Solve {l}{30x+60y=72}{10x+30y=360}

$$\left\{ \begin{array} { l } { 30 x + 60 y = 72 } \\ { 10 x + 30 y = 360 } \end{array} \right.$$

Solve for x, y

$$x = -\frac{324}{5} = -64\frac{4}{5} = -64.8$$

$$y = \frac{168}{5} = 33\frac{3}{5} = 33.6$$

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.

$$30x+60y=72,10x+30y=360$$

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

$$30x+60y=72$$

Subtract 60y from both sides of the equation.

$$30x=-60y+72$$

Divide both sides by 30.

$$x=\frac{1}{30}\left(-60y+72\right)$$

Multiply \frac{1}{30} times -60y+72.

$$x=-2y+\frac{12}{5}$$

Substitute -2y+\frac{12}{5} for x in the other equation, 10x+30y=360.

$$10\left(-2y+\frac{12}{5}\right)+30y=360$$

Multiply 10 times -2y+\frac{12}{5}.

$$-20y+24+30y=360$$

Add -20y to 30y.

$$10y+24=360$$

Subtract 24 from both sides of the equation.

$$10y=336$$

Divide both sides by 10.

$$y=\frac{168}{5}$$

Substitute \frac{168}{5} for y in x=-2y+\frac{12}{5}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-2\times \frac{168}{5}+\frac{12}{5}$$

Multiply -2 times \frac{168}{5}.

$$x=\frac{-336+12}{5}$$

Add \frac{12}{5} to -\frac{336}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{324}{5}$$

The system is now solved.

$$x=-\frac{324}{5},y=\frac{168}{5}$$

Steps Using Matrices

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

$$30x+60y=72,10x+30y=360$$

Write the equations in matrix form.

$$\left(\begin{matrix}30&60\\10&30\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}72\\360\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}30&60\\10&30\end{matrix}\right).

$$inverse(\left(\begin{matrix}30&60\\10&30\end{matrix}\right))\left(\begin{matrix}30&60\\10&30\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}30&60\\10&30\end{matrix}\right))\left(\begin{matrix}72\\360\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}30&60\\10&30\end{matrix}\right))\left(\begin{matrix}72\\360\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}30&60\\10&30\end{matrix}\right))\left(\begin{matrix}72\\360\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{30}{30\times 30-60\times 10}&-\frac{60}{30\times 30-60\times 10}\\-\frac{10}{30\times 30-60\times 10}&\frac{30}{30\times 30-60\times 10}\end{matrix}\right)\left(\begin{matrix}72\\360\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{10}&-\frac{1}{5}\\-\frac{1}{30}&\frac{1}{10}\end{matrix}\right)\left(\begin{matrix}72\\360\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{10}\times 72-\frac{1}{5}\times 360\\-\frac{1}{30}\times 72+\frac{1}{10}\times 360\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{324}{5}\\\frac{168}{5}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-\frac{324}{5},y=\frac{168}{5}$$

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.

$$30x+60y=72,10x+30y=360$$

To make 30x and 10x equal, multiply all terms on each side of the first equation by 10 and all terms on each side of the second by 30.

$$10\times 30x+10\times 60y=10\times 72,30\times 10x+30\times 30y=30\times 360$$

Simplify.

$$300x+600y=720,300x+900y=10800$$

Subtract 300x+900y=10800 from 300x+600y=720 by subtracting like terms on each side of the equal sign.

$$300x-300x+600y-900y=720-10800$$

Add 300x to -300x. Terms 300x and -300x cancel out, leaving an equation with only one variable that can be solved.

$$600y-900y=720-10800$$

Add 600y to -900y.

$$-300y=720-10800$$

Add 720 to -10800.

$$-300y=-10080$$

Divide both sides by -300.

$$y=\frac{168}{5}$$

Substitute \frac{168}{5} for y in 10x+30y=360. Because the resulting equation contains only one variable, you can solve for x directly.

$$10x+30\times \frac{168}{5}=360$$

Multiply 30 times \frac{168}{5}.

$$10x+1008=360$$

Subtract 1008 from both sides of the equation.

$$10x=-648$$

Divide both sides by 10.

$$x=-\frac{324}{5}$$

The system is now solved.

$$x=-\frac{324}{5},y=\frac{168}{5}$$