Solve {l}{20x+30y=10200}{30x+40y=14400(2)}

$$\left. \begin{array} { l } { 20 x + 30 y = 10200 } \\ { 30 x + 40 y = 14400 ( 2 ) } \end{array} \right.$$

Solve for x, y

$$x=4560$$

$$y=-2700$$

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.

$$20x+30y=10200,30x+40y=28800$$

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

$$20x+30y=10200$$

Subtract 30y from both sides of the equation.

$$20x=-30y+10200$$

Divide both sides by 20.

$$x=\frac{1}{20}\left(-30y+10200\right)$$

Multiply \frac{1}{20} times -30y+10200.

$$x=-\frac{3}{2}y+510$$

Substitute -\frac{3y}{2}+510 for x in the other equation, 30x+40y=28800.

$$30\left(-\frac{3}{2}y+510\right)+40y=28800$$

Multiply 30 times -\frac{3y}{2}+510.

$$-45y+15300+40y=28800$$

Add -45y to 40y.

$$-5y+15300=28800$$

Subtract 15300 from both sides of the equation.

$$-5y=13500$$

Divide both sides by -5.

$$y=-2700$$

Substitute -2700 for y in x=-\frac{3}{2}y+510. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{3}{2}\left(-2700\right)+510$$

Multiply -\frac{3}{2} times -2700.

$$x=4050+510$$

Add 510 to 4050.

$$x=4560$$

The system is now solved.

$$x=4560,y=-2700$$

Steps Using Matrices

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

$$20x+30y=10200,30x+40y=28800$$

Write the equations in matrix form.

$$\left(\begin{matrix}20&30\\30&40\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}10200\\28800\end{matrix}\right)$$

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

$$inverse(\left(\begin{matrix}20&30\\30&40\end{matrix}\right))\left(\begin{matrix}20&30\\30&40\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}20&30\\30&40\end{matrix}\right))\left(\begin{matrix}10200\\28800\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}20&30\\30&40\end{matrix}\right))\left(\begin{matrix}10200\\28800\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}20&30\\30&40\end{matrix}\right))\left(\begin{matrix}10200\\28800\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{40}{20\times 40-30\times 30}&-\frac{30}{20\times 40-30\times 30}\\-\frac{30}{20\times 40-30\times 30}&\frac{20}{20\times 40-30\times 30}\end{matrix}\right)\left(\begin{matrix}10200\\28800\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{5}&\frac{3}{10}\\\frac{3}{10}&-\frac{1}{5}\end{matrix}\right)\left(\begin{matrix}10200\\28800\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{5}\times 10200+\frac{3}{10}\times 28800\\\frac{3}{10}\times 10200-\frac{1}{5}\times 28800\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}4560\\-2700\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=4560,y=-2700$$

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.

$$20x+30y=10200,30x+40y=28800$$

To make 20x and 30x equal, multiply all terms on each side of the first equation by 30 and all terms on each side of the second by 20.

$$30\times 20x+30\times 30y=30\times 10200,20\times 30x+20\times 40y=20\times 28800$$

Simplify.

$$600x+900y=306000,600x+800y=576000$$

Subtract 600x+800y=576000 from 600x+900y=306000 by subtracting like terms on each side of the equal sign.

$$600x-600x+900y-800y=306000-576000$$

Add 600x to -600x. Terms 600x and -600x cancel out, leaving an equation with only one variable that can be solved.

$$900y-800y=306000-576000$$

Add 900y to -800y.

$$100y=306000-576000$$

Add 306000 to -576000.

$$100y=-270000$$

Divide both sides by 100.

$$y=-2700$$

Substitute -2700 for y in 30x+40y=28800. Because the resulting equation contains only one variable, you can solve for x directly.

$$30x+40\left(-2700\right)=28800$$

Multiply 40 times -2700.

$$30x-108000=28800$$

Add 108000 to both sides of the equation.

$$30x=136800$$

Divide both sides by 30.

$$x=4560$$

The system is now solved.

$$x=4560,y=-2700$$