Solve {l}{x+y=120}{x-25=3(y+25)}

$$\left\{ \begin{array} { l } { x + y = 120 } \\ { x - 25 = 3 ( y + 25 ) } \end{array} \right.$$

Solve for x, y

$$x=115$$

$$y=5$$

Steps Using Substitution

Consider the second equation. Use the distributive property to multiply 3 by y+25.

$$x-25=3y+75$$

Subtract 3y from both sides.

$$x-25-3y=75$$

Add 25 to both sides.

$$x-3y=75+25$$

Add 75 and 25 to get 100.

$$x-3y=100$$

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.

$$x+y=120,x-3y=100$$

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

$$x+y=120$$

Subtract y from both sides of the equation.

$$x=-y+120$$

Substitute -y+120 for x in the other equation, x-3y=100.

$$-y+120-3y=100$$

Add -y to -3y.

$$-4y+120=100$$

Subtract 120 from both sides of the equation.

$$-4y=-20$$

Divide both sides by -4.

$$y=5$$

Substitute 5 for y in x=-y+120. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-5+120$$

Add 120 to -5.

$$x=115$$

The system is now solved.

$$x=115,y=5$$

Steps Using Matrices

Consider the second equation. Use the distributive property to multiply 3 by y+25.

$$x-25=3y+75$$

Subtract 3y from both sides.

$$x-25-3y=75$$

Add 25 to both sides.

$$x-3y=75+25$$

Add 75 and 25 to get 100.

$$x-3y=100$$

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

$$x+y=120,x-3y=100$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\1&-3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}120\\100\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\1&-3\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\1&-3\end{matrix}\right))\left(\begin{matrix}1&1\\1&-3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\1&-3\end{matrix}\right))\left(\begin{matrix}120\\100\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\1&-3\end{matrix}\right))\left(\begin{matrix}120\\100\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\1&-3\end{matrix}\right))\left(\begin{matrix}120\\100\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{-3-1}&-\frac{1}{-3-1}\\-\frac{1}{-3-1}&\frac{1}{-3-1}\end{matrix}\right)\left(\begin{matrix}120\\100\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{4}&\frac{1}{4}\\\frac{1}{4}&-\frac{1}{4}\end{matrix}\right)\left(\begin{matrix}120\\100\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{4}\times 120+\frac{1}{4}\times 100\\\frac{1}{4}\times 120-\frac{1}{4}\times 100\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}115\\5\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=115,y=5$$

Steps Using Elimination

Consider the second equation. Use the distributive property to multiply 3 by y+25.

$$x-25=3y+75$$

Subtract 3y from both sides.

$$x-25-3y=75$$

Add 25 to both sides.

$$x-3y=75+25$$

Add 75 and 25 to get 100.

$$x-3y=100$$

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.

$$x+y=120,x-3y=100$$

Subtract x-3y=100 from x+y=120 by subtracting like terms on each side of the equal sign.

$$x-x+y+3y=120-100$$

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

$$y+3y=120-100$$

Add y to 3y.

$$4y=120-100$$

Add 120 to -100.

$$4y=20$$

Divide both sides by 4.

$$y=5$$

Substitute 5 for y in x-3y=100. Because the resulting equation contains only one variable, you can solve for x directly.

$$x-3\times 5=100$$

Multiply -3 times 5.

$$x-15=100$$

Add 15 to both sides of the equation.

$$x=115$$

The system is now solved.

$$x=115,y=5$$