Solve {c}{x+3/4+y-1/3=1}{2x-y=12}

$$\left. \begin{array} { c } { \frac { x + 3 } { 4 } + \frac { y - 1 } { 3 } = 1 } \\ { 2 x - y = 12 } \end{array} \right.$$

Solve for x, y

$$x=5$$

$$y=-2$$

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.

$$\frac{1}{4}\left(x+3\right)+\frac{1}{3}\left(y-1\right)=1,2x-y=12$$

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

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

Multiply \frac{1}{4} times x+3.

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

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

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

Add \frac{3}{4} to -\frac{1}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$\frac{1}{4}x+\frac{1}{3}y+\frac{5}{12}=1$$

Subtract \frac{5}{12} from both sides of the equation.

$$\frac{1}{4}x+\frac{1}{3}y=\frac{7}{12}$$

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

$$\frac{1}{4}x=-\frac{1}{3}y+\frac{7}{12}$$

Multiply both sides by 4.

$$x=4\left(-\frac{1}{3}y+\frac{7}{12}\right)$$

Multiply 4 times -\frac{y}{3}+\frac{7}{12}.

$$x=-\frac{4}{3}y+\frac{7}{3}$$

Substitute \frac{-4y+7}{3} for x in the other equation, 2x-y=12.

$$2\left(-\frac{4}{3}y+\frac{7}{3}\right)-y=12$$

Multiply 2 times \frac{-4y+7}{3}.

$$-\frac{8}{3}y+\frac{14}{3}-y=12$$

Add -\frac{8y}{3} to -y.

$$-\frac{11}{3}y+\frac{14}{3}=12$$

Subtract \frac{14}{3} from both sides of the equation.

$$-\frac{11}{3}y=\frac{22}{3}$$

Divide both sides of the equation by -\frac{11}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=-2$$

Substitute -2 for y in x=-\frac{4}{3}y+\frac{7}{3}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{4}{3}\left(-2\right)+\frac{7}{3}$$

Multiply -\frac{4}{3} times -2.

$$x=\frac{8+7}{3}$$

Add \frac{7}{3} to \frac{8}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=5$$

The system is now solved.

$$x=5,y=-2$$

Steps Using Matrices

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

$$\frac{1}{4}\left(x+3\right)+\frac{1}{3}\left(y-1\right)=1,2x-y=12$$

Simplify the first equation to put it in standard form.

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

Multiply \frac{1}{4} times x+3.

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

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

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

Add \frac{3}{4} to -\frac{1}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$\frac{1}{4}x+\frac{1}{3}y+\frac{5}{12}=1$$

Subtract \frac{5}{12} from both sides of the equation.

$$\frac{1}{4}x+\frac{1}{3}y=\frac{7}{12}$$

Write the equations in matrix form.

$$\left(\begin{matrix}\frac{1}{4}&\frac{1}{3}\\2&-1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{7}{12}\\12\end{matrix}\right)$$

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

$$inverse(\left(\begin{matrix}\frac{1}{4}&\frac{1}{3}\\2&-1\end{matrix}\right))\left(\begin{matrix}\frac{1}{4}&\frac{1}{3}\\2&-1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{4}&\frac{1}{3}\\2&-1\end{matrix}\right))\left(\begin{matrix}\frac{7}{12}\\12\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{4}&\frac{1}{3}\\2&-1\end{matrix}\right))\left(\begin{matrix}\frac{7}{12}\\12\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{4}&\frac{1}{3}\\2&-1\end{matrix}\right))\left(\begin{matrix}\frac{7}{12}\\12\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{\frac{1}{4}\left(-1\right)-\frac{1}{3}\times 2}&-\frac{\frac{1}{3}}{\frac{1}{4}\left(-1\right)-\frac{1}{3}\times 2}\\-\frac{2}{\frac{1}{4}\left(-1\right)-\frac{1}{3}\times 2}&\frac{\frac{1}{4}}{\frac{1}{4}\left(-1\right)-\frac{1}{3}\times 2}\end{matrix}\right)\left(\begin{matrix}\frac{7}{12}\\12\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{12}{11}&\frac{4}{11}\\\frac{24}{11}&-\frac{3}{11}\end{matrix}\right)\left(\begin{matrix}\frac{7}{12}\\12\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{12}{11}\times \frac{7}{12}+\frac{4}{11}\times 12\\\frac{24}{11}\times \frac{7}{12}-\frac{3}{11}\times 12\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}5\\-2\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=5,y=-2$$