Question

Solve the following for x and y using the method of substitution:
4x + 3y = 12; x - 6y = 9

• A. x = $\frac{33}{9}$, y = $-\frac{8}{9}$
• B. x = $\frac{33}{9}$, y = $-\frac{7}{9}$
• C. x = $\frac{30}{9}$, y = $-\frac{8}{9}$
• D. x = $\frac{30}{9}$, y = $-\frac{7}{9}$

Answer 1

The correct option is A x = $\frac{33}{9}$, y = $-\frac{8}{9}$

Given equations are:
4x + 3y = 12 --------(i)

x - 6y = 9 --------(ii)

From equation (ii),

x = 6y + 9 --------(iii)

Put this value in equation (i)

$⟹$ 4(6y + 9) + 3y = 12

$⟹$24y + 36 + 3y = 12

$⟹$27y = -24

$⟹\text{y}=-\frac{8}{9}$

Put the value of y in equation (iii):
x = 6y + 9
= $-\frac{48}{9}$ + 9
= $\frac{33}{9}$

$\therefore$ The values of x and y are $\frac{33}{9}$ and -$\frac{8}{9}$ respectively.