Solve {l}{20x+30y=1200}{40x+10y=900}

$$\left\{ \begin{array} { l } { 20 x + 30 y = 1200 } \\ { 40 x + 10 y = 900 } \end{array} \right.$$

Solve for x, y

$$x=15$$

$$y=30$$

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.

$$20x+30y=1200,40x+10y=900$$

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

$$20x+30y=1200$$

Subtract 30y from both sides of the equation.

$$20x=-30y+1200$$

Divide both sides by 20.

$$x=\frac{1}{20}\left(-30y+1200\right)$$

Multiply \frac{1}{20} times -30y+1200.

$$x=-\frac{3}{2}y+60$$

Substitute -\frac{3y}{2}+60 for x in the other equation, 40x+10y=900.

$$40\left(-\frac{3}{2}y+60\right)+10y=900$$

Multiply 40 times -\frac{3y}{2}+60.

$$-60y+2400+10y=900$$

Add -60y to 10y.

$$-50y+2400=900$$

Subtract 2400 from both sides of the equation.

$$-50y=-1500$$

Divide both sides by -50.

$$y=30$$

Substitute 30 for y in x=-\frac{3}{2}y+60. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{3}{2}\times 30+60$$

Multiply -\frac{3}{2} times 30.

$$x=-45+60$$

Add 60 to -45.

$$x=15$$

The system is now solved.

$$x=15,y=30$$

Steps Using Matrices

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

$$20x+30y=1200,40x+10y=900$$

Write the equations in matrix form.

$$\left(\begin{matrix}20&30\\40&10\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1200\\900\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}20&30\\40&10\end{matrix}\right).

$$inverse(\left(\begin{matrix}20&30\\40&10\end{matrix}\right))\left(\begin{matrix}20&30\\40&10\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}20&30\\40&10\end{matrix}\right))\left(\begin{matrix}1200\\900\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}20&30\\40&10\end{matrix}\right))\left(\begin{matrix}1200\\900\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}20&30\\40&10\end{matrix}\right))\left(\begin{matrix}1200\\900\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{10}{20\times 10-30\times 40}&-\frac{30}{20\times 10-30\times 40}\\-\frac{40}{20\times 10-30\times 40}&\frac{20}{20\times 10-30\times 40}\end{matrix}\right)\left(\begin{matrix}1200\\900\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{100}&\frac{3}{100}\\\frac{1}{25}&-\frac{1}{50}\end{matrix}\right)\left(\begin{matrix}1200\\900\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{100}\times 1200+\frac{3}{100}\times 900\\\frac{1}{25}\times 1200-\frac{1}{50}\times 900\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}15\\30\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=15,y=30$$

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.

$$20x+30y=1200,40x+10y=900$$

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.

$$40\times 20x+40\times 30y=40\times 1200,20\times 40x+20\times 10y=20\times 900$$

Simplify.

$$800x+1200y=48000,800x+200y=18000$$

Subtract 800x+200y=18000 from 800x+1200y=48000 by subtracting like terms on each side of the equal sign.

$$800x-800x+1200y-200y=48000-18000$$

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

$$1200y-200y=48000-18000$$

Add 1200y to -200y.

$$1000y=48000-18000$$

Add 48000 to -18000.

$$1000y=30000$$

Divide both sides by 1000.

$$y=30$$

Substitute 30 for y in 40x+10y=900. Because the resulting equation contains only one variable, you can solve for x directly.

$$40x+10\times 30=900$$

Multiply 10 times 30.

$$40x+300=900$$

Subtract 300 from both sides of the equation.

$$40x=600$$

Divide both sides by 40.

$$x=15$$

The system is now solved.

$$x=15,y=30$$