Solve {l}{50x+20y=590}{150x+40y=1430}

$$\left. \begin{array} { l } { 50 x + 20 y = 590 } \\ { 150 x + 40 y = 1430 } \end{array} \right.$$

Solve for x, y

$$x=5$$

$$y=17$$

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.

$$50x+20y=590,150x+40y=1430$$

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

$$50x+20y=590$$

Subtract 20y from both sides of the equation.

$$50x=-20y+590$$

Divide both sides by 50.

$$x=\frac{1}{50}\left(-20y+590\right)$$

Multiply \frac{1}{50} times -20y+590.

$$x=-\frac{2}{5}y+\frac{59}{5}$$

Substitute \frac{-2y+59}{5} for x in the other equation, 150x+40y=1430.

$$150\left(-\frac{2}{5}y+\frac{59}{5}\right)+40y=1430$$

Multiply 150 times \frac{-2y+59}{5}.

$$-60y+1770+40y=1430$$

Add -60y to 40y.

$$-20y+1770=1430$$

Subtract 1770 from both sides of the equation.

$$-20y=-340$$

Divide both sides by -20.

$$y=17$$

Substitute 17 for y in x=-\frac{2}{5}y+\frac{59}{5}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{2}{5}\times 17+\frac{59}{5}$$

Multiply -\frac{2}{5} times 17.

$$x=\frac{-34+59}{5}$$

Add \frac{59}{5} to -\frac{34}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=5$$

The system is now solved.

$$x=5,y=17$$

Steps Using Matrices

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

$$50x+20y=590,150x+40y=1430$$

Write the equations in matrix form.

$$\left(\begin{matrix}50&20\\150&40\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}590\\1430\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}50&20\\150&40\end{matrix}\right).

$$inverse(\left(\begin{matrix}50&20\\150&40\end{matrix}\right))\left(\begin{matrix}50&20\\150&40\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}50&20\\150&40\end{matrix}\right))\left(\begin{matrix}590\\1430\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}50&20\\150&40\end{matrix}\right))\left(\begin{matrix}590\\1430\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}50&20\\150&40\end{matrix}\right))\left(\begin{matrix}590\\1430\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{40}{50\times 40-20\times 150}&-\frac{20}{50\times 40-20\times 150}\\-\frac{150}{50\times 40-20\times 150}&\frac{50}{50\times 40-20\times 150}\end{matrix}\right)\left(\begin{matrix}590\\1430\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{25}&\frac{1}{50}\\\frac{3}{20}&-\frac{1}{20}\end{matrix}\right)\left(\begin{matrix}590\\1430\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{25}\times 590+\frac{1}{50}\times 1430\\\frac{3}{20}\times 590-\frac{1}{20}\times 1430\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}5\\17\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=5,y=17$$

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.

$$50x+20y=590,150x+40y=1430$$

To make 50x and 150x equal, multiply all terms on each side of the first equation by 150 and all terms on each side of the second by 50.

$$150\times 50x+150\times 20y=150\times 590,50\times 150x+50\times 40y=50\times 1430$$

Simplify.

$$7500x+3000y=88500,7500x+2000y=71500$$

Subtract 7500x+2000y=71500 from 7500x+3000y=88500 by subtracting like terms on each side of the equal sign.

$$7500x-7500x+3000y-2000y=88500-71500$$

Add 7500x to -7500x. Terms 7500x and -7500x cancel out, leaving an equation with only one variable that can be solved.

$$3000y-2000y=88500-71500$$

Add 3000y to -2000y.

$$1000y=88500-71500$$

Add 88500 to -71500.

$$1000y=17000$$

Divide both sides by 1000.

$$y=17$$

Substitute 17 for y in 150x+40y=1430. Because the resulting equation contains only one variable, you can solve for x directly.

$$150x+40\times 17=1430$$

Multiply 40 times 17.

$$150x+680=1430$$

Subtract 680 from both sides of the equation.

$$150x=750$$

Divide both sides by 150.

$$x=5$$

The system is now solved.

$$x=5,y=17$$