Solve {l}{x+y=300}{24(x+20)+50y=12880}

$$\left. \begin{array}{l}{ x + y = 300 }\\{ 24 ( x + 20 ) + 50 y = 12880 }\end{array} \right.$$

Solve for x, y

$$x=100$$

$$y=200$$

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.

$$x+y=300,24\left(x+20\right)+50y=12880$$

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

$$x+y=300$$

Subtract y from both sides of the equation.

$$x=-y+300$$

Substitute -y+300 for x in the other equation, 24\left(x+20\right)+50y=12880.

$$24\left(-y+300+20\right)+50y=12880$$

Add 300 to 20.

$$24\left(-y+320\right)+50y=12880$$

Multiply 24 times -y+320.

$$-24y+7680+50y=12880$$

Add -24y to 50y.

$$26y+7680=12880$$

Subtract 7680 from both sides of the equation.

$$26y=5200$$

Divide both sides by 26.

$$y=200$$

Substitute 200 for y in x=-y+300. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-200+300$$

Add 300 to -200.

$$x=100$$

The system is now solved.

$$x=100,y=200$$

Steps Using Matrices

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

$$x+y=300,24\left(x+20\right)+50y=12880$$

Simplify the second equation to put it in standard form.

$$24\left(x+20\right)+50y=12880$$

Multiply 24 times x+20.

$$24x+480+50y=12880$$

Subtract 480 from both sides of the equation.

$$24x+50y=12400$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\24&50\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}300\\12400\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\24&50\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\24&50\end{matrix}\right))\left(\begin{matrix}1&1\\24&50\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\24&50\end{matrix}\right))\left(\begin{matrix}300\\12400\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\24&50\end{matrix}\right))\left(\begin{matrix}300\\12400\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\24&50\end{matrix}\right))\left(\begin{matrix}300\\12400\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{50}{50-24}&-\frac{1}{50-24}\\-\frac{24}{50-24}&\frac{1}{50-24}\end{matrix}\right)\left(\begin{matrix}300\\12400\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{25}{13}&-\frac{1}{26}\\-\frac{12}{13}&\frac{1}{26}\end{matrix}\right)\left(\begin{matrix}300\\12400\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{25}{13}\times 300-\frac{1}{26}\times 12400\\-\frac{12}{13}\times 300+\frac{1}{26}\times 12400\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}100\\200\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=100,y=200$$