# 10th CBSE maths solution for exercise 3.4 part 1

This page 10th cbse maths solution for exercise 3.4 part 1is going to provide you solution for every problems that you find in the exercise no 3.4

## 10th CBSE maths solution for Exercise 3.4 part 1

(1) Solve the following pairs of linear equations by the elimination method and the substitution method

(i) x + y = 5 and 2 x – 3 y = 4

(i)   x + y = 5 --------(1)

2 x – 3 y = 4 --------(2)

Elimination method:

Multiply first equation by 3=>

3 x + 3 y =15

Now we are going to add these two equations to eliminate y Apply x = 19/5 in first or second equation to get the value of y

(19/5) + y = 5

y = 5 - (19/5)

y = (25 – 19)/5

y = 6/5

Therefore solution is x = 19/5 and y = 6/5

Substitution method:

x + y = 5 --------(1)

2 x – 3 y = 4 --------(2)

Step 1:

Find the value of one variable in terms of another variable

y = 5 – x

Step 2:

Substitute this value of y in the other equation, and reduce it to an equation in one variable.

2 x – 3 (5 – x) = 4

2 x – 15 + 3 x = 4

5 x = 4 – 15

5 x = 19

x = 19/5

Step 3:

Apply x = 19/5 in the equation y = 5 – x

y = 5 – (19/5)

y = (25 – 19)/5

y = 6/5

Therefore solution is x = 19/5 and y = 6/5

(ii) 3 x + 4 y = 10 and 2 x – 2 y = 2

3 x + 4 y = 10 ---------(1)

2 x – 2 y = 2 ---------(2)

Elimination method:

Multiply the second equation by 2=>

4 x - 4 y = 4

Now we are going to add these two equations to eliminate y 10th cbse maths solution for exercise 3.4 part 1 10th cbse maths solution for exercise 3.4 part 1

x = 2

Apply x = 2 in first or second equation to get the value of y

2 (2) – 2 y = 2

4 – 2 y = 2

-2 y = 2 – 4

- 2 y = -2

y = (-2)/(-2)

y = 1

Therefore solution is x = 2 and y = 1

Substitution method:

3 x + 4 y = 10 ---------(1)

2 x – 2 y = 2 ---------(2)

Step 1:

Find the value of one variable in terms of another variable

-2y = 2 – 2 x

y = (2 x – 2)/2

y = x - 1

Step 2:

Substitute this value of y in the other equation, and reduce it to an equation in one variable.

3 x + 4 (x - 1) = 10

3 x + 4 x – 4 = 10

7 x – 4 = 10

7 x = 10 + 4

7 x = 14

x = 14/7

x = 2

Step 3:

Apply x = 2 in the equation y = x - 1

y = 2 - 1

y = 1

Therefore solution is x = 2 and y = 1

(iii) 3 x – 5 y – 4 = 0 and 9 x = 2 y + 7

3 x – 5 y = 4 ---------(1)

9 x - 2 y = 7---------(2)

Elimination method:

Multiply the first equation by 3=>

9 x -15 y =12

Now we are going to subtract these two equations to eliminate y 10th cbse maths solution for exercise 3.4 part 1

y = -5/13

Apply y = -5/13 in first or second equation to get the value of y

3 x – 5 (-5/13) = 4

3 x + (25/13) = 4

(39 x + 25)/13 = 4

39 x + 25 = 4(13)

39 x + 25 = 52

39 x = 52 – 25

39 x = 27

x = 27/39

x = 9/13

Therefore solution is x = 9/13 and y = -5/13

Substitution method:

3 x – 5 y = 4 ---------(1)

9 x - 2 y = 7---------(2)

Step 1:

Find the value of one variable in terms of another variable

-2y = 7 – 9 x

y = (9 x – 7)/2

Step 2:

Substitute this value of y in the other equation, and reduce it to an equation in one variable.

3 x – 5(9 x – 7)/2 = 4

(6 x - 45x + 35)/2 = 4

-39 x + 35 = 4(2)

-39 x = 8 - 35

-39 x = - 27

x = -27/(-39)

x = 9/13

Step 3:

Apply x = 9/13 in the equation y =(9 x – 7)/2

y = [9(9/13) – 7]/2

y = [(81/13) – 7]/2

y = [(81 –91)/26]

y = -10/26

y = -5/13

Therefore solution is x = 9/13 and y = -5/13

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