Solve by Substitution 4x+13y=40 4x+3y=-40

$4 x + 13 y = 40$ $4 x + 3 y = - 40$

Step 1

Solve for $x$ in $4 x + 13 y = 40$ .

Subtract $13 y$ from both sides of the equation.

$4 x = 40 - 13 y$

$4 x + 3 y = - 40$

Divide each term in $4 x = 40 - 13 y$ by $4$ and simplify.

Divide each term in $4 x = 40 - 13 y$ by $4$ .

$\frac{4 x}{4} = \frac{40}{4} + \frac{- 13 y}{4}$

$4 x + 3 y = - 40$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{40}{4} + \frac{- 13 y}{4}$

$4 x + 3 y = - 40$

Divide $x$ by $1$ .

$x = \frac{40}{4} + \frac{- 13 y}{4}$

$4 x + 3 y = - 40$

$x = \frac{40}{4} + \frac{- 13 y}{4}$

$4 x + 3 y = - 40$

$x = \frac{40}{4} + \frac{- 13 y}{4}$

$4 x + 3 y = - 40$

Simplify the right side.

Simplify each term.

Divide $40$ by $4$ .

$x = 10 + \frac{- 13 y}{4}$

$4 x + 3 y = - 40$

Move the negative in front of the fraction.

$x = 10 - \frac{13 y}{4}$

$4 x + 3 y = - 40$

$x = 10 - \frac{13 y}{4}$

$4 x + 3 y = - 40$

$x = 10 - \frac{13 y}{4}$

$4 x + 3 y = - 40$

$x = 10 - \frac{13 y}{4}$

$4 x + 3 y = - 40$

$x = 10 - \frac{13 y}{4}$

$4 x + 3 y = - 40$

Step 2

Replace all occurrences of $x$ with $10 - \frac{13 y}{4}$ in each equation.

Replace all occurrences of $x$ in $4 x + 3 y = - 40$ with $10 - \frac{13 y}{4}$ .

$4 (10 - \frac{13 y}{4}) + 3 y = - 40$

$x = 10 - \frac{13 y}{4}$

Simplify the left side.

Simplify $4 (10 - \frac{13 y}{4}) + 3 y$ .

Simplify each term.

Apply the distributive property.

$4 \cdot 10 + 4 (- \frac{13 y}{4}) + 3 y = - 40$

$x = 10 - \frac{13 y}{4}$

Multiply $4$ by $10$ .

$40 + 4 (- \frac{13 y}{4}) + 3 y = - 40$

$x = 10 - \frac{13 y}{4}$

Cancel the common factor of $4$ .

Move the leading negative in $- \frac{13 y}{4}$ into the numerator.

$40 + 4 (\frac{- 13 y}{4}) + 3 y = - 40$

$x = 10 - \frac{13 y}{4}$

Cancel the common factor.

$40 + 4 (\frac{- 13 y}{4}) + 3 y = - 40$

$x = 10 - \frac{13 y}{4}$

Rewrite the expression.

$40 - 13 y + 3 y = - 40$

$x = 10 - \frac{13 y}{4}$

$40 - 13 y + 3 y = - 40$

$x = 10 - \frac{13 y}{4}$

$40 - 13 y + 3 y = - 40$

$x = 10 - \frac{13 y}{4}$

Add $- 13 y$ and $3 y$ .

$40 - 10 y = - 40$

$x = 10 - \frac{13 y}{4}$

$40 - 10 y = - 40$

$x = 10 - \frac{13 y}{4}$

$40 - 10 y = - 40$

$x = 10 - \frac{13 y}{4}$

$40 - 10 y = - 40$

$x = 10 - \frac{13 y}{4}$

Step 3

Solve for $y$ in $40 - 10 y = - 40$ .

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

Subtract $40$ from both sides of the equation.

$- 10 y = - 40 - 40$

$x = 10 - \frac{13 y}{4}$

Subtract $40$ from $- 40$ .

$- 10 y = - 80$

$x = 10 - \frac{13 y}{4}$

$- 10 y = - 80$

$x = 10 - \frac{13 y}{4}$

Divide each term in $- 10 y = - 80$ by $- 10$ and simplify.

Divide each term in $- 10 y = - 80$ by $- 10$ .

$\frac{- 10 y}{- 10} = \frac{- 80}{- 10}$

$x = 10 - \frac{13 y}{4}$

Simplify the left side.

Cancel the common factor of $- 10$ .

Cancel the common factor.

$\frac{- 10 y}{- 10} = \frac{- 80}{- 10}$

$x = 10 - \frac{13 y}{4}$

Divide $y$ by $1$ .

$y = \frac{- 80}{- 10}$

$x = 10 - \frac{13 y}{4}$

$y = \frac{- 80}{- 10}$

$x = 10 - \frac{13 y}{4}$

$y = \frac{- 80}{- 10}$

$x = 10 - \frac{13 y}{4}$

Simplify the right side.

Divide $- 80$ by $- 10$ .

$y = 8$

$x = 10 - \frac{13 y}{4}$

$y = 8$

$x = 10 - \frac{13 y}{4}$

$y = 8$

$x = 10 - \frac{13 y}{4}$

$y = 8$

$x = 10 - \frac{13 y}{4}$

Step 4

Replace all occurrences of $y$ with $8$ in each equation.

Replace all occurrences of $y$ in $x = 10 - \frac{13 y}{4}$ with $8$ .

$x = 10 - \frac{13 (8)}{4}$

$y = 8$

Simplify the right side.

Simplify $10 - \frac{13 (8)}{4}$ .

Simplify each term.

Cancel the common factor of $8$ and $4$ .

Factor $4$ out of $13 (8)$ .

$x = 10 - \frac{4 (13 \cdot (2))}{4}$

$y = 8$

Cancel the common factors.

Factor $4$ out of $4$ .

$x = 10 - \frac{4 (13 \cdot (2))}{4 (1)}$

$y = 8$

Cancel the common factor.

$x = 10 - \frac{4 (13 \cdot (2))}{4 \cdot 1}$

$y = 8$

Rewrite the expression.

$x = 10 - \frac{13 \cdot (2)}{1}$

$y = 8$

Divide $13 \cdot (2)$ by $1$ .

$x = 10 - (13 \cdot (2))$

$y = 8$

$x = 10 - (13 \cdot (2))$

$y = 8$

$x = 10 - (13 \cdot (2))$

$y = 8$

Multiply $- (13 \cdot (2))$ .

Multiply $13$ by $2$ .

$x = 10 - 1 \cdot 26$

$y = 8$

Multiply $- 1$ by $26$ .

$x = 10 - 26$

$y = 8$

$x = 10 - 26$

$y = 8$

$x = 10 - 26$

$y = 8$

Subtract $26$ from $10$ .

$x = - 16$

$y = 8$

$x = - 16$

$y = 8$

$x = - 16$

$y = 8$

$x = - 16$

$y = 8$

Step 5

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

$(- 16, 8)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(- 16, 8)$

Equation Form:

$x = - 16, y = 8$

Step 7

Solve by Substitution 4x+13y=40 4x+3y=-40

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