Solve by Substitution y=-1/4x-1 , 2x-4y=16

$y = - \frac{1}{4} x - 1$ , $2 x - 4 y = 16$

Step 1

Combine $x$ and $\frac{1}{4}$ .

$y = - \frac{x}{4} - 1$

$2 x - 4 y = 16$

Step 2

Replace all occurrences of $y$ with $- \frac{x}{4} - 1$ in each equation.

Replace all occurrences of $y$ in $2 x - 4 y = 16$ with $- \frac{x}{4} - 1$ .

$2 x - 4 (- \frac{x}{4} - 1) = 16$

$y = - \frac{x}{4} - 1$

Simplify the left side.

Simplify $2 x - 4 (- \frac{x}{4} - 1)$ .

Simplify each term.

Apply the distributive property.

$2 x - 4 (- \frac{x}{4}) - 4 \cdot - 1 = 16$

$y = - \frac{x}{4} - 1$

Cancel the common factor of $4$ .

Move the leading negative in $- \frac{x}{4}$ into the numerator.

$2 x - 4 \frac{- x}{4} - 4 \cdot - 1 = 16$

$y = - \frac{x}{4} - 1$

Factor $4$ out of $- 4$ .

$2 x + 4 (- 1) (\frac{- x}{4}) - 4 \cdot - 1 = 16$

$y = - \frac{x}{4} - 1$

Cancel the common factor.

$2 x + 4 \cdot (- 1 \frac{- x}{4}) - 4 \cdot - 1 = 16$

$y = - \frac{x}{4} - 1$

Rewrite the expression.

$2 x - 1 (- x) - 4 \cdot - 1 = 16$

$y = - \frac{x}{4} - 1$

$2 x - 1 (- x) - 4 \cdot - 1 = 16$

$y = - \frac{x}{4} - 1$

Multiply $- 1$ by $- 1$ .

$2 x + 1 x - 4 \cdot - 1 = 16$

$y = - \frac{x}{4} - 1$

Multiply $x$ by $1$ .

$2 x + x - 4 \cdot - 1 = 16$

$y = - \frac{x}{4} - 1$

Multiply $- 4$ by $- 1$ .

$2 x + x + 4 = 16$

$y = - \frac{x}{4} - 1$

$2 x + x + 4 = 16$

$y = - \frac{x}{4} - 1$

Add $2 x$ and $x$ .

$3 x + 4 = 16$

$y = - \frac{x}{4} - 1$

$3 x + 4 = 16$

$y = - \frac{x}{4} - 1$

$3 x + 4 = 16$

$y = - \frac{x}{4} - 1$

$3 x + 4 = 16$

$y = - \frac{x}{4} - 1$

Step 3

Solve for $x$ in $3 x + 4 = 16$ .

Move all terms not containing $x$ to the right side of the equation.

Subtract $4$ from both sides of the equation.

$3 x = 16 - 4$

$y = - \frac{x}{4} - 1$

Subtract $4$ from $16$ .

$3 x = 12$

$y = - \frac{x}{4} - 1$

$3 x = 12$

$y = - \frac{x}{4} - 1$

Divide each term in $3 x = 12$ by $3$ and simplify.

Divide each term in $3 x = 12$ by $3$ .

$\frac{3 x}{3} = \frac{12}{3}$

$y = - \frac{x}{4} - 1$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{12}{3}$

$y = - \frac{x}{4} - 1$

Divide $x$ by $1$ .

$x = \frac{12}{3}$

$y = - \frac{x}{4} - 1$

$x = \frac{12}{3}$

$y = - \frac{x}{4} - 1$

$x = \frac{12}{3}$

$y = - \frac{x}{4} - 1$

Simplify the right side.

Divide $12$ by $3$ .

$x = 4$

$y = - \frac{x}{4} - 1$

$x = 4$

$y = - \frac{x}{4} - 1$

$x = 4$

$y = - \frac{x}{4} - 1$

$x = 4$

$y = - \frac{x}{4} - 1$

Step 4

Replace all occurrences of $x$ with $4$ in each equation.

Replace all occurrences of $x$ in $y = - \frac{x}{4} - 1$ with $4$ .

$y = - \frac{4}{4} - 1$

$x = 4$

Simplify the right side.

Simplify $- \frac{4}{4} - 1$ .

Simplify each term.

Cancel the common factor of $4$ .

Cancel the common factor.

$y = - \frac{4}{4} - 1$

$x = 4$

Rewrite the expression.

$y = - 1 \cdot 1 - 1$

$x = 4$

$y = - 1 \cdot 1 - 1$

$x = 4$

Multiply $- 1$ by $1$ .

$y = - 1 - 1$

$x = 4$

$y = - 1 - 1$

$x = 4$

Subtract $1$ from $- 1$ .

$y = - 2$

$x = 4$

$y = - 2$

$x = 4$

$y = - 2$

$x = 4$

$y = - 2$

$x = 4$

Step 5

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(4, - 2)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(4, - 2)$

Equation Form:

$x = 4, y = - 2$

Step 7

Solve by Substitution y=-1/4x-1 , 2x-4y=16

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