Solve {l}{8x+9y=-13}{39x+28y=16}

$$\left. \begin{array} { l } { 8 x + 9 y = - 13 } \\ { 39 x + 28 y = 16 } \end{array} \right.$$

Solve for x, y

$$x=4$$

$$y=-5$$

Steps Using Substitution

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.

$$8x+9y=-13,39x+28y=16$$

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

$$8x+9y=-13$$

Subtract 9y from both sides of the equation.

$$8x=-9y-13$$

Divide both sides by 8.

$$x=\frac{1}{8}\left(-9y-13\right)$$

Multiply \frac{1}{8} times -9y-13.

$$x=-\frac{9}{8}y-\frac{13}{8}$$

Substitute \frac{-9y-13}{8} for x in the other equation, 39x+28y=16.

$$39\left(-\frac{9}{8}y-\frac{13}{8}\right)+28y=16$$

Multiply 39 times \frac{-9y-13}{8}.

$$-\frac{351}{8}y-\frac{507}{8}+28y=16$$

Add -\frac{351y}{8} to 28y.

$$-\frac{127}{8}y-\frac{507}{8}=16$$

Add \frac{507}{8} to both sides of the equation.

$$-\frac{127}{8}y=\frac{635}{8}$$

Divide both sides of the equation by -\frac{127}{8}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=-5$$

Substitute -5 for y in x=-\frac{9}{8}y-\frac{13}{8}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{9}{8}\left(-5\right)-\frac{13}{8}$$

Multiply -\frac{9}{8} times -5.

$$x=\frac{45-13}{8}$$

Add -\frac{13}{8} to \frac{45}{8} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=4$$

The system is now solved.

$$x=4,y=-5$$

Steps Using Matrices

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

$$8x+9y=-13,39x+28y=16$$

Write the equations in matrix form.

$$\left(\begin{matrix}8&9\\39&28\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-13\\16\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}8&9\\39&28\end{matrix}\right).

$$inverse(\left(\begin{matrix}8&9\\39&28\end{matrix}\right))\left(\begin{matrix}8&9\\39&28\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&9\\39&28\end{matrix}\right))\left(\begin{matrix}-13\\16\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&9\\39&28\end{matrix}\right))\left(\begin{matrix}-13\\16\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&9\\39&28\end{matrix}\right))\left(\begin{matrix}-13\\16\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{28}{8\times 28-9\times 39}&-\frac{9}{8\times 28-9\times 39}\\-\frac{39}{8\times 28-9\times 39}&\frac{8}{8\times 28-9\times 39}\end{matrix}\right)\left(\begin{matrix}-13\\16\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{28}{127}&\frac{9}{127}\\\frac{39}{127}&-\frac{8}{127}\end{matrix}\right)\left(\begin{matrix}-13\\16\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{28}{127}\left(-13\right)+\frac{9}{127}\times 16\\\frac{39}{127}\left(-13\right)-\frac{8}{127}\times 16\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}4\\-5\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=4,y=-5$$

Steps Using Elimination

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.

$$8x+9y=-13,39x+28y=16$$

To make 8x and 39x equal, multiply all terms on each side of the first equation by 39 and all terms on each side of the second by 8.

$$39\times 8x+39\times 9y=39\left(-13\right),8\times 39x+8\times 28y=8\times 16$$

Simplify.

$$312x+351y=-507,312x+224y=128$$

Subtract 312x+224y=128 from 312x+351y=-507 by subtracting like terms on each side of the equal sign.

$$312x-312x+351y-224y=-507-128$$

Add 312x to -312x. Terms 312x and -312x cancel out, leaving an equation with only one variable that can be solved.

$$351y-224y=-507-128$$

Add 351y to -224y.

$$127y=-507-128$$

Add -507 to -128.

$$127y=-635$$

Divide both sides by 127.

$$y=-5$$

Substitute -5 for y in 39x+28y=16. Because the resulting equation contains only one variable, you can solve for x directly.

$$39x+28\left(-5\right)=16$$

Multiply 28 times -5.

$$39x-140=16$$

Add 140 to both sides of the equation.

$$39x=156$$

Divide both sides by 39.

$$x=4$$

The system is now solved.

$$x=4,y=-5$$