Question

solve the system of equations:
$12x - 5y = 40$
$12x - 11y = 88$
$x = square$
$y = square$

Explanation:

Step1: Label the equations

Let:
$$12x - 5y = 40 \tag{1}$$
$$12x - 11y = 88 \tag{2}$$

Step2: Eliminate $x$ via subtraction

Subtract equation (2) from equation (1) to eliminate $12x$:
$$(12x - 5y) - (12x - 11y) = 40 - 88$$
$$12x -5y -12x +11y = -48$$
$$6y = -48$$

Step3: Solve for $y$

Divide both sides by 6:
$$y = \frac{-48}{6} = -8$$

Step4: Substitute $y=-8$ into (1)

Plug $y=-8$ into equation (1) to solve for $x$:
$$12x - 5(-8) = 40$$
$$12x + 40 = 40$$

Step5: Solve for $x$

Subtract 40 from both sides, then divide by 12:
$$12x = 40 - 40 = 0$$
$$x = \frac{0}{12} = 0$$

Answer:

$x = 0$
$y = -8$