Solve {l}{x+2x+50+y=180}{y+2x+50=180}

$$\left. \begin{array} { l } { x + 2 x + 50 + y = 180 } \\ { y + 2 x + 50 = 180 } \end{array} \right.$$

Solve for x, y

$$x=0$$

$$y=130$$

Steps Using Substitution

Consider the first equation. Combine x and 2x to get 3x.

$$3x+50+y=180$$

Subtract 50 from both sides.

$$3x+y=180-50$$

Subtract 50 from 180 to get 130.

$$3x+y=130$$

Consider the second equation. Subtract 50 from both sides.

$$y+2x=180-50$$

Subtract 50 from 180 to get 130.

$$y+2x=130$$

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.

$$3x+y=130,2x+y=130$$

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

$$3x+y=130$$

Subtract y from both sides of the equation.

$$3x=-y+130$$

Divide both sides by 3.

$$x=\frac{1}{3}\left(-y+130\right)$$

Multiply \frac{1}{3} times -y+130.

$$x=-\frac{1}{3}y+\frac{130}{3}$$

Substitute \frac{-y+130}{3} for x in the other equation, 2x+y=130.

$$2\left(-\frac{1}{3}y+\frac{130}{3}\right)+y=130$$

Multiply 2 times \frac{-y+130}{3}.

$$-\frac{2}{3}y+\frac{260}{3}+y=130$$

Add -\frac{2y}{3} to y.

$$\frac{1}{3}y+\frac{260}{3}=130$$

Subtract \frac{260}{3} from both sides of the equation.

$$\frac{1}{3}y=\frac{130}{3}$$

Multiply both sides by 3.

$$y=130$$

Substitute 130 for y in x=-\frac{1}{3}y+\frac{130}{3}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{1}{3}\times 130+\frac{130}{3}$$

Multiply -\frac{1}{3} times 130.

$$x=\frac{-130+130}{3}$$

Add \frac{130}{3} to -\frac{130}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=0$$

The system is now solved.

$$x=0,y=130$$

Steps Using Matrices

Consider the first equation. Combine x and 2x to get 3x.

$$3x+50+y=180$$

Subtract 50 from both sides.

$$3x+y=180-50$$

Subtract 50 from 180 to get 130.

$$3x+y=130$$

Consider the second equation. Subtract 50 from both sides.

$$y+2x=180-50$$

Subtract 50 from 180 to get 130.

$$y+2x=130$$

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

$$3x+y=130,2x+y=130$$

Write the equations in matrix form.

$$\left(\begin{matrix}3&1\\2&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}130\\130\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}3&1\\2&1\end{matrix}\right).

$$inverse(\left(\begin{matrix}3&1\\2&1\end{matrix}\right))\left(\begin{matrix}3&1\\2&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&1\\2&1\end{matrix}\right))\left(\begin{matrix}130\\130\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&1\\2&1\end{matrix}\right))\left(\begin{matrix}130\\130\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&1\\2&1\end{matrix}\right))\left(\begin{matrix}130\\130\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{3-2}&-\frac{1}{3-2}\\-\frac{2}{3-2}&\frac{3}{3-2}\end{matrix}\right)\left(\begin{matrix}130\\130\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1&-1\\-2&3\end{matrix}\right)\left(\begin{matrix}130\\130\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}130-130\\-2\times 130+3\times 130\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\130\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=0,y=130$$

Steps Using Elimination

Consider the first equation. Combine x and 2x to get 3x.

$$3x+50+y=180$$

Subtract 50 from both sides.

$$3x+y=180-50$$

Subtract 50 from 180 to get 130.

$$3x+y=130$$

Consider the second equation. Subtract 50 from both sides.

$$y+2x=180-50$$

Subtract 50 from 180 to get 130.

$$y+2x=130$$

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.

$$3x+y=130,2x+y=130$$

Subtract 2x+y=130 from 3x+y=130 by subtracting like terms on each side of the equal sign.

$$3x-2x+y-y=130-130$$

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

$$3x-2x=130-130$$

Add 3x to -2x.

$$x=130-130$$

Add 130 to -130.

$$x=0$$

Substitute 0 for x in 2x+y=130. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=130$$

The system is now solved.

$$x=0,y=130$$