Solve {l}{8x+20y=11400}{10x+30y=22500}

$$\left\{ \begin{array} { l } { 8 x + 20 y = 11400 } \\ { 10 x + 30 y = 22500 } \end{array} \right.$$

Solve for x, y

$$x=-2700$$

$$y=1650$$

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.

$$8x+20y=11400,10x+30y=22500$$

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

$$8x+20y=11400$$

Subtract 20y from both sides of the equation.

$$8x=-20y+11400$$

Divide both sides by 8.

$$x=\frac{1}{8}\left(-20y+11400\right)$$

Multiply \frac{1}{8} times -20y+11400.

$$x=-\frac{5}{2}y+1425$$

Substitute -\frac{5y}{2}+1425 for x in the other equation, 10x+30y=22500.

$$10\left(-\frac{5}{2}y+1425\right)+30y=22500$$

Multiply 10 times -\frac{5y}{2}+1425.

$$-25y+14250+30y=22500$$

Add -25y to 30y.

$$5y+14250=22500$$

Subtract 14250 from both sides of the equation.

$$5y=8250$$

Divide both sides by 5.

$$y=1650$$

Substitute 1650 for y in x=-\frac{5}{2}y+1425. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{5}{2}\times 1650+1425$$

Multiply -\frac{5}{2} times 1650.

$$x=-4125+1425$$

Add 1425 to -4125.

$$x=-2700$$

The system is now solved.

$$x=-2700,y=1650$$

Steps Using Matrices

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

$$8x+20y=11400,10x+30y=22500$$

Write the equations in matrix form.

$$\left(\begin{matrix}8&20\\10&30\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}11400\\22500\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}8&20\\10&30\end{matrix}\right).

$$inverse(\left(\begin{matrix}8&20\\10&30\end{matrix}\right))\left(\begin{matrix}8&20\\10&30\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&20\\10&30\end{matrix}\right))\left(\begin{matrix}11400\\22500\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&20\\10&30\end{matrix}\right))\left(\begin{matrix}11400\\22500\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&20\\10&30\end{matrix}\right))\left(\begin{matrix}11400\\22500\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{30}{8\times 30-20\times 10}&-\frac{20}{8\times 30-20\times 10}\\-\frac{10}{8\times 30-20\times 10}&\frac{8}{8\times 30-20\times 10}\end{matrix}\right)\left(\begin{matrix}11400\\22500\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{4}&-\frac{1}{2}\\-\frac{1}{4}&\frac{1}{5}\end{matrix}\right)\left(\begin{matrix}11400\\22500\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{4}\times 11400-\frac{1}{2}\times 22500\\-\frac{1}{4}\times 11400+\frac{1}{5}\times 22500\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-2700\\1650\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-2700,y=1650$$

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.

$$8x+20y=11400,10x+30y=22500$$

To make 8x and 10x equal, multiply all terms on each side of the first equation by 10 and all terms on each side of the second by 8.

$$10\times 8x+10\times 20y=10\times 11400,8\times 10x+8\times 30y=8\times 22500$$

Simplify.

$$80x+200y=114000,80x+240y=180000$$

Subtract 80x+240y=180000 from 80x+200y=114000 by subtracting like terms on each side of the equal sign.

$$80x-80x+200y-240y=114000-180000$$

Add 80x to -80x. Terms 80x and -80x cancel out, leaving an equation with only one variable that can be solved.

$$200y-240y=114000-180000$$

Add 200y to -240y.

$$-40y=114000-180000$$

Add 114000 to -180000.

$$-40y=-66000$$

Divide both sides by -40.

$$y=1650$$

Substitute 1650 for y in 10x+30y=22500. Because the resulting equation contains only one variable, you can solve for x directly.

$$10x+30\times 1650=22500$$

Multiply 30 times 1650.

$$10x+49500=22500$$

Subtract 49500 from both sides of the equation.

$$10x=-27000$$

Divide both sides by 10.

$$x=-2700$$

The system is now solved.

$$x=-2700,y=1650$$