Solve {l}{(x+10)(y+3)=+500+xy}{(x-15)(y-9)=xy+(-600)}

$$\left\{ \begin{array} { l } { ( x + 10 ) ( y + 3 ) = + 500 + x y } \\ { ( x - 15 ) ( y - 9 ) = x y + ( - 600 ) } \end{array} \right.$$

Solve for x, y

$$x = \frac{20}{3} = 6\frac{2}{3} \approx 6.666666667$$

$$y=45$$

Steps Using Substitution

Consider the first equation. Use the distributive property to multiply x+10 by y+3.

$$xy+3x+10y+30=500+xy$$

Subtract xy from both sides.

$$xy+3x+10y+30-xy=500$$

Combine xy and -xy to get 0.

$$3x+10y+30=500$$

Subtract 30 from both sides.

$$3x+10y=500-30$$

Subtract 30 from 500 to get 470.

$$3x+10y=470$$

Consider the second equation. Use the distributive property to multiply x-15 by y-9.

$$xy-9x-15y+135=xy-600$$

Subtract xy from both sides.

$$xy-9x-15y+135-xy=-600$$

Combine xy and -xy to get 0.

$$-9x-15y+135=-600$$

Subtract 135 from both sides.

$$-9x-15y=-600-135$$

Subtract 135 from -600 to get -735.

$$-9x-15y=-735$$

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.

$$3x+10y=470,-9x-15y=-735$$

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

$$3x+10y=470$$

Subtract 10y from both sides of the equation.

$$3x=-10y+470$$

Divide both sides by 3.

$$x=\frac{1}{3}\left(-10y+470\right)$$

Multiply \frac{1}{3} times -10y+470.

$$x=-\frac{10}{3}y+\frac{470}{3}$$

Substitute \frac{-10y+470}{3} for x in the other equation, -9x-15y=-735.

$$-9\left(-\frac{10}{3}y+\frac{470}{3}\right)-15y=-735$$

Multiply -9 times \frac{-10y+470}{3}.

$$30y-1410-15y=-735$$

Add 30y to -15y.

$$15y-1410=-735$$

Add 1410 to both sides of the equation.

$$15y=675$$

Divide both sides by 15.

$$y=45$$

Substitute 45 for y in x=-\frac{10}{3}y+\frac{470}{3}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{10}{3}\times 45+\frac{470}{3}$$

Multiply -\frac{10}{3} times 45.

$$x=-150+\frac{470}{3}$$

Add \frac{470}{3} to -150.

$$x=\frac{20}{3}$$

The system is now solved.

$$x=\frac{20}{3},y=45$$

Steps Using Matrices

Consider the first equation. Use the distributive property to multiply x+10 by y+3.

$$xy+3x+10y+30=500+xy$$

Subtract xy from both sides.

$$xy+3x+10y+30-xy=500$$

Combine xy and -xy to get 0.

$$3x+10y+30=500$$

Subtract 30 from both sides.

$$3x+10y=500-30$$

Subtract 30 from 500 to get 470.

$$3x+10y=470$$

Consider the second equation. Use the distributive property to multiply x-15 by y-9.

$$xy-9x-15y+135=xy-600$$

Subtract xy from both sides.

$$xy-9x-15y+135-xy=-600$$

Combine xy and -xy to get 0.

$$-9x-15y+135=-600$$

Subtract 135 from both sides.

$$-9x-15y=-600-135$$

Subtract 135 from -600 to get -735.

$$-9x-15y=-735$$

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

$$3x+10y=470,-9x-15y=-735$$

Write the equations in matrix form.

$$\left(\begin{matrix}3&10\\-9&-15\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}470\\-735\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}3&10\\-9&-15\end{matrix}\right).

$$inverse(\left(\begin{matrix}3&10\\-9&-15\end{matrix}\right))\left(\begin{matrix}3&10\\-9&-15\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&10\\-9&-15\end{matrix}\right))\left(\begin{matrix}470\\-735\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&10\\-9&-15\end{matrix}\right))\left(\begin{matrix}470\\-735\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&10\\-9&-15\end{matrix}\right))\left(\begin{matrix}470\\-735\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{15}{3\left(-15\right)-10\left(-9\right)}&-\frac{10}{3\left(-15\right)-10\left(-9\right)}\\-\frac{-9}{3\left(-15\right)-10\left(-9\right)}&\frac{3}{3\left(-15\right)-10\left(-9\right)}\end{matrix}\right)\left(\begin{matrix}470\\-735\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{3}&-\frac{2}{9}\\\frac{1}{5}&\frac{1}{15}\end{matrix}\right)\left(\begin{matrix}470\\-735\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{3}\times 470-\frac{2}{9}\left(-735\right)\\\frac{1}{5}\times 470+\frac{1}{15}\left(-735\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{20}{3}\\45\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{20}{3},y=45$$

Steps Using Elimination

Consider the first equation. Use the distributive property to multiply x+10 by y+3.

$$xy+3x+10y+30=500+xy$$

Subtract xy from both sides.

$$xy+3x+10y+30-xy=500$$

Combine xy and -xy to get 0.

$$3x+10y+30=500$$

Subtract 30 from both sides.

$$3x+10y=500-30$$

Subtract 30 from 500 to get 470.

$$3x+10y=470$$

Consider the second equation. Use the distributive property to multiply x-15 by y-9.

$$xy-9x-15y+135=xy-600$$

Subtract xy from both sides.

$$xy-9x-15y+135-xy=-600$$

Combine xy and -xy to get 0.

$$-9x-15y+135=-600$$

Subtract 135 from both sides.

$$-9x-15y=-600-135$$

Subtract 135 from -600 to get -735.

$$-9x-15y=-735$$

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.

$$3x+10y=470,-9x-15y=-735$$

To make 3x and -9x equal, multiply all terms on each side of the first equation by -9 and all terms on each side of the second by 3.

$$-9\times 3x-9\times 10y=-9\times 470,3\left(-9\right)x+3\left(-15\right)y=3\left(-735\right)$$

Simplify.

$$-27x-90y=-4230,-27x-45y=-2205$$

Subtract -27x-45y=-2205 from -27x-90y=-4230 by subtracting like terms on each side of the equal sign.

$$-27x+27x-90y+45y=-4230+2205$$

Add -27x to 27x. Terms -27x and 27x cancel out, leaving an equation with only one variable that can be solved.

$$-90y+45y=-4230+2205$$

Add -90y to 45y.

$$-45y=-4230+2205$$

Add -4230 to 2205.

$$-45y=-2025$$

Divide both sides by -45.

$$y=45$$

Substitute 45 for y in -9x-15y=-735. Because the resulting equation contains only one variable, you can solve for x directly.

$$-9x-15\times 45=-735$$

Multiply -15 times 45.

$$-9x-675=-735$$

Add 675 to both sides of the equation.

$$-9x=-60$$

Divide both sides by -9.

$$x=\frac{20}{3}$$

The system is now solved.

$$x=\frac{20}{3},y=45$$