Solve the System of Equations 4x=-12y+16 and x+3y=4

$4 x = - 12 y + 16$ and $x + 3 y = 4$

Step 1

Divide each term in $4 x = - 12 y + 16$ by $4$ and simplify.

Divide each term in $4 x = - 12 y + 16$ by $4$ .

$\frac{4 x}{4} = \frac{- 12 y}{4} + \frac{16}{4}$

$x + 3 y = 4$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{- 12 y}{4} + \frac{16}{4}$

$x + 3 y = 4$

Divide $x$ by $1$ .

$x = \frac{- 12 y}{4} + \frac{16}{4}$

$x + 3 y = 4$

$x = \frac{- 12 y}{4} + \frac{16}{4}$

$x + 3 y = 4$

$x = \frac{- 12 y}{4} + \frac{16}{4}$

$x + 3 y = 4$

Simplify the right side.

Simplify each term.

Cancel the common factor of $- 12$ and $4$ .

Factor $4$ out of $- 12 y$ .

$x = \frac{4 (- 3 y)}{4} + \frac{16}{4}$

$x + 3 y = 4$

Cancel the common factors.

Factor $4$ out of $4$ .

$x = \frac{4 (- 3 y)}{4 (1)} + \frac{16}{4}$

$x + 3 y = 4$

Cancel the common factor.

$x = \frac{4 (- 3 y)}{4 \cdot 1} + \frac{16}{4}$

$x + 3 y = 4$

Rewrite the expression.

$x = \frac{- 3 y}{1} + \frac{16}{4}$

$x + 3 y = 4$

Divide $- 3 y$ by $1$ .

$x = - 3 y + \frac{16}{4}$

$x + 3 y = 4$

$x = - 3 y + \frac{16}{4}$

$x + 3 y = 4$

$x = - 3 y + \frac{16}{4}$

$x + 3 y = 4$

Divide $16$ by $4$ .

$x = - 3 y + 4$

$x + 3 y = 4$

$x = - 3 y + 4$

$x + 3 y = 4$

$x = - 3 y + 4$

$x + 3 y = 4$

$x = - 3 y + 4$

$x + 3 y = 4$

Step 2

Replace all occurrences of $x$ with $- 3 y + 4$ in each equation.

Replace all occurrences of $x$ in $x + 3 y = 4$ with $- 3 y + 4$ .

$(- 3 y + 4) + 3 y = 4$

$x = - 3 y + 4$

Simplify the left side.

Combine the opposite terms in $(- 3 y + 4) + 3 y$ .

Add $- 3 y$ and $3 y$ .

$0 + 4 = 4$

$x = - 3 y + 4$

Add $0$ and $4$ .

$4 = 4$

$x = - 3 y + 4$

$4 = 4$

$x = - 3 y + 4$

$4 = 4$

$x = - 3 y + 4$

$4 = 4$

$x = - 3 y + 4$

Step 3

Remove any equations from the system that are always true.

$x = - 3 y + 4$

Step 4

Solve the System of Equations 4x=-12y+16 and x+3y=4

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