Solve {l}{120x+100y=36000}{(138-120)x+(120-100)y=6000}

$$\left\{ \begin{array} { l } { 120 x + 100 y = 36000 } \\ { ( 138 - 120 ) x + ( 120 - 100 ) y = 6000 } \end{array} \right.$$

Solve for x, y

$$x=200$$

$$y=120$$

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.

$$120x+100y=36000,18x+20y=6000$$

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

$$120x+100y=36000$$

Subtract 100y from both sides of the equation.

$$120x=-100y+36000$$

Divide both sides by 120.

$$x=\frac{1}{120}\left(-100y+36000\right)$$

Multiply \frac{1}{120} times -100y+36000.

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

Substitute -\frac{5y}{6}+300 for x in the other equation, 18x+20y=6000.

$$18\left(-\frac{5}{6}y+300\right)+20y=6000$$

Multiply 18 times -\frac{5y}{6}+300.

$$-15y+5400+20y=6000$$

Add -15y to 20y.

$$5y+5400=6000$$

Subtract 5400 from both sides of the equation.

$$5y=600$$

Divide both sides by 5.

$$y=120$$

Substitute 120 for y in x=-\frac{5}{6}y+300. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{5}{6}\times 120+300$$

Multiply -\frac{5}{6} times 120.

$$x=-100+300$$

Add 300 to -100.

$$x=200$$

The system is now solved.

$$x=200,y=120$$

Steps Using Matrices

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

$$120x+100y=36000,18x+20y=6000$$

Write the equations in matrix form.

$$\left(\begin{matrix}120&100\\18&20\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}36000\\6000\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}120&100\\18&20\end{matrix}\right).

$$inverse(\left(\begin{matrix}120&100\\18&20\end{matrix}\right))\left(\begin{matrix}120&100\\18&20\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}120&100\\18&20\end{matrix}\right))\left(\begin{matrix}36000\\6000\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}120&100\\18&20\end{matrix}\right))\left(\begin{matrix}36000\\6000\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}120&100\\18&20\end{matrix}\right))\left(\begin{matrix}36000\\6000\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{20}{120\times 20-100\times 18}&-\frac{100}{120\times 20-100\times 18}\\-\frac{18}{120\times 20-100\times 18}&\frac{120}{120\times 20-100\times 18}\end{matrix}\right)\left(\begin{matrix}36000\\6000\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{30}&-\frac{1}{6}\\-\frac{3}{100}&\frac{1}{5}\end{matrix}\right)\left(\begin{matrix}36000\\6000\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{30}\times 36000-\frac{1}{6}\times 6000\\-\frac{3}{100}\times 36000+\frac{1}{5}\times 6000\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}200\\120\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=200,y=120$$

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.

$$120x+100y=36000,18x+20y=6000$$

To make 120x and 18x equal, multiply all terms on each side of the first equation by 18 and all terms on each side of the second by 120.

$$18\times 120x+18\times 100y=18\times 36000,120\times 18x+120\times 20y=120\times 6000$$

Simplify.

$$2160x+1800y=648000,2160x+2400y=720000$$

Subtract 2160x+2400y=720000 from 2160x+1800y=648000 by subtracting like terms on each side of the equal sign.

$$2160x-2160x+1800y-2400y=648000-720000$$

Add 2160x to -2160x. Terms 2160x and -2160x cancel out, leaving an equation with only one variable that can be solved.

$$1800y-2400y=648000-720000$$

Add 1800y to -2400y.

$$-600y=648000-720000$$

Add 648000 to -720000.

$$-600y=-72000$$

Divide both sides by -600.

$$y=120$$

Substitute 120 for y in 18x+20y=6000. Because the resulting equation contains only one variable, you can solve for x directly.

$$18x+20\times 120=6000$$

Multiply 20 times 120.

$$18x+2400=6000$$

Subtract 2400 from both sides of the equation.

$$18x=3600$$

Divide both sides by 18.

$$x=200$$

The system is now solved.

$$x=200,y=120$$