Solve {l}{5(x+y)=100}{(5+5)(x-y)=100}

$$\left\{ \begin{array} { l } { 5 ( x + y ) = 100 } \\ { ( 5 + 5 ) ( x - y ) = 100 } \end{array} \right.$$

Solve for x, y

$$x=15$$

$$y=5$$

Steps Using Substitution

Consider the first equation. Divide both sides by 5.

$$x+y=\frac{100}{5}$$

Divide 100 by 5 to get 20.

$$x+y=20$$

Consider the second equation. Add 5 and 5 to get 10.

$$10\left(x-y\right)=100$$

Use the distributive property to multiply 10 by x-y.

$$10x-10y=100$$

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.

$$x+y=20,10x-10y=100$$

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

$$x+y=20$$

Subtract y from both sides of the equation.

$$x=-y+20$$

Substitute -y+20 for x in the other equation, 10x-10y=100.

$$10\left(-y+20\right)-10y=100$$

Multiply 10 times -y+20.

$$-10y+200-10y=100$$

Add -10y to -10y.

$$-20y+200=100$$

Subtract 200 from both sides of the equation.

$$-20y=-100$$

Divide both sides by -20.

$$y=5$$

Substitute 5 for y in x=-y+20. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-5+20$$

Add 20 to -5.

$$x=15$$

The system is now solved.

$$x=15,y=5$$

Steps Using Matrices

Consider the first equation. Divide both sides by 5.

$$x+y=\frac{100}{5}$$

Divide 100 by 5 to get 20.

$$x+y=20$$

Consider the second equation. Add 5 and 5 to get 10.

$$10\left(x-y\right)=100$$

Use the distributive property to multiply 10 by x-y.

$$10x-10y=100$$

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

$$x+y=20,10x-10y=100$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\10&-10\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}20\\100\end{matrix}\right)$$

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

$$inverse(\left(\begin{matrix}1&1\\10&-10\end{matrix}\right))\left(\begin{matrix}1&1\\10&-10\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\10&-10\end{matrix}\right))\left(\begin{matrix}20\\100\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\10&-10\end{matrix}\right))\left(\begin{matrix}20\\100\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\10&-10\end{matrix}\right))\left(\begin{matrix}20\\100\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{10}{-10-10}&-\frac{1}{-10-10}\\-\frac{10}{-10-10}&\frac{1}{-10-10}\end{matrix}\right)\left(\begin{matrix}20\\100\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2}&\frac{1}{20}\\\frac{1}{2}&-\frac{1}{20}\end{matrix}\right)\left(\begin{matrix}20\\100\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2}\times 20+\frac{1}{20}\times 100\\\frac{1}{2}\times 20-\frac{1}{20}\times 100\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}15\\5\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=15,y=5$$

Steps Using Elimination

Consider the first equation. Divide both sides by 5.

$$x+y=\frac{100}{5}$$

Divide 100 by 5 to get 20.

$$x+y=20$$

Consider the second equation. Add 5 and 5 to get 10.

$$10\left(x-y\right)=100$$

Use the distributive property to multiply 10 by x-y.

$$10x-10y=100$$

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.

$$x+y=20,10x-10y=100$$

To make x and 10x equal, multiply all terms on each side of the first equation by 10 and all terms on each side of the second by 1.

$$10x+10y=10\times 20,10x-10y=100$$

Simplify.

$$10x+10y=200,10x-10y=100$$

Subtract 10x-10y=100 from 10x+10y=200 by subtracting like terms on each side of the equal sign.

$$10x-10x+10y+10y=200-100$$

Add 10x to -10x. Terms 10x and -10x cancel out, leaving an equation with only one variable that can be solved.

$$10y+10y=200-100$$

Add 10y to 10y.

$$20y=200-100$$

Add 200 to -100.

$$20y=100$$

Divide both sides by 20.

$$y=5$$

Substitute 5 for y in 10x-10y=100. Because the resulting equation contains only one variable, you can solve for x directly.

$$10x-10\times 5=100$$

Multiply -10 times 5.

$$10x-50=100$$

Add 50 to both sides of the equation.

$$10x=150$$

Divide both sides by 10.

$$x=15$$

The system is now solved.

$$x=15,y=5$$