Solve {l}{6x+3y=660}{40x+30y=5200}

$$\left\{ \begin{array} { l } { 6 x + 3 y = 660 } \\ { 40 x + 30 y = 5200 } \end{array} \right.$$

Solve for x, y

$$x=70$$

$$y=80$$

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.

$$6x+3y=660,40x+30y=5200$$

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

$$6x+3y=660$$

Subtract 3y from both sides of the equation.

$$6x=-3y+660$$

Divide both sides by 6.

$$x=\frac{1}{6}\left(-3y+660\right)$$

Multiply \frac{1}{6} times -3y+660.

$$x=-\frac{1}{2}y+110$$

Substitute -\frac{y}{2}+110 for x in the other equation, 40x+30y=5200.

$$40\left(-\frac{1}{2}y+110\right)+30y=5200$$

Multiply 40 times -\frac{y}{2}+110.

$$-20y+4400+30y=5200$$

Add -20y to 30y.

$$10y+4400=5200$$

Subtract 4400 from both sides of the equation.

$$10y=800$$

Divide both sides by 10.

$$y=80$$

Substitute 80 for y in x=-\frac{1}{2}y+110. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{1}{2}\times 80+110$$

Multiply -\frac{1}{2} times 80.

$$x=-40+110$$

Add 110 to -40.

$$x=70$$

The system is now solved.

$$x=70,y=80$$

Steps Using Matrices

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

$$6x+3y=660,40x+30y=5200$$

Write the equations in matrix form.

$$\left(\begin{matrix}6&3\\40&30\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}660\\5200\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}6&3\\40&30\end{matrix}\right).

$$inverse(\left(\begin{matrix}6&3\\40&30\end{matrix}\right))\left(\begin{matrix}6&3\\40&30\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}6&3\\40&30\end{matrix}\right))\left(\begin{matrix}660\\5200\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}6&3\\40&30\end{matrix}\right))\left(\begin{matrix}660\\5200\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}6&3\\40&30\end{matrix}\right))\left(\begin{matrix}660\\5200\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{30}{6\times 30-3\times 40}&-\frac{3}{6\times 30-3\times 40}\\-\frac{40}{6\times 30-3\times 40}&\frac{6}{6\times 30-3\times 40}\end{matrix}\right)\left(\begin{matrix}660\\5200\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2}&-\frac{1}{20}\\-\frac{2}{3}&\frac{1}{10}\end{matrix}\right)\left(\begin{matrix}660\\5200\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2}\times 660-\frac{1}{20}\times 5200\\-\frac{2}{3}\times 660+\frac{1}{10}\times 5200\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}70\\80\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=70,y=80$$

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.

$$6x+3y=660,40x+30y=5200$$

To make 6x and 40x equal, multiply all terms on each side of the first equation by 40 and all terms on each side of the second by 6.

$$40\times 6x+40\times 3y=40\times 660,6\times 40x+6\times 30y=6\times 5200$$

Simplify.

$$240x+120y=26400,240x+180y=31200$$

Subtract 240x+180y=31200 from 240x+120y=26400 by subtracting like terms on each side of the equal sign.

$$240x-240x+120y-180y=26400-31200$$

Add 240x to -240x. Terms 240x and -240x cancel out, leaving an equation with only one variable that can be solved.

$$120y-180y=26400-31200$$

Add 120y to -180y.

$$-60y=26400-31200$$

Add 26400 to -31200.

$$-60y=-4800$$

Divide both sides by -60.

$$y=80$$

Substitute 80 for y in 40x+30y=5200. Because the resulting equation contains only one variable, you can solve for x directly.

$$40x+30\times 80=5200$$

Multiply 30 times 80.

$$40x+2400=5200$$

Subtract 2400 from both sides of the equation.

$$40x=2800$$

Divide both sides by 40.

$$x=70$$

The system is now solved.

$$x=70,y=80$$