Solve {l}{3x+3y=600}{6x=9y}

$$\left\{ \begin{array} { l } { 3 x + 3 y = 600 } \\ { 6 x = 9 y } \end{array} \right.$$

Solve for x, y

$$x=120$$

$$y=80$$

Steps Using Substitution

Consider the second equation. Subtract 9y from both sides.

$$6x-9y=0$$

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.

$$3x+3y=600,6x-9y=0$$

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

$$3x+3y=600$$

Subtract 3y from both sides of the equation.

$$3x=-3y+600$$

Divide both sides by 3.

$$x=\frac{1}{3}\left(-3y+600\right)$$

Multiply \frac{1}{3} times -3y+600.

$$x=-y+200$$

Substitute -y+200 for x in the other equation, 6x-9y=0.

$$6\left(-y+200\right)-9y=0$$

Multiply 6 times -y+200.

$$-6y+1200-9y=0$$

Add -6y to -9y.

$$-15y+1200=0$$

Subtract 1200 from both sides of the equation.

$$-15y=-1200$$

Divide both sides by -15.

$$y=80$$

Substitute 80 for y in x=-y+200. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-80+200$$

Add 200 to -80.

$$x=120$$

The system is now solved.

$$x=120,y=80$$

Steps Using Matrices

Consider the second equation. Subtract 9y from both sides.

$$6x-9y=0$$

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

$$3x+3y=600,6x-9y=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}3&3\\6&-9\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}600\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}3&3\\6&-9\end{matrix}\right).

$$inverse(\left(\begin{matrix}3&3\\6&-9\end{matrix}\right))\left(\begin{matrix}3&3\\6&-9\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&3\\6&-9\end{matrix}\right))\left(\begin{matrix}600\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&3\\6&-9\end{matrix}\right))\left(\begin{matrix}600\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&3\\6&-9\end{matrix}\right))\left(\begin{matrix}600\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{9}{3\left(-9\right)-3\times 6}&-\frac{3}{3\left(-9\right)-3\times 6}\\-\frac{6}{3\left(-9\right)-3\times 6}&\frac{3}{3\left(-9\right)-3\times 6}\end{matrix}\right)\left(\begin{matrix}600\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{5}&\frac{1}{15}\\\frac{2}{15}&-\frac{1}{15}\end{matrix}\right)\left(\begin{matrix}600\\0\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{5}\times 600\\\frac{2}{15}\times 600\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}120\\80\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=120,y=80$$

Steps Using Elimination

Consider the second equation. Subtract 9y from both sides.

$$6x-9y=0$$

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.

$$3x+3y=600,6x-9y=0$$

To make 3x and 6x equal, multiply all terms on each side of the first equation by 6 and all terms on each side of the second by 3.

$$6\times 3x+6\times 3y=6\times 600,3\times 6x+3\left(-9\right)y=0$$

Simplify.

$$18x+18y=3600,18x-27y=0$$

Subtract 18x-27y=0 from 18x+18y=3600 by subtracting like terms on each side of the equal sign.

$$18x-18x+18y+27y=3600$$

Add 18x to -18x. Terms 18x and -18x cancel out, leaving an equation with only one variable that can be solved.

$$18y+27y=3600$$

Add 18y to 27y.

$$45y=3600$$

Divide both sides by 45.

$$y=80$$

Substitute 80 for y in 6x-9y=0. Because the resulting equation contains only one variable, you can solve for x directly.

$$6x-9\times 80=0$$

Multiply -9 times 80.

$$6x-720=0$$

Add 720 to both sides of the equation.

$$6x=720$$

Divide both sides by 6.

$$x=120$$

The system is now solved.

$$x=120,y=80$$