Solve {l}{y=300-300x}{y=200-100x}

$$\left. \begin{array} { l } { y = 300 - 300 x } \\ { y = 200 - 100 x } \end{array} \right.$$

Solve for y, x

$$x=\frac{1}{2}=0.5$$

$$y=150$$

Steps Using Substitution

Consider the first equation. Add 300x to both sides.

$$y+300x=300$$

Consider the second equation. Add 100x to both sides.

$$y+100x=200$$

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.

$$y+300x=300,y+100x=200$$

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

$$y+300x=300$$

Subtract 300x from both sides of the equation.

$$y=-300x+300$$

Substitute -300x+300 for y in the other equation, y+100x=200.

$$-300x+300+100x=200$$

Add -300x to 100x.

$$-200x+300=200$$

Subtract 300 from both sides of the equation.

$$-200x=-100$$

Divide both sides by -200.

$$x=\frac{1}{2}$$

Substitute \frac{1}{2} for x in y=-300x+300. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=-300\times \frac{1}{2}+300$$

Multiply -300 times \frac{1}{2}.

$$y=-150+300$$

Add 300 to -150.

$$y=150$$

The system is now solved.

$$y=150,x=\frac{1}{2}$$

Steps Using Matrices

Consider the first equation. Add 300x to both sides.

$$y+300x=300$$

Consider the second equation. Add 100x to both sides.

$$y+100x=200$$

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

$$y+300x=300,y+100x=200$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&300\\1&100\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}300\\200\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&300\\1&100\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&300\\1&100\end{matrix}\right))\left(\begin{matrix}1&300\\1&100\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}1&300\\1&100\end{matrix}\right))\left(\begin{matrix}300\\200\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}1&300\\1&100\end{matrix}\right))\left(\begin{matrix}300\\200\end{matrix}\right)$$

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

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}1&300\\1&100\end{matrix}\right))\left(\begin{matrix}300\\200\end{matrix}\right)$$

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.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{100}{100-300}&-\frac{300}{100-300}\\-\frac{1}{100-300}&\frac{1}{100-300}\end{matrix}\right)\left(\begin{matrix}300\\200\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{2}&\frac{3}{2}\\\frac{1}{200}&-\frac{1}{200}\end{matrix}\right)\left(\begin{matrix}300\\200\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{2}\times 300+\frac{3}{2}\times 200\\\frac{1}{200}\times 300-\frac{1}{200}\times 200\end{matrix}\right)$$

Do the arithmetic.

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

Extract the matrix elements y and x.

$$y=150,x=\frac{1}{2}$$

Steps Using Elimination

Consider the first equation. Add 300x to both sides.

$$y+300x=300$$

Consider the second equation. Add 100x to both sides.

$$y+100x=200$$

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.

$$y+300x=300,y+100x=200$$

Subtract y+100x=200 from y+300x=300 by subtracting like terms on each side of the equal sign.

$$y-y+300x-100x=300-200$$

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

$$300x-100x=300-200$$

Add 300x to -100x.

$$200x=300-200$$

Add 300 to -200.

$$200x=100$$

Divide both sides by 200.

$$x=\frac{1}{2}$$

Substitute \frac{1}{2} for x in y+100x=200. Because the resulting equation contains only one variable, you can solve for y directly.

$$y+100\times \frac{1}{2}=200$$

Multiply 100 times \frac{1}{2}.

$$y+50=200$$

Subtract 50 from both sides of the equation.

$$y=150$$

The system is now solved.

$$y=150,x=\frac{1}{2}$$