Solve {l}{x/3=y-2}{x-9/2=y}

$$\left\{ \begin{array} { l } { \frac { x } { 3 } = y - 2 } \\ { \frac { x - 9 } { 2 } = y } \end{array} \right.$$

Solve for x, y

$$x=39$$

$$y=15$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 3.

$$x=3y-6$$

Substitute -6+3y for x in the other equation, \frac{1}{2}x-y=\frac{9}{2}.

$$\frac{1}{2}\left(3y-6\right)-y=\frac{9}{2}$$

Multiply \frac{1}{2} times -6+3y.

$$\frac{3}{2}y-3-y=\frac{9}{2}$$

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

$$\frac{1}{2}y-3=\frac{9}{2}$$

Add 3 to both sides of the equation.

$$\frac{1}{2}y=\frac{15}{2}$$

Multiply both sides by 2.

$$y=15$$

Substitute 15 for y in x=3y-6. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=3\times 15-6$$

Multiply 3 times 15.

$$x=45-6$$

Add -6 to 45.

$$x=39$$

The system is now solved.

$$x=39,y=15$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 3.

$$x=3y-6$$

Subtract 3y from both sides.

$$x-3y=-6$$

Consider the second equation. Divide each term of x-9 by 2 to get \frac{1}{2}x-\frac{9}{2}.

$$\frac{1}{2}x-\frac{9}{2}=y$$

Subtract y from both sides.

$$\frac{1}{2}x-\frac{9}{2}-y=0$$

Add \frac{9}{2} to both sides. Anything plus zero gives itself.

$$\frac{1}{2}x-y=\frac{9}{2}$$

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

$$x-3y=-6,\frac{1}{2}x-y=\frac{9}{2}$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&-3\\\frac{1}{2}&-1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-6\\\frac{9}{2}\end{matrix}\right)$$

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

$$inverse(\left(\begin{matrix}1&-3\\\frac{1}{2}&-1\end{matrix}\right))\left(\begin{matrix}1&-3\\\frac{1}{2}&-1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-3\\\frac{1}{2}&-1\end{matrix}\right))\left(\begin{matrix}-6\\\frac{9}{2}\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-3\\\frac{1}{2}&-1\end{matrix}\right))\left(\begin{matrix}-6\\\frac{9}{2}\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-3\\\frac{1}{2}&-1\end{matrix}\right))\left(\begin{matrix}-6\\\frac{9}{2}\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{-1-\left(-3\times \frac{1}{2}\right)}&-\frac{-3}{-1-\left(-3\times \frac{1}{2}\right)}\\-\frac{\frac{1}{2}}{-1-\left(-3\times \frac{1}{2}\right)}&\frac{1}{-1-\left(-3\times \frac{1}{2}\right)}\end{matrix}\right)\left(\begin{matrix}-6\\\frac{9}{2}\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-2&6\\-1&2\end{matrix}\right)\left(\begin{matrix}-6\\\frac{9}{2}\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-2\left(-6\right)+6\times \frac{9}{2}\\-\left(-6\right)+2\times \frac{9}{2}\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}39\\15\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=39,y=15$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 3.

$$x=3y-6$$

Subtract 3y from both sides.

$$x-3y=-6$$

Consider the second equation. Divide each term of x-9 by 2 to get \frac{1}{2}x-\frac{9}{2}.

$$\frac{1}{2}x-\frac{9}{2}=y$$

Subtract y from both sides.

$$\frac{1}{2}x-\frac{9}{2}-y=0$$

Add \frac{9}{2} to both sides. Anything plus zero gives itself.

$$\frac{1}{2}x-y=\frac{9}{2}$$

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.

$$x-3y=-6,\frac{1}{2}x-y=\frac{9}{2}$$

To make x and \frac{x}{2} equal, multiply all terms on each side of the first equation by \frac{1}{2} and all terms on each side of the second by 1.

$$\frac{1}{2}x+\frac{1}{2}\left(-3\right)y=\frac{1}{2}\left(-6\right),\frac{1}{2}x-y=\frac{9}{2}$$

Simplify.

$$\frac{1}{2}x-\frac{3}{2}y=-3,\frac{1}{2}x-y=\frac{9}{2}$$

Subtract \frac{1}{2}x-y=\frac{9}{2} from \frac{1}{2}x-\frac{3}{2}y=-3 by subtracting like terms on each side of the equal sign.

$$\frac{1}{2}x-\frac{1}{2}x-\frac{3}{2}y+y=-3-\frac{9}{2}$$

Add \frac{x}{2} to -\frac{x}{2}. Terms \frac{x}{2} and -\frac{x}{2} cancel out, leaving an equation with only one variable that can be solved.

$$-\frac{3}{2}y+y=-3-\frac{9}{2}$$

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

$$-\frac{1}{2}y=-3-\frac{9}{2}$$

Add -3 to -\frac{9}{2}.

$$-\frac{1}{2}y=-\frac{15}{2}$$

Multiply both sides by -2.

$$y=15$$

Substitute 15 for y in \frac{1}{2}x-y=\frac{9}{2}. Because the resulting equation contains only one variable, you can solve for x directly.

$$\frac{1}{2}x-15=\frac{9}{2}$$

Add 15 to both sides of the equation.

$$\frac{1}{2}x=\frac{39}{2}$$

Multiply both sides by 2.

$$x=39$$

The system is now solved.

$$x=39,y=15$$