Solve {l}{x+3y=8}{x-1/2-y=1}

$$\left. \begin{array} { l } { x + 3 y = 8 } \\ { \frac { x - 1 } { 2 } - y = 1 } \end{array} \right.$$

Solve for x, y

$$x=5$$

$$y=1$$

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.

$$x+3y=8,\frac{1}{2}\left(x-1\right)-y=1$$

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

$$x+3y=8$$

Subtract 3y from both sides of the equation.

$$x=-3y+8$$

Substitute -3y+8 for x in the other equation, \frac{1}{2}\left(x-1\right)-y=1.

$$\frac{1}{2}\left(-3y+8-1\right)-y=1$$

Add 8 to -1.

$$\frac{1}{2}\left(-3y+7\right)-y=1$$

Multiply \frac{1}{2} times -3y+7.

$$-\frac{3}{2}y+\frac{7}{2}-y=1$$

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

$$-\frac{5}{2}y+\frac{7}{2}=1$$

Subtract \frac{7}{2} from both sides of the equation.

$$-\frac{5}{2}y=-\frac{5}{2}$$

Divide both sides of the equation by -\frac{5}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=1$$

Substitute 1 for y in x=-3y+8. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-3+8$$

Add 8 to -3.

$$x=5$$

The system is now solved.

$$x=5,y=1$$

Steps Using Matrices

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

$$x+3y=8,\frac{1}{2}\left(x-1\right)-y=1$$

Simplify the second equation to put it in standard form.

$$\frac{1}{2}\left(x-1\right)-y=1$$

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

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

Add \frac{1}{2} to both sides of the equation.

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

Write the equations in matrix form.

$$\left(\begin{matrix}1&3\\\frac{1}{2}&-1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}8\\\frac{3}{2}\end{matrix}\right)$$

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

$$inverse(\left(\begin{matrix}1&3\\\frac{1}{2}&-1\end{matrix}\right))\left(\begin{matrix}1&3\\\frac{1}{2}&-1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&3\\\frac{1}{2}&-1\end{matrix}\right))\left(\begin{matrix}8\\\frac{3}{2}\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&3\\\frac{1}{2}&-1\end{matrix}\right))\left(\begin{matrix}8\\\frac{3}{2}\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&3\\\frac{1}{2}&-1\end{matrix}\right))\left(\begin{matrix}8\\\frac{3}{2}\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{-1-3\times \frac{1}{2}}&-\frac{3}{-1-3\times \frac{1}{2}}\\-\frac{\frac{1}{2}}{-1-3\times \frac{1}{2}}&\frac{1}{-1-3\times \frac{1}{2}}\end{matrix}\right)\left(\begin{matrix}8\\\frac{3}{2}\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{5}&\frac{6}{5}\\\frac{1}{5}&-\frac{2}{5}\end{matrix}\right)\left(\begin{matrix}8\\\frac{3}{2}\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{5}\times 8+\frac{6}{5}\times \frac{3}{2}\\\frac{1}{5}\times 8-\frac{2}{5}\times \frac{3}{2}\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}5\\1\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=5,y=1$$