ………….(2)

Substituting the value of x in equation (2),

11400 –
18y + 35y = 12250

17y = 850

y = 50

Substituting
the value of y in equation (1),

x = 500

Hence cost of one bat = ₹ 500 and cost of one ball = ₹ 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 ****₹****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?**

**Solution****:**

Let the fixed charge of taxi = ₹x

And the cost of each additional km = ₹ y

According to the first condition,

x + 10y =
105 …………………(1)

according to the second condition,

x + 15y =
155……………………(2)

From equation (1),

x + 10y =
105

x = 105 –
10y

Substituting the value of x in equation
(2),

x + 15y =
155

(105 –
10y) + 15y = 155

105 + 5y =
155

5y = 50

y = 10

Substituting
the value of y in equation (1),

x
+ 10y = 105

x
+ 10×10 = 105

x
= 105 – 100

x
= 5

Hence, the fixed charge
of the taxi = ₹ 5 and the cost for each additional km = ₹
10.

for 10 km,

x + 10y

5
+ 10×10 = 105

Similarly charges for 25 km = x + 25y

= 5 + 25×10

= 5 + 250

= ₹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 numerator = x

and denominator = y

Therefore, rational number = x/y

according to the first condition,

11x + 22 =
9y + 18

according to the second condition,

6x + 18 =
5y + 15

6x – 5y =
-3………………(2)

Substituting the value of x in Equation (1),

54y – 24 –
55y = -33

y = 9

Substituting the value of y in equation (2),,

6x – 5×9 =
-3

6x = 45 –
3

6x = 42

x = 7

Hence the fractional number = x/y = 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****?**

**Solution****:**

Let Jacob's present age =
x years

and present age of son = y
years

5
years later,

Jacob's age = x + 5 years

Son's age = y + 5 years

According to Question,

x + 5 = 3(y + 5)

x + 5 = 3y + 15

x = 3y + 10………………(1)

5 years ago,

Jacob's age = x - 5 years

Son's age = x - 5 years

According to Question,

x – 5 = 7(y-5)

x – 5 = 7y – 35

x – 7y = -30…………………(2)

Substituting the value of
x in equation (2),,

3y + 10 – 7y = -30

-4y = -40

y = 10

Substituting the value of
y in equation (1),

x = 3×10 + 10

x = 30 + 10

x = 40

Hence, the age of Jacob is
40 years and present age of
son is 10 years.

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