Solve {l}{120x+80y=2800}{48x+40y=1280}

$$\left\{ \begin{array} { l } { 120 x + 80 y = 2800 } \\ { 48 x + 40 y = 1280 } \end{array} \right.$$

Solve for x, y

$$x=10$$

$$y=20$$

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.

$$120x+80y=2800,48x+40y=1280$$

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

$$120x+80y=2800$$

Subtract 80y from both sides of the equation.

$$120x=-80y+2800$$

Divide both sides by 120.

$$x=\frac{1}{120}\left(-80y+2800\right)$$

Multiply \frac{1}{120} times -80y+2800.

$$x=-\frac{2}{3}y+\frac{70}{3}$$

Substitute \frac{-2y+70}{3} for x in the other equation, 48x+40y=1280.

$$48\left(-\frac{2}{3}y+\frac{70}{3}\right)+40y=1280$$

Multiply 48 times \frac{-2y+70}{3}.

$$-32y+1120+40y=1280$$

Add -32y to 40y.

$$8y+1120=1280$$

Subtract 1120 from both sides of the equation.

$$8y=160$$

Divide both sides by 8.

$$y=20$$

Substitute 20 for y in x=-\frac{2}{3}y+\frac{70}{3}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{2}{3}\times 20+\frac{70}{3}$$

Multiply -\frac{2}{3} times 20.

$$x=\frac{-40+70}{3}$$

Add \frac{70}{3} to -\frac{40}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=10$$

The system is now solved.

$$x=10,y=20$$

Steps Using Matrices

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

$$120x+80y=2800,48x+40y=1280$$

Write the equations in matrix form.

$$\left(\begin{matrix}120&80\\48&40\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2800\\1280\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}120&80\\48&40\end{matrix}\right).

$$inverse(\left(\begin{matrix}120&80\\48&40\end{matrix}\right))\left(\begin{matrix}120&80\\48&40\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}120&80\\48&40\end{matrix}\right))\left(\begin{matrix}2800\\1280\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}120&80\\48&40\end{matrix}\right))\left(\begin{matrix}2800\\1280\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}120&80\\48&40\end{matrix}\right))\left(\begin{matrix}2800\\1280\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{40}{120\times 40-80\times 48}&-\frac{80}{120\times 40-80\times 48}\\-\frac{48}{120\times 40-80\times 48}&\frac{120}{120\times 40-80\times 48}\end{matrix}\right)\left(\begin{matrix}2800\\1280\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{24}&-\frac{1}{12}\\-\frac{1}{20}&\frac{1}{8}\end{matrix}\right)\left(\begin{matrix}2800\\1280\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{24}\times 2800-\frac{1}{12}\times 1280\\-\frac{1}{20}\times 2800+\frac{1}{8}\times 1280\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}10\\20\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=10,y=20$$

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.

$$120x+80y=2800,48x+40y=1280$$

To make 120x and 48x equal, multiply all terms on each side of the first equation by 48 and all terms on each side of the second by 120.

$$48\times 120x+48\times 80y=48\times 2800,120\times 48x+120\times 40y=120\times 1280$$

Simplify.

$$5760x+3840y=134400,5760x+4800y=153600$$

Subtract 5760x+4800y=153600 from 5760x+3840y=134400 by subtracting like terms on each side of the equal sign.

$$5760x-5760x+3840y-4800y=134400-153600$$

Add 5760x to -5760x. Terms 5760x and -5760x cancel out, leaving an equation with only one variable that can be solved.

$$3840y-4800y=134400-153600$$

Add 3840y to -4800y.

$$-960y=134400-153600$$

Add 134400 to -153600.

$$-960y=-19200$$

Divide both sides by -960.

$$y=20$$

Substitute 20 for y in 48x+40y=1280. Because the resulting equation contains only one variable, you can solve for x directly.

$$48x+40\times 20=1280$$

Multiply 40 times 20.

$$48x+800=1280$$

Subtract 800 from both sides of the equation.

$$48x=480$$

Divide both sides by 48.

$$x=10$$

The system is now solved.

$$x=10,y=20$$