Solve {l}{200(30-x)+50(30-y)=1800}{x+y=40}

$$\left\{ \begin{array} { l } { 200 ( 30 - x ) + 50 ( 30 - y ) = 1800 } \\ { x + y = 40 } \end{array} \right.$$

Solve for x, y

$$x = \frac{74}{3} = 24\frac{2}{3} \approx 24.666666667$$

$$y = \frac{46}{3} = 15\frac{1}{3} \approx 15.333333333$$

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.

$$200\left(-x+30\right)+50\left(-y+30\right)=1800,x+y=40$$

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

$$200\left(-x+30\right)+50\left(-y+30\right)=1800$$

Multiply 200 times -x+30.

$$-200x+6000+50\left(-y+30\right)=1800$$

Multiply 50 times -y+30.

$$-200x+6000-50y+1500=1800$$

Add 6000 to 1500.

$$-200x-50y+7500=1800$$

Subtract 7500 from both sides of the equation.

$$-200x-50y=-5700$$

Add 50y to both sides of the equation.

$$-200x=50y-5700$$

Divide both sides by -200.

$$x=-\frac{1}{200}\left(50y-5700\right)$$

Multiply -\frac{1}{200} times -5700+50y.

$$x=-\frac{1}{4}y+\frac{57}{2}$$

Substitute -\frac{y}{4}+\frac{57}{2} for x in the other equation, x+y=40.

$$-\frac{1}{4}y+\frac{57}{2}+y=40$$

Add -\frac{y}{4} to y.

$$\frac{3}{4}y+\frac{57}{2}=40$$

Subtract \frac{57}{2} from both sides of the equation.

$$\frac{3}{4}y=\frac{23}{2}$$

Divide both sides of the equation by \frac{3}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=\frac{46}{3}$$

Substitute \frac{46}{3} for y in x=-\frac{1}{4}y+\frac{57}{2}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{1}{4}\times \frac{46}{3}+\frac{57}{2}$$

Multiply -\frac{1}{4} times \frac{46}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{23}{6}+\frac{57}{2}$$

Add \frac{57}{2} to -\frac{23}{6} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{74}{3}$$

The system is now solved.

$$x=\frac{74}{3},y=\frac{46}{3}$$

Steps Using Matrices

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

$$200\left(-x+30\right)+50\left(-y+30\right)=1800,x+y=40$$

Simplify the first equation to put it in standard form.

$$200\left(-x+30\right)+50\left(-y+30\right)=1800$$

Multiply 200 times -x+30.

$$-200x+6000+50\left(-y+30\right)=1800$$

Multiply 50 times -y+30.

$$-200x+6000-50y+1500=1800$$

Add 6000 to 1500.

$$-200x-50y+7500=1800$$

Subtract 7500 from both sides of the equation.

$$-200x-50y=-5700$$

Write the equations in matrix form.

$$\left(\begin{matrix}-200&-50\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-5700\\40\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}-200&-50\\1&1\end{matrix}\right).

$$inverse(\left(\begin{matrix}-200&-50\\1&1\end{matrix}\right))\left(\begin{matrix}-200&-50\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-200&-50\\1&1\end{matrix}\right))\left(\begin{matrix}-5700\\40\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-200&-50\\1&1\end{matrix}\right))\left(\begin{matrix}-5700\\40\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-200&-50\\1&1\end{matrix}\right))\left(\begin{matrix}-5700\\40\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{-200-\left(-50\right)}&-\frac{-50}{-200-\left(-50\right)}\\-\frac{1}{-200-\left(-50\right)}&-\frac{200}{-200-\left(-50\right)}\end{matrix}\right)\left(\begin{matrix}-5700\\40\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{150}&-\frac{1}{3}\\\frac{1}{150}&\frac{4}{3}\end{matrix}\right)\left(\begin{matrix}-5700\\40\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{150}\left(-5700\right)-\frac{1}{3}\times 40\\\frac{1}{150}\left(-5700\right)+\frac{4}{3}\times 40\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{74}{3}\\\frac{46}{3}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{74}{3},y=\frac{46}{3}$$