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

(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

**Exercise 3.3 Part 1****Exercise 3.3 Part 2****Exercise 3.3 Part 3****Exercise 3.3 Part 4****Exercise 3.3 Part 5****Go to exercise 3.4**

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