Solve by Substitution 5x-6y=-38 3x+4y=0

$5 x - 6 y = - 38$ $3 x + 4 y = 0$

Solve for $x$ in the first equation.

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Add $6 y$ to both sides of the equation.

$5 x = - 38 + 6 y$

$3 x + 4 y = 0$

Divide each term by $5$ and simplify.

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Divide each term in $5 x = - 38 + 6 y$ by $5$ .

$\frac{5 x}{5} = \frac{- 38}{5} + \frac{6 y}{5}$

$3 x + 4 y = 0$

Cancel the common factor of $5$ .

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Cancel the common factor.

$\frac{5 x}{5} = \frac{- 38}{5} + \frac{6 y}{5}$

$3 x + 4 y = 0$

Divide $x$ by $1$ .

$x = \frac{- 38}{5} + \frac{6 y}{5}$

$3 x + 4 y = 0$

$x = \frac{- 38}{5} + \frac{6 y}{5}$

$3 x + 4 y = 0$

Move the negative in front of the fraction.

$x = - \frac{38}{5} + \frac{6 y}{5}$

$3 x + 4 y = 0$

$x = - \frac{38}{5} + \frac{6 y}{5}$

$3 x + 4 y = 0$

$x = - \frac{38}{5} + \frac{6 y}{5}$

$3 x + 4 y = 0$

Replace all occurrences of $x$ with $- \frac{38}{5} + \frac{6 y}{5}$ in each equation.

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Replace all occurrences of $x$ in $3 x + 4 y = 0$ with $- \frac{38}{5} + \frac{6 y}{5}$ .

$3 (- \frac{38}{5} + \frac{6 y}{5}) + 4 y = 0$

$x = - \frac{38}{5} + \frac{6 y}{5}$

Simplify $3 (- \frac{38}{5} + \frac{6 y}{5}) + 4 y$ .

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Simplify each term.

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Apply the distributive property.

$3 (- \frac{38}{5}) + 3 (\frac{6 y}{5}) + 4 y = 0$

$x = - \frac{38}{5} + \frac{6 y}{5}$

Multiply $3 (- \frac{38}{5})$ .

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Multiply $- 1$ by $3$ .

$- 3 (\frac{38}{5}) + 3 (\frac{6 y}{5}) + 4 y = 0$

$x = - \frac{38}{5} + \frac{6 y}{5}$

Combine $- 3$ and $\frac{38}{5}$ .

$\frac{- 3 \cdot 38}{5} + 3 (\frac{6 y}{5}) + 4 y = 0$

$x = - \frac{38}{5} + \frac{6 y}{5}$

Multiply $- 3$ by $38$ .

$\frac{- 114}{5} + 3 (\frac{6 y}{5}) + 4 y = 0$

$x = - \frac{38}{5} + \frac{6 y}{5}$

$\frac{- 114}{5} + 3 (\frac{6 y}{5}) + 4 y = 0$

$x = - \frac{38}{5} + \frac{6 y}{5}$

Multiply $3 \frac{6 y}{5}$ .

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Combine $3$ and $\frac{6 y}{5}$ .

$\frac{- 114}{5} + \frac{3 (6 y)}{5} + 4 y = 0$

$x = - \frac{38}{5} + \frac{6 y}{5}$

Multiply $6$ by $3$ .

$\frac{- 114}{5} + \frac{18 y}{5} + 4 y = 0$

$x = - \frac{38}{5} + \frac{6 y}{5}$

$\frac{- 114}{5} + \frac{18 y}{5} + 4 y = 0$

$x = - \frac{38}{5} + \frac{6 y}{5}$

Move the negative in front of the fraction.

$- \frac{114}{5} + \frac{18 y}{5} + 4 y = 0$

$x = - \frac{38}{5} + \frac{6 y}{5}$

$- \frac{114}{5} + \frac{18 y}{5} + 4 y = 0$

$x = - \frac{38}{5} + \frac{6 y}{5}$

To write $4 y$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$- \frac{114}{5} + \frac{18 y}{5} + 4 y \cdot \frac{5}{5} = 0$

$x = - \frac{38}{5} + \frac{6 y}{5}$

Combine $4 y$ and $\frac{5}{5}$ .

$- \frac{114}{5} + \frac{18 y}{5} + \frac{4 y \cdot 5}{5} = 0$

$x = - \frac{38}{5} + \frac{6 y}{5}$

Combine the numerators over the common denominator.

$- \frac{114}{5} + \frac{18 y + 4 y \cdot 5}{5} = 0$

$x = - \frac{38}{5} + \frac{6 y}{5}$

Combine the numerators over the common denominator.

$\frac{- 114 + 18 y + 4 y \cdot 5}{5} = 0$

$x = - \frac{38}{5} + \frac{6 y}{5}$

Multiply $5$ by $4$ .

$\frac{- 114 + 18 y + 20 y}{5} = 0$

$x = - \frac{38}{5} + \frac{6 y}{5}$

Add $18 y$ and $20 y$ .

$\frac{- 114 + 38 y}{5} = 0$

$x = - \frac{38}{5} + \frac{6 y}{5}$

Factor $38$ out of $- 114 + 38 y$ .

