Solve {l}{2x-3y-10=0}{7y=-17-8x}

$$\left\{ \begin{array} { l } { 2 x - 3 y - 10 = 0 } \\ { 7 y = - 17 - 8 x } \end{array} \right.$$

Solve for x, y

$$x=\frac{1}{2}=0.5$$

$$y=-3$$

Steps Using Substitution

Consider the first equation. Add 10 to both sides. Anything plus zero gives itself.

$$2x-3y=10$$

Consider the second equation. Add 8x to both sides.

$$7y+8x=-17$$

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$$2x-3y=10,8x+7y=-17$$

Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.

$$2x-3y=10$$

Add 3y to both sides of the equation.

$$2x=3y+10$$

Divide both sides by 2.

$$x=\frac{1}{2}\left(3y+10\right)$$

Multiply \frac{1}{2} times 3y+10.

$$x=\frac{3}{2}y+5$$

Substitute \frac{3y}{2}+5 for x in the other equation, 8x+7y=-17.

$$8\left(\frac{3}{2}y+5\right)+7y=-17$$

Multiply 8 times \frac{3y}{2}+5.

$$12y+40+7y=-17$$

Add 12y to 7y.

$$19y+40=-17$$

Subtract 40 from both sides of the equation.

$$19y=-57$$

Divide both sides by 19.

$$y=-3$$

Substitute -3 for y in x=\frac{3}{2}y+5. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{3}{2}\left(-3\right)+5$$

Multiply \frac{3}{2} times -3.

$$x=-\frac{9}{2}+5$$

Add 5 to -\frac{9}{2}.

$$x=\frac{1}{2}$$

The system is now solved.

$$x=\frac{1}{2},y=-3$$

Steps Using Matrices

Consider the first equation. Add 10 to both sides. Anything plus zero gives itself.

$$2x-3y=10$$

Consider the second equation. Add 8x to both sides.

$$7y+8x=-17$$

Put the equations in standard form and then use matrices to solve the system of equations.

$$2x-3y=10,8x+7y=-17$$

Write the equations in matrix form.

$$\left(\begin{matrix}2&-3\\8&7\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}10\\-17\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&-3\\8&7\end{matrix}\right).

$$inverse(\left(\begin{matrix}2&-3\\8&7\end{matrix}\right))\left(\begin{matrix}2&-3\\8&7\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-3\\8&7\end{matrix}\right))\left(\begin{matrix}10\\-17\end{matrix}\right)$$

The product of a matrix and its inverse is the identity matrix.

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-3\\8&7\end{matrix}\right))\left(\begin{matrix}10\\-17\end{matrix}\right)$$

Multiply the matrices on the left hand side of the equal sign.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-3\\8&7\end{matrix}\right))\left(\begin{matrix}10\\-17\end{matrix}\right)$$

For the 2\times 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the inverse matrix is \left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right), so the matrix equation can be rewritten as a matrix multiplication problem.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{7}{2\times 7-\left(-3\times 8\right)}&-\frac{-3}{2\times 7-\left(-3\times 8\right)}\\-\frac{8}{2\times 7-\left(-3\times 8\right)}&\frac{2}{2\times 7-\left(-3\times 8\right)}\end{matrix}\right)\left(\begin{matrix}10\\-17\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{7}{38}&\frac{3}{38}\\-\frac{4}{19}&\frac{1}{19}\end{matrix}\right)\left(\begin{matrix}10\\-17\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{7}{38}\times 10+\frac{3}{38}\left(-17\right)\\-\frac{4}{19}\times 10+\frac{1}{19}\left(-17\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2}\\-3\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{1}{2},y=-3$$

Steps Using Elimination

Consider the first equation. Add 10 to both sides. Anything plus zero gives itself.

$$2x-3y=10$$

Consider the second equation. Add 8x to both sides.

$$7y+8x=-17$$

In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.

$$2x-3y=10,8x+7y=-17$$

To make 2x and 8x equal, multiply all terms on each side of the first equation by 8 and all terms on each side of the second by 2.

$$8\times 2x+8\left(-3\right)y=8\times 10,2\times 8x+2\times 7y=2\left(-17\right)$$

Simplify.

$$16x-24y=80,16x+14y=-34$$

Subtract 16x+14y=-34 from 16x-24y=80 by subtracting like terms on each side of the equal sign.

$$16x-16x-24y-14y=80+34$$

Add 16x to -16x. Terms 16x and -16x cancel out, leaving an equation with only one variable that can be solved.

$$-24y-14y=80+34$$

Add -24y to -14y.

$$-38y=80+34$$

Add 80 to 34.

$$-38y=114$$

Divide both sides by -38.

$$y=-3$$

Substitute -3 for y in 8x+7y=-17. Because the resulting equation contains only one variable, you can solve for x directly.

$$8x+7\left(-3\right)=-17$$

Multiply 7 times -3.

$$8x-21=-17$$

Add 21 to both sides of the equation.

$$8x=4$$

Divide both sides by 8.

$$x=\frac{1}{2}$$

The system is now solved.

$$x=\frac{1}{2},y=-3$$