Solve {l}{13y+2x=32}{5x+4y=80}

$$\left. \begin{array} { l } { 13 y + 2 x = 32 } \\ { 5 x + 4 y = 80 } \end{array} \right.$$

Solve for y, x

$$x=16$$

$$y=0$$

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.

$$13y+2x=32,4y+5x=80$$

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

$$13y+2x=32$$

Subtract 2x from both sides of the equation.

$$13y=-2x+32$$

Divide both sides by 13.

$$y=\frac{1}{13}\left(-2x+32\right)$$

Multiply \frac{1}{13} times -2x+32.

$$y=-\frac{2}{13}x+\frac{32}{13}$$

Substitute \frac{-2x+32}{13} for y in the other equation, 4y+5x=80.

$$4\left(-\frac{2}{13}x+\frac{32}{13}\right)+5x=80$$

Multiply 4 times \frac{-2x+32}{13}.

$$-\frac{8}{13}x+\frac{128}{13}+5x=80$$

Add -\frac{8x}{13} to 5x.

$$\frac{57}{13}x+\frac{128}{13}=80$$

Subtract \frac{128}{13} from both sides of the equation.

$$\frac{57}{13}x=\frac{912}{13}$$

Divide both sides of the equation by \frac{57}{13}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$x=16$$

Substitute 16 for x in y=-\frac{2}{13}x+\frac{32}{13}. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=-\frac{2}{13}\times 16+\frac{32}{13}$$

Multiply -\frac{2}{13} times 16.

$$y=\frac{-32+32}{13}$$

Add \frac{32}{13} to -\frac{32}{13} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=0$$

The system is now solved.

$$y=0,x=16$$

Steps Using Matrices

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

$$13y+2x=32,4y+5x=80$$

Write the equations in matrix form.

$$\left(\begin{matrix}13&2\\4&5\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}32\\80\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}13&2\\4&5\end{matrix}\right).

$$inverse(\left(\begin{matrix}13&2\\4&5\end{matrix}\right))\left(\begin{matrix}13&2\\4&5\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}13&2\\4&5\end{matrix}\right))\left(\begin{matrix}32\\80\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}13&2\\4&5\end{matrix}\right))\left(\begin{matrix}32\\80\end{matrix}\right)$$

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

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}13&2\\4&5\end{matrix}\right))\left(\begin{matrix}32\\80\end{matrix}\right)$$

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.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{5}{13\times 5-2\times 4}&-\frac{2}{13\times 5-2\times 4}\\-\frac{4}{13\times 5-2\times 4}&\frac{13}{13\times 5-2\times 4}\end{matrix}\right)\left(\begin{matrix}32\\80\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{5}{57}&-\frac{2}{57}\\-\frac{4}{57}&\frac{13}{57}\end{matrix}\right)\left(\begin{matrix}32\\80\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{5}{57}\times 32-\frac{2}{57}\times 80\\-\frac{4}{57}\times 32+\frac{13}{57}\times 80\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}0\\16\end{matrix}\right)$$

Extract the matrix elements y and x.

$$y=0,x=16$$

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.

$$13y+2x=32,4y+5x=80$$

To make 13y and 4y equal, multiply all terms on each side of the first equation by 4 and all terms on each side of the second by 13.

$$4\times 13y+4\times 2x=4\times 32,13\times 4y+13\times 5x=13\times 80$$

Simplify.

$$52y+8x=128,52y+65x=1040$$

Subtract 52y+65x=1040 from 52y+8x=128 by subtracting like terms on each side of the equal sign.

$$52y-52y+8x-65x=128-1040$$

Add 52y to -52y. Terms 52y and -52y cancel out, leaving an equation with only one variable that can be solved.

$$8x-65x=128-1040$$

Add 8x to -65x.

$$-57x=128-1040$$

Add 128 to -1040.

$$-57x=-912$$

Divide both sides by -57.

$$x=16$$

Substitute 16 for x in 4y+5x=80. Because the resulting equation contains only one variable, you can solve for y directly.

$$4y+5\times 16=80$$

Multiply 5 times 16.

$$4y+80=80$$

Subtract 80 from both sides of the equation.

$$4y=0$$

Divide both sides by 4.

$$y=0$$

The system is now solved.

$$y=0,x=16$$