Solve {l}{x+y=150}{10x+20y=20}

$$\left\{ \begin{array} { l } { x + y = 150 } \\ { 10 x + 20 y = 20 } \end{array} \right.$$

Solve for x, y

$$x=298$$

$$y=-148$$

Steps Using Substitution

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.

$$x+y=150,10x+20y=20$$

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

$$x+y=150$$

Subtract y from both sides of the equation.

$$x=-y+150$$

Substitute -y+150 for x in the other equation, 10x+20y=20.

$$10\left(-y+150\right)+20y=20$$

Multiply 10 times -y+150.

$$-10y+1500+20y=20$$

Add -10y to 20y.

$$10y+1500=20$$

Subtract 1500 from both sides of the equation.

$$10y=-1480$$

Divide both sides by 10.

$$y=-148$$

Substitute -148 for y in x=-y+150. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\left(-148\right)+150$$

Multiply -1 times -148.

$$x=148+150$$

Add 150 to 148.

$$x=298$$

The system is now solved.

$$x=298,y=-148$$

Steps Using Matrices

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

$$x+y=150,10x+20y=20$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\10&20\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}150\\20\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\10&20\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\10&20\end{matrix}\right))\left(\begin{matrix}1&1\\10&20\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\10&20\end{matrix}\right))\left(\begin{matrix}150\\20\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\10&20\end{matrix}\right))\left(\begin{matrix}150\\20\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\10&20\end{matrix}\right))\left(\begin{matrix}150\\20\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{20}{20-10}&-\frac{1}{20-10}\\-\frac{10}{20-10}&\frac{1}{20-10}\end{matrix}\right)\left(\begin{matrix}150\\20\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2&-\frac{1}{10}\\-1&\frac{1}{10}\end{matrix}\right)\left(\begin{matrix}150\\20\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2\times 150-\frac{1}{10}\times 20\\-150+\frac{1}{10}\times 20\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}298\\-148\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=298,y=-148$$

Steps Using Elimination

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.

$$x+y=150,10x+20y=20$$

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.

$$10x+10y=10\times 150,10x+20y=20$$

Simplify.

$$10x+10y=1500,10x+20y=20$$

Subtract 10x+20y=20 from 10x+10y=1500 by subtracting like terms on each side of the equal sign.

$$10x-10x+10y-20y=1500-20$$

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

$$10y-20y=1500-20$$

Add 10y to -20y.

$$-10y=1500-20$$

Add 1500 to -20.

$$-10y=1480$$

Divide both sides by -10.

$$y=-148$$

Substitute -148 for y in 10x+20y=20. Because the resulting equation contains only one variable, you can solve for x directly.

$$10x+20\left(-148\right)=20$$

Multiply 20 times -148.

$$10x-2960=20$$

Add 2960 to both sides of the equation.

$$10x=2980$$

Divide both sides by 10.

$$x=298$$

The system is now solved.

$$x=298,y=-148$$