Question

3x - 5y - 4 = 0 and 9x = 2y + 7 by substitution method​

Answer 1

Answer:

There are two equations provided here.

3x – 5y – 4 = 0 or 3x -5y = 4 ……. (1).

9x = 2y + 7 or 9x -2y = 7 ……(2).

Now only focusing on eq. numbered 1 and 2 …

3x – 5y = 4………(1)

9x – 2y = 7 ……(2)

Multiply eq. (1) by 3, we get….

3 (3x – 5y = 4) or 9x – 15y = 12…… (3).

Now we can easily subtract (2) and (3) to get…..

9x – 2y = 7

– 9x – 15y = 12.

– +13 y = -5

Or we get 13 y = -5 or y = -5/13.

Now we substitute the value of Y = -5/13 in equation (1) we get the value of x.

3x – 5(-5/13) = 4

3x +25/13 = 4

39x + 25 = 52

Or 39x = 52 – 25

39x = 27

Or x = 27/39 = 9 / 13

So x = 9/13 and y = -5/13.

Answer 2

Solution :-

3x - 5y - 4 = 0 → 3x - 5y = 4 (1)

9x = 2y + 7 → 9x - 2y = 7 (2)

From equation (1) :-

3x - 5y = 4

3x = 4 + 5y

x = (4 + 5y)/3

Substituting x = (4 + 5y)/3 in equation (2) :-

9x - 2y = 7

9 × (4 + 5y)/3 - 2y = 7

9 × (4 + 5y) - 6y = 21

36 + 45y - 6y = 21

39y = 21 - 36

y = -15/39

Substituting the value of y in equation (1) :-

3x - 5y = 4

3x - 5 × (-15/39) = 4

3x + 75/39 = 4

3x = 4 - 75/39

3x = 156 - 75 / 39

x = 81/39 × 1/3

x = 27/39

Proof :-

Put the value of x and y in any of the equation :-

In equation (1) :-

3x - 5y = 4

L.H.S. → 3 × 27/39 - 5 × -15/39

→ 81/39 + 75/39

→ 81 + 75 / 39

→ 156/39 = 4 = R.H.S. ✓✓

In equation (2) :-

9x - 2y = 7

L.H.S. → 9 × 27/39 - 2 × -15/39

→ 243/39 + 30/39

→ 243 + 30 / 39

→ 273/39 = 7 = R.H.S. ✓✓

Hence, proved.