Solve {l}{1200x+1000y=360000}{180x+200y=6000}

$$\left\{ \begin{array} { l } { 1200 x + 1000 y = 360000 } \\ { 180 x + 200 y = 6000 } \end{array} \right.$$

Solve for x, y

$$x=1100$$

$$y=-960$$

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.

$$1200x+1000y=360000,180x+200y=6000$$

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

$$1200x+1000y=360000$$

Subtract 1000y from both sides of the equation.

$$1200x=-1000y+360000$$

Divide both sides by 1200.

$$x=\frac{1}{1200}\left(-1000y+360000\right)$$

Multiply \frac{1}{1200} times -1000y+360000.

$$x=-\frac{5}{6}y+300$$

Substitute -\frac{5y}{6}+300 for x in the other equation, 180x+200y=6000.

$$180\left(-\frac{5}{6}y+300\right)+200y=6000$$

Multiply 180 times -\frac{5y}{6}+300.

$$-150y+54000+200y=6000$$

Add -150y to 200y.

$$50y+54000=6000$$

Subtract 54000 from both sides of the equation.

$$50y=-48000$$

Divide both sides by 50.

$$y=-960$$

Substitute -960 for y in x=-\frac{5}{6}y+300. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{5}{6}\left(-960\right)+300$$

Multiply -\frac{5}{6} times -960.

$$x=800+300$$

Add 300 to 800.

$$x=1100$$

The system is now solved.

$$x=1100,y=-960$$

Steps Using Matrices

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

$$1200x+1000y=360000,180x+200y=6000$$

Write the equations in matrix form.

$$\left(\begin{matrix}1200&1000\\180&200\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}360000\\6000\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1200&1000\\180&200\end{matrix}\right).

$$inverse(\left(\begin{matrix}1200&1000\\180&200\end{matrix}\right))\left(\begin{matrix}1200&1000\\180&200\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1200&1000\\180&200\end{matrix}\right))\left(\begin{matrix}360000\\6000\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1200&1000\\180&200\end{matrix}\right))\left(\begin{matrix}360000\\6000\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1200&1000\\180&200\end{matrix}\right))\left(\begin{matrix}360000\\6000\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{200}{1200\times 200-1000\times 180}&-\frac{1000}{1200\times 200-1000\times 180}\\-\frac{180}{1200\times 200-1000\times 180}&\frac{1200}{1200\times 200-1000\times 180}\end{matrix}\right)\left(\begin{matrix}360000\\6000\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{300}&-\frac{1}{60}\\-\frac{3}{1000}&\frac{1}{50}\end{matrix}\right)\left(\begin{matrix}360000\\6000\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{300}\times 360000-\frac{1}{60}\times 6000\\-\frac{3}{1000}\times 360000+\frac{1}{50}\times 6000\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1100\\-960\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=1100,y=-960$$

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.

$$1200x+1000y=360000,180x+200y=6000$$

To make 1200x and 180x equal, multiply all terms on each side of the first equation by 180 and all terms on each side of the second by 1200.

$$180\times 1200x+180\times 1000y=180\times 360000,1200\times 180x+1200\times 200y=1200\times 6000$$

Simplify.

$$216000x+180000y=64800000,216000x+240000y=7200000$$

Subtract 216000x+240000y=7200000 from 216000x+180000y=64800000 by subtracting like terms on each side of the equal sign.

$$216000x-216000x+180000y-240000y=64800000-7200000$$

Add 216000x to -216000x. Terms 216000x and -216000x cancel out, leaving an equation with only one variable that can be solved.

$$180000y-240000y=64800000-7200000$$

Add 180000y to -240000y.

$$-60000y=64800000-7200000$$

Add 64800000 to -7200000.

$$-60000y=57600000$$

Divide both sides by -60000.

$$y=-960$$

Substitute -960 for y in 180x+200y=6000. Because the resulting equation contains only one variable, you can solve for x directly.

$$180x+200\left(-960\right)=6000$$

Multiply 200 times -960.

$$180x-192000=6000$$

Add 192000 to both sides of the equation.

$$180x=198000$$

Divide both sides by 180.

$$x=1100$$

The system is now solved.

$$x=1100,y=-960$$