Question

solve the following system of equations.
$2x + 4y = 8$
$3x + 4y = 16$
$x = \square$
$y = \square$

Explanation:

Step1: Subtract the first equation from the second equation to eliminate \( y \).

The two equations are:
\( 2x + 4y = 8 \) (Equation 1)
\( 3x + 4y = 16 \) (Equation 2)

Subtract Equation 1 from Equation 2:
\( (3x + 4y) - (2x + 4y) = 16 - 8 \)
Simplify the left side: \( 3x + 4y - 2x - 4y = x \)
Simplify the right side: \( 8 \)
So, \( x = 8 \)

Step2: Substitute \( x = 8 \) into Equation 1 to find \( y \).

Substitute \( x = 8 \) into \( 2x + 4y = 8 \):
\( 2(8) + 4y = 8 \)
\( 16 + 4y = 8 \)
Subtract 16 from both sides: \( 4y = 8 - 16 \)
\( 4y = -8 \)
Divide both sides by 4: \( y = \frac{-8}{4} = -2 \)

Answer:

\( x = 8 \)
\( y = -2 \)