Solve {l}{30x+20y=680}{50x+40y=880}

$$\left\{ \begin{array} { l } { 30 x + 20 y = 680 } \\ { 50 x + 40 y = 880 } \end{array} \right.$$

Solve for x, y

$$x=48$$

$$y=-38$$

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.

$$30x+20y=680,50x+40y=880$$

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

$$30x+20y=680$$

Subtract 20y from both sides of the equation.

$$30x=-20y+680$$

Divide both sides by 30.

$$x=\frac{1}{30}\left(-20y+680\right)$$

Multiply \frac{1}{30} times -20y+680.

$$x=-\frac{2}{3}y+\frac{68}{3}$$

Substitute \frac{-2y+68}{3} for x in the other equation, 50x+40y=880.

$$50\left(-\frac{2}{3}y+\frac{68}{3}\right)+40y=880$$

Multiply 50 times \frac{-2y+68}{3}.

$$-\frac{100}{3}y+\frac{3400}{3}+40y=880$$

Add -\frac{100y}{3} to 40y.

$$\frac{20}{3}y+\frac{3400}{3}=880$$

Subtract \frac{3400}{3} from both sides of the equation.

$$\frac{20}{3}y=-\frac{760}{3}$$

Divide both sides of the equation by \frac{20}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=-38$$

Substitute -38 for y in x=-\frac{2}{3}y+\frac{68}{3}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{2}{3}\left(-38\right)+\frac{68}{3}$$

Multiply -\frac{2}{3} times -38.

$$x=\frac{76+68}{3}$$

Add \frac{68}{3} to \frac{76}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=48$$

The system is now solved.

$$x=48,y=-38$$

Steps Using Matrices

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

$$30x+20y=680,50x+40y=880$$

Write the equations in matrix form.

$$\left(\begin{matrix}30&20\\50&40\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}680\\880\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}30&20\\50&40\end{matrix}\right).

$$inverse(\left(\begin{matrix}30&20\\50&40\end{matrix}\right))\left(\begin{matrix}30&20\\50&40\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}30&20\\50&40\end{matrix}\right))\left(\begin{matrix}680\\880\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}30&20\\50&40\end{matrix}\right))\left(\begin{matrix}680\\880\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}30&20\\50&40\end{matrix}\right))\left(\begin{matrix}680\\880\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{40}{30\times 40-20\times 50}&-\frac{20}{30\times 40-20\times 50}\\-\frac{50}{30\times 40-20\times 50}&\frac{30}{30\times 40-20\times 50}\end{matrix}\right)\left(\begin{matrix}680\\880\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{5}&-\frac{1}{10}\\-\frac{1}{4}&\frac{3}{20}\end{matrix}\right)\left(\begin{matrix}680\\880\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{5}\times 680-\frac{1}{10}\times 880\\-\frac{1}{4}\times 680+\frac{3}{20}\times 880\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}48\\-38\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=48,y=-38$$

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.

$$30x+20y=680,50x+40y=880$$

To make 30x and 50x equal, multiply all terms on each side of the first equation by 50 and all terms on each side of the second by 30.

$$50\times 30x+50\times 20y=50\times 680,30\times 50x+30\times 40y=30\times 880$$

Simplify.

$$1500x+1000y=34000,1500x+1200y=26400$$

Subtract 1500x+1200y=26400 from 1500x+1000y=34000 by subtracting like terms on each side of the equal sign.

$$1500x-1500x+1000y-1200y=34000-26400$$

Add 1500x to -1500x. Terms 1500x and -1500x cancel out, leaving an equation with only one variable that can be solved.

$$1000y-1200y=34000-26400$$

Add 1000y to -1200y.

$$-200y=34000-26400$$

Add 34000 to -26400.

$$-200y=7600$$

Divide both sides by -200.

$$y=-38$$

Substitute -38 for y in 50x+40y=880. Because the resulting equation contains only one variable, you can solve for x directly.

$$50x+40\left(-38\right)=880$$

Multiply 40 times -38.

$$50x-1520=880$$

Add 1520 to both sides of the equation.

$$50x=2400$$

Divide both sides by 50.

$$x=48$$

The system is now solved.

$$x=48,y=-38$$