Solve {l}{x+1/3+y/2=1}{x/2-1/4y=2}

$$\left\{ \begin{array} { l } { \frac { x + 1 } { 3 } + \frac { y } { 2 } = 1 } \\ { \frac { x } { 2 } - \frac { 1 } { 4 } y = 2 } \end{array} \right.$$

Solve for x, y

$$x = \frac{7}{2} = 3\frac{1}{2} = 3.5$$

$$y=-1$$

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.

$$\frac{1}{3}\left(x+1\right)+\frac{1}{2}y=1,\frac{1}{2}x-\frac{1}{4}y=2$$

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

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

Multiply \frac{1}{3} times x+1.

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

Subtract \frac{1}{3} from both sides of the equation.

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

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

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

Multiply both sides by 3.

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

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

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

Substitute -\frac{3y}{2}+2 for x in the other equation, \frac{1}{2}x-\frac{1}{4}y=2.

$$\frac{1}{2}\left(-\frac{3}{2}y+2\right)-\frac{1}{4}y=2$$

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

$$-\frac{3}{4}y+1-\frac{1}{4}y=2$$

Add -\frac{3y}{4} to -\frac{y}{4}.

$$-y+1=2$$

Subtract 1 from both sides of the equation.

$$-y=1$$

Divide both sides by -1.

$$y=-1$$

Substitute -1 for y in x=-\frac{3}{2}y+2. Because the resulting equation contains only one variable, you can solve for x directly.

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

Multiply -\frac{3}{2} times -1.

$$x=\frac{3}{2}+2$$

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

$$x=\frac{7}{2}$$

The system is now solved.

$$x=\frac{7}{2},y=-1$$

Steps Using Matrices

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

$$\frac{1}{3}\left(x+1\right)+\frac{1}{2}y=1,\frac{1}{2}x-\frac{1}{4}y=2$$

Simplify the first equation to put it in standard form.

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

Multiply \frac{1}{3} times x+1.

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

Subtract \frac{1}{3} from both sides of the equation.

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

Write the equations in matrix form.

$$\left(\begin{matrix}\frac{1}{3}&\frac{1}{2}\\\frac{1}{2}&-\frac{1}{4}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{3}\\2\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}\frac{1}{3}&\frac{1}{2}\\\frac{1}{2}&-\frac{1}{4}\end{matrix}\right).

$$inverse(\left(\begin{matrix}\frac{1}{3}&\frac{1}{2}\\\frac{1}{2}&-\frac{1}{4}\end{matrix}\right))\left(\begin{matrix}\frac{1}{3}&\frac{1}{2}\\\frac{1}{2}&-\frac{1}{4}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{3}&\frac{1}{2}\\\frac{1}{2}&-\frac{1}{4}\end{matrix}\right))\left(\begin{matrix}\frac{2}{3}\\2\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{3}&\frac{1}{2}\\\frac{1}{2}&-\frac{1}{4}\end{matrix}\right))\left(\begin{matrix}\frac{2}{3}\\2\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{3}&\frac{1}{2}\\\frac{1}{2}&-\frac{1}{4}\end{matrix}\right))\left(\begin{matrix}\frac{2}{3}\\2\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{\frac{1}{4}}{\frac{1}{3}\left(-\frac{1}{4}\right)-\frac{1}{2}\times \frac{1}{2}}&-\frac{\frac{1}{2}}{\frac{1}{3}\left(-\frac{1}{4}\right)-\frac{1}{2}\times \frac{1}{2}}\\-\frac{\frac{1}{2}}{\frac{1}{3}\left(-\frac{1}{4}\right)-\frac{1}{2}\times \frac{1}{2}}&\frac{\frac{1}{3}}{\frac{1}{3}\left(-\frac{1}{4}\right)-\frac{1}{2}\times \frac{1}{2}}\end{matrix}\right)\left(\begin{matrix}\frac{2}{3}\\2\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{4}&\frac{3}{2}\\\frac{3}{2}&-1\end{matrix}\right)\left(\begin{matrix}\frac{2}{3}\\2\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{4}\times \frac{2}{3}+\frac{3}{2}\times 2\\\frac{3}{2}\times \frac{2}{3}-2\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{7}{2}\\-1\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{7}{2},y=-1$$