Solve {l}{x-1/4+y+2/4=5/12}{6x-y=1}

$$\left\{ \begin{array} { l } { \frac { x - 1 } { 4 } + \frac { y + 2 } { 4 } = \frac { 5 } { 12 } } \\ { 6 x - y = 1 } \end{array} \right.$$

Solve for x, y

$$x=\frac{5}{21}\approx 0.238095238$$

$$y=\frac{3}{7}\approx 0.428571429$$

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.

$$\frac{1}{4}\left(x-1\right)+\frac{1}{4}\left(y+2\right)=\frac{5}{12},6x-y=1$$

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

$$\frac{1}{4}\left(x-1\right)+\frac{1}{4}\left(y+2\right)=\frac{5}{12}$$

Multiply \frac{1}{4} times x-1.

$$\frac{1}{4}x-\frac{1}{4}+\frac{1}{4}\left(y+2\right)=\frac{5}{12}$$

Multiply \frac{1}{4} times y+2.

$$\frac{1}{4}x-\frac{1}{4}+\frac{1}{4}y+\frac{1}{2}=\frac{5}{12}$$

Add -\frac{1}{4} to \frac{1}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Subtract \frac{1}{4} from both sides of the equation.

$$\frac{1}{4}x+\frac{1}{4}y=\frac{1}{6}$$

Subtract \frac{y}{4} from both sides of the equation.

$$\frac{1}{4}x=-\frac{1}{4}y+\frac{1}{6}$$

Multiply both sides by 4.

$$x=4\left(-\frac{1}{4}y+\frac{1}{6}\right)$$

Multiply 4 times -\frac{y}{4}+\frac{1}{6}.

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

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

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

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

$$-6y+4-y=1$$

Add -6y to -y.

$$-7y+4=1$$

Subtract 4 from both sides of the equation.

$$-7y=-3$$

Divide both sides by -7.

$$y=\frac{3}{7}$$

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

$$x=-\frac{3}{7}+\frac{2}{3}$$

Add \frac{2}{3} to -\frac{3}{7} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{5}{21}$$

The system is now solved.

$$x=\frac{5}{21},y=\frac{3}{7}$$

Steps Using Matrices

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

$$\frac{1}{4}\left(x-1\right)+\frac{1}{4}\left(y+2\right)=\frac{5}{12},6x-y=1$$

Simplify the first equation to put it in standard form.

$$\frac{1}{4}\left(x-1\right)+\frac{1}{4}\left(y+2\right)=\frac{5}{12}$$

Multiply \frac{1}{4} times x-1.

$$\frac{1}{4}x-\frac{1}{4}+\frac{1}{4}\left(y+2\right)=\frac{5}{12}$$

Multiply \frac{1}{4} times y+2.

$$\frac{1}{4}x-\frac{1}{4}+\frac{1}{4}y+\frac{1}{2}=\frac{5}{12}$$

Add -\frac{1}{4} to \frac{1}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Subtract \frac{1}{4} from both sides of the equation.

$$\frac{1}{4}x+\frac{1}{4}y=\frac{1}{6}$$

Write the equations in matrix form.

$$\left(\begin{matrix}\frac{1}{4}&\frac{1}{4}\\6&-1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{6}\\1\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}\frac{1}{4}&\frac{1}{4}\\6&-1\end{matrix}\right).

$$inverse(\left(\begin{matrix}\frac{1}{4}&\frac{1}{4}\\6&-1\end{matrix}\right))\left(\begin{matrix}\frac{1}{4}&\frac{1}{4}\\6&-1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{4}&\frac{1}{4}\\6&-1\end{matrix}\right))\left(\begin{matrix}\frac{1}{6}\\1\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{4}&\frac{1}{4}\\6&-1\end{matrix}\right))\left(\begin{matrix}\frac{1}{6}\\1\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{4}&\frac{1}{4}\\6&-1\end{matrix}\right))\left(\begin{matrix}\frac{1}{6}\\1\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{\frac{1}{4}\left(-1\right)-\frac{1}{4}\times 6}&-\frac{\frac{1}{4}}{\frac{1}{4}\left(-1\right)-\frac{1}{4}\times 6}\\-\frac{6}{\frac{1}{4}\left(-1\right)-\frac{1}{4}\times 6}&\frac{\frac{1}{4}}{\frac{1}{4}\left(-1\right)-\frac{1}{4}\times 6}\end{matrix}\right)\left(\begin{matrix}\frac{1}{6}\\1\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{4}{7}&\frac{1}{7}\\\frac{24}{7}&-\frac{1}{7}\end{matrix}\right)\left(\begin{matrix}\frac{1}{6}\\1\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{4}{7}\times \frac{1}{6}+\frac{1}{7}\\\frac{24}{7}\times \frac{1}{6}-\frac{1}{7}\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{21}\\\frac{3}{7}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{5}{21},y=\frac{3}{7}$$