Solve {l}{x+y=100}{18x+10y=75(x+4)}

$$\left\{ \begin{array} { l } { x + y = 100 } \\ { 18 x + 10 y = 75 ( x + 4 ) } \end{array} \right.$$

Solve for x, y

$$x = \frac{700}{67} = 10\frac{30}{67} \approx 10.447761194$$

$$y = \frac{6000}{67} = 89\frac{37}{67} \approx 89.552238806$$

Steps Using Substitution

Consider the second equation. Use the distributive property to multiply 75 by x+4.

$$18x+10y=75x+300$$

Subtract 75x from both sides.

$$18x+10y-75x=300$$

Combine 18x and -75x to get -57x.

$$-57x+10y=300$$

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.

$$x+y=100,-57x+10y=300$$

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

$$x+y=100$$

Subtract y from both sides of the equation.

$$x=-y+100$$

Substitute -y+100 for x in the other equation, -57x+10y=300.

$$-57\left(-y+100\right)+10y=300$$

Multiply -57 times -y+100.

$$57y-5700+10y=300$$

Add 57y to 10y.

$$67y-5700=300$$

Add 5700 to both sides of the equation.

$$67y=6000$$

Divide both sides by 67.

$$y=\frac{6000}{67}$$

Substitute \frac{6000}{67} for y in x=-y+100. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{6000}{67}+100$$

Add 100 to -\frac{6000}{67}.

$$x=\frac{700}{67}$$

The system is now solved.

$$x=\frac{700}{67},y=\frac{6000}{67}$$

Steps Using Matrices

Consider the second equation. Use the distributive property to multiply 75 by x+4.

$$18x+10y=75x+300$$

Subtract 75x from both sides.

$$18x+10y-75x=300$$

Combine 18x and -75x to get -57x.

$$-57x+10y=300$$

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

$$x+y=100,-57x+10y=300$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\-57&10\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}100\\300\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\-57&10\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\-57&10\end{matrix}\right))\left(\begin{matrix}1&1\\-57&10\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\-57&10\end{matrix}\right))\left(\begin{matrix}100\\300\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\-57&10\end{matrix}\right))\left(\begin{matrix}100\\300\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\-57&10\end{matrix}\right))\left(\begin{matrix}100\\300\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{10}{10-\left(-57\right)}&-\frac{1}{10-\left(-57\right)}\\-\frac{-57}{10-\left(-57\right)}&\frac{1}{10-\left(-57\right)}\end{matrix}\right)\left(\begin{matrix}100\\300\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{10}{67}&-\frac{1}{67}\\\frac{57}{67}&\frac{1}{67}\end{matrix}\right)\left(\begin{matrix}100\\300\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{10}{67}\times 100-\frac{1}{67}\times 300\\\frac{57}{67}\times 100+\frac{1}{67}\times 300\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{700}{67}\\\frac{6000}{67}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{700}{67},y=\frac{6000}{67}$$

Steps Using Elimination

Consider the second equation. Use the distributive property to multiply 75 by x+4.

$$18x+10y=75x+300$$

Subtract 75x from both sides.

$$18x+10y-75x=300$$

Combine 18x and -75x to get -57x.

$$-57x+10y=300$$

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.

$$x+y=100,-57x+10y=300$$

To make x and -57x equal, multiply all terms on each side of the first equation by -57 and all terms on each side of the second by 1.

$$-57x-57y=-57\times 100,-57x+10y=300$$

Simplify.

$$-57x-57y=-5700,-57x+10y=300$$

Subtract -57x+10y=300 from -57x-57y=-5700 by subtracting like terms on each side of the equal sign.

$$-57x+57x-57y-10y=-5700-300$$

Add -57x to 57x. Terms -57x and 57x cancel out, leaving an equation with only one variable that can be solved.

$$-57y-10y=-5700-300$$

Add -57y to -10y.

$$-67y=-5700-300$$

Add -5700 to -300.

$$-67y=-6000$$

Divide both sides by -67.

$$y=\frac{6000}{67}$$

Substitute \frac{6000}{67} for y in -57x+10y=300. Because the resulting equation contains only one variable, you can solve for x directly.

$$-57x+10\times \frac{6000}{67}=300$$

Multiply 10 times \frac{6000}{67}.

$$-57x+\frac{60000}{67}=300$$

Subtract \frac{60000}{67} from both sides of the equation.

$$-57x=-\frac{39900}{67}$$

Divide both sides by -57.

$$x=\frac{700}{67}$$

The system is now solved.

$$x=\frac{700}{67},y=\frac{6000}{67}$$