Solve {l}{40x+30y=500}{60x+15y=600}

$$\left. \begin{array} { l } { 40 x + 30 y = 500 } \\ { 60 x + 15 y = 600 } \end{array} \right.$$

Solve for x, y

$$x = \frac{35}{4} = 8\frac{3}{4} = 8.75$$

$$y=5$$

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.

$$40x+30y=500,60x+15y=600$$

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

$$40x+30y=500$$

Subtract 30y from both sides of the equation.

$$40x=-30y+500$$

Divide both sides by 40.

$$x=\frac{1}{40}\left(-30y+500\right)$$

Multiply \frac{1}{40} times -30y+500.

$$x=-\frac{3}{4}y+\frac{25}{2}$$

Substitute -\frac{3y}{4}+\frac{25}{2} for x in the other equation, 60x+15y=600.

$$60\left(-\frac{3}{4}y+\frac{25}{2}\right)+15y=600$$

Multiply 60 times -\frac{3y}{4}+\frac{25}{2}.

$$-45y+750+15y=600$$

Add -45y to 15y.

$$-30y+750=600$$

Subtract 750 from both sides of the equation.

$$-30y=-150$$

Divide both sides by -30.

$$y=5$$

Substitute 5 for y in x=-\frac{3}{4}y+\frac{25}{2}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{3}{4}\times 5+\frac{25}{2}$$

Multiply -\frac{3}{4} times 5.

$$x=-\frac{15}{4}+\frac{25}{2}$$

Add \frac{25}{2} to -\frac{15}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{35}{4}$$

The system is now solved.

$$x=\frac{35}{4},y=5$$

Steps Using Matrices

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

$$40x+30y=500,60x+15y=600$$

Write the equations in matrix form.

$$\left(\begin{matrix}40&30\\60&15\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}500\\600\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}40&30\\60&15\end{matrix}\right).

$$inverse(\left(\begin{matrix}40&30\\60&15\end{matrix}\right))\left(\begin{matrix}40&30\\60&15\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}40&30\\60&15\end{matrix}\right))\left(\begin{matrix}500\\600\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}40&30\\60&15\end{matrix}\right))\left(\begin{matrix}500\\600\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}40&30\\60&15\end{matrix}\right))\left(\begin{matrix}500\\600\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{15}{40\times 15-30\times 60}&-\frac{30}{40\times 15-30\times 60}\\-\frac{60}{40\times 15-30\times 60}&\frac{40}{40\times 15-30\times 60}\end{matrix}\right)\left(\begin{matrix}500\\600\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{80}&\frac{1}{40}\\\frac{1}{20}&-\frac{1}{30}\end{matrix}\right)\left(\begin{matrix}500\\600\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{80}\times 500+\frac{1}{40}\times 600\\\frac{1}{20}\times 500-\frac{1}{30}\times 600\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{35}{4}\\5\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{35}{4},y=5$$

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.

$$40x+30y=500,60x+15y=600$$

To make 40x and 60x equal, multiply all terms on each side of the first equation by 60 and all terms on each side of the second by 40.

$$60\times 40x+60\times 30y=60\times 500,40\times 60x+40\times 15y=40\times 600$$

Simplify.

$$2400x+1800y=30000,2400x+600y=24000$$

Subtract 2400x+600y=24000 from 2400x+1800y=30000 by subtracting like terms on each side of the equal sign.

$$2400x-2400x+1800y-600y=30000-24000$$

Add 2400x to -2400x. Terms 2400x and -2400x cancel out, leaving an equation with only one variable that can be solved.

$$1800y-600y=30000-24000$$

Add 1800y to -600y.

$$1200y=30000-24000$$

Add 30000 to -24000.

$$1200y=6000$$

Divide both sides by 1200.

$$y=5$$

Substitute 5 for y in 60x+15y=600. Because the resulting equation contains only one variable, you can solve for x directly.

$$60x+15\times 5=600$$

Multiply 15 times 5.

$$60x+75=600$$

Subtract 75 from both sides of the equation.

$$60x=525$$

Divide both sides by 60.

$$x=\frac{35}{4}$$

The system is now solved.

$$x=\frac{35}{4},y=5$$