Solve by Addition/Elimination 3x-8y=9 -3x+10y=-15

$3 x - 8 y = 9$ $- 3 x + 10 y = - 15$

Add the two equations together to eliminate $x$ from the system.

Divide each term in $2 y = - 6$ by $2$ and simplify.

Divide each term in $2 y = - 6$ by $2$ .

$\frac{2 y}{2} = \frac{- 6}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 y}{2} = \frac{- 6}{2}$

Divide $y$ by $1$ .

$y = \frac{- 6}{2}$

Simplify the right side.

Divide $- 6$ by $2$ .

$y = - 3$

Substitute the value found for $y$ into one of the original equations, then solve for $x$ .

Substitute the value found for $y$ into one of the original equations to solve for $x$ .

$3 x - 8 \cdot - 3 = 9$

Multiply $- 8$ by $- 3$ .

$3 x + 24 = 9$

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

Subtract $24$ from both sides of the equation.

$3 x = 9 - 24$

Subtract $24$ from $9$ .

$3 x = - 15$

Divide each term in $3 x = - 15$ by $3$ and simplify.

Divide each term in $3 x = - 15$ by $3$ .

$\frac{3 x}{3} = \frac{- 15}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 15}{3}$

Divide $x$ by $1$ .

$x = \frac{- 15}{3}$

Simplify the right side.

Divide $- 15$ by $3$ .

$x = - 5$

The solution to the independent system of equations can be represented as a point.

$(- 5, - 3)$

The result can be shown in multiple forms.

Point Form:

$(- 5, - 3)$

Equation Form:

$x = - 5, y = - 3$