Solve the System of Equations 7x-4y=-8 y=3/4x-3

$7 x - 4 y = - 8$ $y = \frac{3}{4} x - 3$

Step 1

Combine $\frac{3}{4}$ and $x$ .

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

$7 x - 4 y = - 8$

Step 2

Replace all occurrences of $y$ with $\frac{3 x}{4} - 3$ in each equation.

Replace all occurrences of $y$ in $7 x - 4 y = - 8$ with $\frac{3 x}{4} - 3$ .

$7 x - 4 (\frac{3 x}{4} - 3) = - 8$

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

Simplify the left side.

Simplify $7 x - 4 (\frac{3 x}{4} - 3)$ .

Simplify each term.

Apply the distributive property.

$7 x - 4 \frac{3 x}{4} - 4 \cdot - 3 = - 8$

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

Cancel the common factor of $4$ .

Factor $4$ out of $- 4$ .

$7 x + 4 (- 1) (\frac{3 x}{4}) - 4 \cdot - 3 = - 8$

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

Cancel the common factor.

$7 x + 4 \cdot (- 1 \frac{3 x}{4}) - 4 \cdot - 3 = - 8$

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

Rewrite the expression.

$7 x - 1 (3 x) - 4 \cdot - 3 = - 8$

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

$7 x - 1 (3 x) - 4 \cdot - 3 = - 8$

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

Multiply $3$ by $- 1$ .

$7 x - 3 x - 4 \cdot - 3 = - 8$

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

Multiply $- 4$ by $- 3$ .

$7 x - 3 x + 12 = - 8$

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

$7 x - 3 x + 12 = - 8$

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

Subtract $3 x$ from $7 x$ .

$4 x + 12 = - 8$

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

$4 x + 12 = - 8$

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

$4 x + 12 = - 8$

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

$4 x + 12 = - 8$

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

Step 3

Solve for $x$ in $4 x + 12 = - 8$ .

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

Subtract $12$ from both sides of the equation.

$4 x = - 8 - 12$

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

Subtract $12$ from $- 8$ .

$4 x = - 20$

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

$4 x = - 20$

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

Divide each term in $4 x = - 20$ by $4$ and simplify.

Divide each term in $4 x = - 20$ by $4$ .

$\frac{4 x}{4} = \frac{- 20}{4}$

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

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{- 20}{4}$

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

Divide $x$ by $1$ .

$x = \frac{- 20}{4}$

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

$x = \frac{- 20}{4}$

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

$x = \frac{- 20}{4}$

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

Simplify the right side.

Divide $- 20$ by $4$ .

$x = - 5$

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

$x = - 5$

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

$x = - 5$

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

$x = - 5$

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

Step 4

Replace all occurrences of $x$ with $- 5$ in each equation.

Replace all occurrences of $x$ in $y = \frac{3 x}{4} - 3$ with $- 5$ .

$y = \frac{3 (- 5)}{4} - 3$

$x = - 5$

Simplify the right side.

Simplify $\frac{3 (- 5)}{4} - 3$ .

Multiply $3$ by $- 5$ .

$y = \frac{- 15}{4} - 3$

$x = - 5$

Move the negative in front of the fraction.

$y = - \frac{15}{4} - 3$

$x = - 5$

To write $- 3$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$y = - \frac{15}{4} - 3 \cdot \frac{4}{4}$

$x = - 5$

Combine $- 3$ and $\frac{4}{4}$ .

$y = - \frac{15}{4} + \frac{- 3 \cdot 4}{4}$

$x = - 5$

Combine the numerators over the common denominator.

$y = \frac{- 15 - 3 \cdot 4}{4}$

$x = - 5$

Simplify the numerator.

Multiply $- 3$ by $4$ .

$y = \frac{- 15 - 12}{4}$

$x = - 5$

Subtract $12$ from $- 15$ .

$y = \frac{- 27}{4}$

$x = - 5$

$y = \frac{- 27}{4}$

$x = - 5$

Move the negative in front of the fraction.

$y = - \frac{27}{4}$

$x = - 5$

$y = - \frac{27}{4}$

$x = - 5$

$y = - \frac{27}{4}$

$x = - 5$

$y = - \frac{27}{4}$

$x = - 5$

Step 5

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

$(- 5, - \frac{27}{4})$

Step 6

The result can be shown in multiple forms.

Point Form:

$(- 5, - \frac{27}{4})$

Equation Form:

$x = - 5, y = - \frac{27}{4}$

Step 7

Solve the System of Equations 7x-4y=-8 y=3/4x-3

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