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Factor $38$ out of $- 114$ .

$\frac{38 \cdot - 3 + 38 y}{5} = 0$

$x = - \frac{38}{5} + \frac{6 y}{5}$

Factor $38$ out of $38 \cdot - 3 + 38 y$ .

$\frac{38 (- 3 + y)}{5} = 0$

$x = - \frac{38}{5} + \frac{6 y}{5}$

$\frac{38 (- 3 + y)}{5} = 0$

$x = - \frac{38}{5} + \frac{6 y}{5}$

Rewrite $- 3$ as $- 1 (3)$ .

$\frac{38 (- 1 \cdot 3 + y)}{5} = 0$

$x = - \frac{38}{5} + \frac{6 y}{5}$

Factor $- 1$ out of $y$ .

$\frac{38 (- 1 \cdot 3 - 1 (- y))}{5} = 0$

$x = - \frac{38}{5} + \frac{6 y}{5}$

Factor $- 1$ out of $- 1 (3) - 1 (- y)$ .

$\frac{38 (- 1 (3 - y))}{5} = 0$

$x = - \frac{38}{5} + \frac{6 y}{5}$

Move the negative in front of the fraction.

$- \frac{38 (3 - y)}{5} = 0$

$x = - \frac{38}{5} + \frac{6 y}{5}$

$- \frac{38 (3 - y)}{5} = 0$

$x = - \frac{38}{5} + \frac{6 y}{5}$

$- \frac{38 (3 - y)}{5} = 0$

$x = - \frac{38}{5} + \frac{6 y}{5}$

Solve for $y$ in the first equation.

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Set the numerator equal to zero.

$38 (3 - y) = 0$

$x = - \frac{38}{5} + \frac{6 y}{5}$

Solve the equation for $y$ .

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Divide each term by $38$ and simplify.

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Divide each term in $38 (3 - y) = 0$ by $38$ .

$\frac{38 (3 - y)}{38} = \frac{0}{38}$

$x = - \frac{38}{5} + \frac{6 y}{5}$

Cancel the common factor of $38$ .

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Cancel the common factor.

$\frac{38 (3 - y)}{38} = \frac{0}{38}$

$x = - \frac{38}{5} + \frac{6 y}{5}$

Divide $3 - y$ by $1$ .

$3 - y = \frac{0}{38}$

$x = - \frac{38}{5} + \frac{6 y}{5}$

$3 - y = \frac{0}{38}$

$x = - \frac{38}{5} + \frac{6 y}{5}$

Divide $0$ by $38$ .

$3 - y = 0$

$x = - \frac{38}{5} + \frac{6 y}{5}$

$3 - y = 0$

$x = - \frac{38}{5} + \frac{6 y}{5}$

Subtract $3$ from both sides of the equation.

$- y = - 3$

$x = - \frac{38}{5} + \frac{6 y}{5}$

Multiply each term in $- y = - 3$ by $- 1$

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Multiply each term in $- y = - 3$ by $- 1$ .

$(- y) \cdot - 1 = (- 3) \cdot - 1$

$x = - \frac{38}{5} + \frac{6 y}{5}$

Multiply $(- y) \cdot - 1$ .

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Multiply $- 1$ by $- 1$ .

$1 y = (- 3) \cdot - 1$

$x = - \frac{38}{5} + \frac{6 y}{5}$

Multiply $y$ by $1$ .

$y = (- 3) \cdot - 1$

$x = - \frac{38}{5} + \frac{6 y}{5}$

$y = (- 3) \cdot - 1$

$x = - \frac{38}{5} + \frac{6 y}{5}$

Multiply $- 3$ by $- 1$ .

$y = 3$

$x = - \frac{38}{5} + \frac{6 y}{5}$

$y = 3$

$x = - \frac{38}{5} + \frac{6 y}{5}$

$y = 3$

$x = - \frac{38}{5} + \frac{6 y}{5}$

$y = 3$

$x = - \frac{38}{5} + \frac{6 y}{5}$

Replace all occurrences of $y$ with $3$ in each equation.

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Replace all occurrences of $y$ in $x = - \frac{38}{5} + \frac{6 y}{5}$ with $3$ .

$x = - \frac{38}{5} + \frac{6 (3)}{5}$

$y = 3$

Simplify $- \frac{38}{5} + \frac{6 (3)}{5}$ .

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Combine the numerators over the common denominator.

$x = \frac{- 38 + 6 (3)}{5}$

$y = 3$

Simplify the expression.

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Multiply $6$ by $3$ .

$x = \frac{- 38 + 18}{5}$

$y = 3$

Add $- 38$ and $18$ .

$x = \frac{- 20}{5}$

$y = 3$

Divide $- 20$ by $5$ .

$x = - 4$

$y = 3$

$x = - 4$

$y = 3$

$x = - 4$

$y = 3$

$x = - 4$

$y = 3$

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

$(- 4, 3)$

The result can be shown in multiple forms.

Point Form:

$(- 4, 3)$

Equation Form:

$x = - 4, y = 3$

Solve by Substitution 5x-6y=-38 3x+4y=0

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