Solve {l}{3600=30x+210y}{7920=48x+552y}

$$\left\{ \begin{array} { l } { 3600 = 30 x + 210 y } \\ { 7920 = 48 x + 552 y } \end{array} \right.$$

Solve for x, y

$$x=50$$

$$y=10$$

Steps Using Substitution

Consider the first equation. Swap sides so that all variable terms are on the left hand side.

$$30x+210y=3600$$

Consider the second equation. Swap sides so that all variable terms are on the left hand side.

$$48x+552y=7920$$

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.

$$30x+210y=3600,48x+552y=7920$$

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

$$30x+210y=3600$$

Subtract 210y from both sides of the equation.

$$30x=-210y+3600$$

Divide both sides by 30.

$$x=\frac{1}{30}\left(-210y+3600\right)$$

Multiply \frac{1}{30} times -210y+3600.

$$x=-7y+120$$

Substitute -7y+120 for x in the other equation, 48x+552y=7920.

$$48\left(-7y+120\right)+552y=7920$$

Multiply 48 times -7y+120.

$$-336y+5760+552y=7920$$

Add -336y to 552y.

$$216y+5760=7920$$

Subtract 5760 from both sides of the equation.

$$216y=2160$$

Divide both sides by 216.

$$y=10$$

Substitute 10 for y in x=-7y+120. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-7\times 10+120$$

Multiply -7 times 10.

$$x=-70+120$$

Add 120 to -70.

$$x=50$$

The system is now solved.

$$x=50,y=10$$

Steps Using Matrices

Consider the first equation. Swap sides so that all variable terms are on the left hand side.

$$30x+210y=3600$$

Consider the second equation. Swap sides so that all variable terms are on the left hand side.

$$48x+552y=7920$$

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

$$30x+210y=3600,48x+552y=7920$$

Write the equations in matrix form.

$$\left(\begin{matrix}30&210\\48&552\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}3600\\7920\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}30&210\\48&552\end{matrix}\right).

$$inverse(\left(\begin{matrix}30&210\\48&552\end{matrix}\right))\left(\begin{matrix}30&210\\48&552\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}30&210\\48&552\end{matrix}\right))\left(\begin{matrix}3600\\7920\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}30&210\\48&552\end{matrix}\right))\left(\begin{matrix}3600\\7920\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}30&210\\48&552\end{matrix}\right))\left(\begin{matrix}3600\\7920\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{552}{30\times 552-210\times 48}&-\frac{210}{30\times 552-210\times 48}\\-\frac{48}{30\times 552-210\times 48}&\frac{30}{30\times 552-210\times 48}\end{matrix}\right)\left(\begin{matrix}3600\\7920\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{23}{270}&-\frac{7}{216}\\-\frac{1}{135}&\frac{1}{216}\end{matrix}\right)\left(\begin{matrix}3600\\7920\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{23}{270}\times 3600-\frac{7}{216}\times 7920\\-\frac{1}{135}\times 3600+\frac{1}{216}\times 7920\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}50\\10\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=50,y=10$$

Steps Using Elimination

Consider the first equation. Swap sides so that all variable terms are on the left hand side.

$$30x+210y=3600$$

Consider the second equation. Swap sides so that all variable terms are on the left hand side.

$$48x+552y=7920$$

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.

$$30x+210y=3600,48x+552y=7920$$

To make 30x and 48x equal, multiply all terms on each side of the first equation by 48 and all terms on each side of the second by 30.

$$48\times 30x+48\times 210y=48\times 3600,30\times 48x+30\times 552y=30\times 7920$$

Simplify.

$$1440x+10080y=172800,1440x+16560y=237600$$

Subtract 1440x+16560y=237600 from 1440x+10080y=172800 by subtracting like terms on each side of the equal sign.

$$1440x-1440x+10080y-16560y=172800-237600$$

Add 1440x to -1440x. Terms 1440x and -1440x cancel out, leaving an equation with only one variable that can be solved.

$$10080y-16560y=172800-237600$$

Add 10080y to -16560y.

$$-6480y=172800-237600$$

Add 172800 to -237600.

$$-6480y=-64800$$

Divide both sides by -6480.

$$y=10$$

Substitute 10 for y in 48x+552y=7920. Because the resulting equation contains only one variable, you can solve for x directly.

$$48x+552\times 10=7920$$

Multiply 552 times 10.

$$48x+5520=7920$$

Subtract 5520 from both sides of the equation.

$$48x=2400$$

Divide both sides by 48.

$$x=50$$

The system is now solved.

$$x=50,y=10$$