Solve {l}{12*5(x+y)=9750}{13(x-y)=9750}

$$\left\{ \begin{array} { l } { 12 \cdot 5 ( x + y ) = 9750 } \\ { 13 ( x - y ) = 9750 } \end{array} \right.$$

Solve for x, y

$$x = \frac{1825}{4} = 456\frac{1}{4} = 456.25$$

$$y = -\frac{1175}{4} = -293\frac{3}{4} = -293.75$$

Steps Using Substitution

Consider the first equation. Multiply 12 and 5 to get 60.

$$60\left(x+y\right)=9750$$

Use the distributive property to multiply 60 by x+y.

$$60x+60y=9750$$

Consider the second equation. Divide both sides by 13.

$$x-y=\frac{9750}{13}$$

Divide 9750 by 13 to get 750.

$$x-y=750$$

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.

$$60x+60y=9750,x-y=750$$

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

$$60x+60y=9750$$

Subtract 60y from both sides of the equation.

$$60x=-60y+9750$$

Divide both sides by 60.

$$x=\frac{1}{60}\left(-60y+9750\right)$$

Multiply \frac{1}{60} times -60y+9750.

$$x=-y+\frac{325}{2}$$

Substitute -y+\frac{325}{2} for x in the other equation, x-y=750.

$$-y+\frac{325}{2}-y=750$$

Add -y to -y.

$$-2y+\frac{325}{2}=750$$

Subtract \frac{325}{2} from both sides of the equation.

$$-2y=\frac{1175}{2}$$

Divide both sides by -2.

$$y=-\frac{1175}{4}$$

Substitute -\frac{1175}{4} for y in x=-y+\frac{325}{2}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\left(-\frac{1175}{4}\right)+\frac{325}{2}$$

Multiply -1 times -\frac{1175}{4}.

$$x=\frac{1175}{4}+\frac{325}{2}$$

Add \frac{325}{2} to \frac{1175}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{1825}{4}$$

The system is now solved.

$$x=\frac{1825}{4},y=-\frac{1175}{4}$$

Steps Using Matrices

Consider the first equation. Multiply 12 and 5 to get 60.

$$60\left(x+y\right)=9750$$

Use the distributive property to multiply 60 by x+y.

$$60x+60y=9750$$

Consider the second equation. Divide both sides by 13.

$$x-y=\frac{9750}{13}$$

Divide 9750 by 13 to get 750.

$$x-y=750$$

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

$$60x+60y=9750,x-y=750$$

Write the equations in matrix form.

$$\left(\begin{matrix}60&60\\1&-1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}9750\\750\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}60&60\\1&-1\end{matrix}\right).

$$inverse(\left(\begin{matrix}60&60\\1&-1\end{matrix}\right))\left(\begin{matrix}60&60\\1&-1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}60&60\\1&-1\end{matrix}\right))\left(\begin{matrix}9750\\750\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}60&60\\1&-1\end{matrix}\right))\left(\begin{matrix}9750\\750\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}60&60\\1&-1\end{matrix}\right))\left(\begin{matrix}9750\\750\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{60\left(-1\right)-60}&-\frac{60}{60\left(-1\right)-60}\\-\frac{1}{60\left(-1\right)-60}&\frac{60}{60\left(-1\right)-60}\end{matrix}\right)\left(\begin{matrix}9750\\750\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{120}&\frac{1}{2}\\\frac{1}{120}&-\frac{1}{2}\end{matrix}\right)\left(\begin{matrix}9750\\750\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{120}\times 9750+\frac{1}{2}\times 750\\\frac{1}{120}\times 9750-\frac{1}{2}\times 750\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1825}{4}\\-\frac{1175}{4}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{1825}{4},y=-\frac{1175}{4}$$

Steps Using Elimination

Consider the first equation. Multiply 12 and 5 to get 60.

$$60\left(x+y\right)=9750$$

Use the distributive property to multiply 60 by x+y.

$$60x+60y=9750$$

Consider the second equation. Divide both sides by 13.

$$x-y=\frac{9750}{13}$$

Divide 9750 by 13 to get 750.

$$x-y=750$$

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.

$$60x+60y=9750,x-y=750$$

To make 60x 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 60.

$$60x+60y=9750,60x+60\left(-1\right)y=60\times 750$$

Simplify.

$$60x+60y=9750,60x-60y=45000$$

Subtract 60x-60y=45000 from 60x+60y=9750 by subtracting like terms on each side of the equal sign.

$$60x-60x+60y+60y=9750-45000$$

Add 60x to -60x. Terms 60x and -60x cancel out, leaving an equation with only one variable that can be solved.

$$60y+60y=9750-45000$$

Add 60y to 60y.

$$120y=9750-45000$$

Add 9750 to -45000.

$$120y=-35250$$

Divide both sides by 120.

$$y=-\frac{1175}{4}$$

Substitute -\frac{1175}{4} for y in x-y=750. Because the resulting equation contains only one variable, you can solve for x directly.

$$x-\left(-\frac{1175}{4}\right)=750$$

Subtract \frac{1175}{4} from both sides of the equation.

$$x=\frac{1825}{4}$$

The system is now solved.

$$x=\frac{1825}{4},y=-\frac{1175}{4}$$