Solve {l}{20x+50y=720}{40x+30y=880}

$$\left\{ \begin{array} { l } { 20 x + 50 y = 720 } \\ { 40 x + 30 y = 880 } \end{array} \right.$$

Solve for x, y

$$x=16$$

$$y=8$$

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.

$$20x+50y=720,40x+30y=880$$

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

$$20x+50y=720$$

Subtract 50y from both sides of the equation.

$$20x=-50y+720$$

Divide both sides by 20.

$$x=\frac{1}{20}\left(-50y+720\right)$$

Multiply \frac{1}{20} times -50y+720.

$$x=-\frac{5}{2}y+36$$

Substitute -\frac{5y}{2}+36 for x in the other equation, 40x+30y=880.

$$40\left(-\frac{5}{2}y+36\right)+30y=880$$

Multiply 40 times -\frac{5y}{2}+36.

$$-100y+1440+30y=880$$

Add -100y to 30y.

$$-70y+1440=880$$

Subtract 1440 from both sides of the equation.

$$-70y=-560$$

Divide both sides by -70.

$$y=8$$

Substitute 8 for y in x=-\frac{5}{2}y+36. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{5}{2}\times 8+36$$

Multiply -\frac{5}{2} times 8.

$$x=-20+36$$

Add 36 to -20.

$$x=16$$

The system is now solved.

$$x=16,y=8$$

Steps Using Matrices

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

$$20x+50y=720,40x+30y=880$$

Write the equations in matrix form.

$$\left(\begin{matrix}20&50\\40&30\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}720\\880\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}20&50\\40&30\end{matrix}\right).

$$inverse(\left(\begin{matrix}20&50\\40&30\end{matrix}\right))\left(\begin{matrix}20&50\\40&30\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}20&50\\40&30\end{matrix}\right))\left(\begin{matrix}720\\880\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}20&50\\40&30\end{matrix}\right))\left(\begin{matrix}720\\880\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}20&50\\40&30\end{matrix}\right))\left(\begin{matrix}720\\880\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{30}{20\times 30-50\times 40}&-\frac{50}{20\times 30-50\times 40}\\-\frac{40}{20\times 30-50\times 40}&\frac{20}{20\times 30-50\times 40}\end{matrix}\right)\left(\begin{matrix}720\\880\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{140}&\frac{1}{28}\\\frac{1}{35}&-\frac{1}{70}\end{matrix}\right)\left(\begin{matrix}720\\880\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{140}\times 720+\frac{1}{28}\times 880\\\frac{1}{35}\times 720-\frac{1}{70}\times 880\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}16\\8\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=16,y=8$$

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.

$$20x+50y=720,40x+30y=880$$

To make 20x and 40x equal, multiply all terms on each side of the first equation by 40 and all terms on each side of the second by 20.

$$40\times 20x+40\times 50y=40\times 720,20\times 40x+20\times 30y=20\times 880$$

Simplify.

$$800x+2000y=28800,800x+600y=17600$$

Subtract 800x+600y=17600 from 800x+2000y=28800 by subtracting like terms on each side of the equal sign.

$$800x-800x+2000y-600y=28800-17600$$

Add 800x to -800x. Terms 800x and -800x cancel out, leaving an equation with only one variable that can be solved.

$$2000y-600y=28800-17600$$

Add 2000y to -600y.

$$1400y=28800-17600$$

Add 28800 to -17600.

$$1400y=11200$$

Divide both sides by 1400.

$$y=8$$

Substitute 8 for y in 40x+30y=880. Because the resulting equation contains only one variable, you can solve for x directly.

$$40x+30\times 8=880$$

Multiply 30 times 8.

$$40x+240=880$$

Subtract 240 from both sides of the equation.

$$40x=640$$

Divide both sides by 40.

$$x=16$$

The system is now solved.

$$x=16,y=8$$