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

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

Solve for x, y

$$x=22$$

$$y=18$$

Steps Using Substitution

Consider the first equation. Use the distributive property to multiply 200 by 30-x.

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

To find the opposite of 30-x, find the opposite of each term.

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

Subtract 30 from 30 to get 0.

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

Use the distributive property to multiply 50 by x-y.

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

Combine -200x and 50x to get -150x.

$$6000-150x-50y=1800$$

Subtract 6000 from both sides.

$$-150x-50y=1800-6000$$

Subtract 6000 from 1800 to get -4200.

$$-150x-50y=-4200$$

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.

$$-150x-50y=-4200,x+y=40$$

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

$$-150x-50y=-4200$$

Add 50y to both sides of the equation.

$$-150x=50y-4200$$

Divide both sides by -150.

$$x=-\frac{1}{150}\left(50y-4200\right)$$

Multiply -\frac{1}{150} times -4200+50y.

$$x=-\frac{1}{3}y+28$$

Substitute -\frac{y}{3}+28 for x in the other equation, x+y=40.

$$-\frac{1}{3}y+28+y=40$$

Add -\frac{y}{3} to y.

$$\frac{2}{3}y+28=40$$

Subtract 28 from both sides of the equation.

$$\frac{2}{3}y=12$$

Divide both sides of the equation by \frac{2}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=18$$

Substitute 18 for y in x=-\frac{1}{3}y+28. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{1}{3}\times 18+28$$

Multiply -\frac{1}{3} times 18.

$$x=-6+28$$

Add 28 to -6.

$$x=22$$

The system is now solved.

$$x=22,y=18$$

Steps Using Matrices

Consider the first equation. Use the distributive property to multiply 200 by 30-x.

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

To find the opposite of 30-x, find the opposite of each term.

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

Subtract 30 from 30 to get 0.

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

Use the distributive property to multiply 50 by x-y.

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

Combine -200x and 50x to get -150x.

$$6000-150x-50y=1800$$

Subtract 6000 from both sides.

$$-150x-50y=1800-6000$$

Subtract 6000 from 1800 to get -4200.

$$-150x-50y=-4200$$

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

$$-150x-50y=-4200,x+y=40$$

Write the equations in matrix form.

$$\left(\begin{matrix}-150&-50\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-4200\\40\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}-150&-50\\1&1\end{matrix}\right).

$$inverse(\left(\begin{matrix}-150&-50\\1&1\end{matrix}\right))\left(\begin{matrix}-150&-50\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-150&-50\\1&1\end{matrix}\right))\left(\begin{matrix}-4200\\40\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-150&-50\\1&1\end{matrix}\right))\left(\begin{matrix}-4200\\40\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-150&-50\\1&1\end{matrix}\right))\left(\begin{matrix}-4200\\40\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{-150-\left(-50\right)}&-\frac{-50}{-150-\left(-50\right)}\\-\frac{1}{-150-\left(-50\right)}&-\frac{150}{-150-\left(-50\right)}\end{matrix}\right)\left(\begin{matrix}-4200\\40\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{100}&-\frac{1}{2}\\\frac{1}{100}&\frac{3}{2}\end{matrix}\right)\left(\begin{matrix}-4200\\40\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{100}\left(-4200\right)-\frac{1}{2}\times 40\\\frac{1}{100}\left(-4200\right)+\frac{3}{2}\times 40\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}22\\18\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=22,y=18$$

Steps Using Elimination

Consider the first equation. Use the distributive property to multiply 200 by 30-x.

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

To find the opposite of 30-x, find the opposite of each term.

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

Subtract 30 from 30 to get 0.

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

Use the distributive property to multiply 50 by x-y.

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

Combine -200x and 50x to get -150x.

$$6000-150x-50y=1800$$

Subtract 6000 from both sides.

$$-150x-50y=1800-6000$$

Subtract 6000 from 1800 to get -4200.

$$-150x-50y=-4200$$

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.

$$-150x-50y=-4200,x+y=40$$

To make -150x and x equal, multiply all terms on each side of the first equation by 1 and all terms on each side of the second by -150.

$$-150x-50y=-4200,-150x-150y=-150\times 40$$

Simplify.

$$-150x-50y=-4200,-150x-150y=-6000$$

Subtract -150x-150y=-6000 from -150x-50y=-4200 by subtracting like terms on each side of the equal sign.

$$-150x+150x-50y+150y=-4200+6000$$

Add -150x to 150x. Terms -150x and 150x cancel out, leaving an equation with only one variable that can be solved.

$$-50y+150y=-4200+6000$$

Add -50y to 150y.

$$100y=-4200+6000$$

Add -4200 to 6000.

$$100y=1800$$

Divide both sides by 100.

$$y=18$$

Substitute 18 for y in x+y=40. Because the resulting equation contains only one variable, you can solve for x directly.

$$x+18=40$$

Subtract 18 from both sides of the equation.

$$x=22$$

The system is now solved.

$$x=22,y=18$$