Solve {l}{(x+2)/5-y=-8}{(y+1)/2+(x-1)/4=2}

$$\left. \begin{array} { l } { ( x + 2 ) / 5 - y = - 8 } \\ { ( y + 1 ) / 2 + ( x - 1 ) / 4 = 2 } \end{array} \right.$$

Solve for x, y

$$x=-7$$

$$y=7$$

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.

$$\frac{1}{5}\left(x+2\right)-y=-8,\frac{1}{4}\left(x-1\right)+\frac{1}{2}\left(y+1\right)=2$$

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

$$\frac{1}{5}\left(x+2\right)-y=-8$$

Multiply \frac{1}{5} times x+2.

$$\frac{1}{5}x+\frac{2}{5}-y=-8$$

Subtract \frac{2}{5} from both sides of the equation.

$$\frac{1}{5}x-y=-\frac{42}{5}$$

Add y to both sides of the equation.

$$\frac{1}{5}x=y-\frac{42}{5}$$

Multiply both sides by 5.

$$x=5\left(y-\frac{42}{5}\right)$$

Multiply 5 times y-\frac{42}{5}.

$$x=5y-42$$

Substitute 5y-42 for x in the other equation, \frac{1}{4}\left(x-1\right)+\frac{1}{2}\left(y+1\right)=2.

$$\frac{1}{4}\left(5y-42-1\right)+\frac{1}{2}\left(y+1\right)=2$$

Add -42 to -1.

$$\frac{1}{4}\left(5y-43\right)+\frac{1}{2}\left(y+1\right)=2$$

Multiply \frac{1}{4} times 5y-43.

$$\frac{5}{4}y-\frac{43}{4}+\frac{1}{2}\left(y+1\right)=2$$

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

$$\frac{5}{4}y-\frac{43}{4}+\frac{1}{2}y+\frac{1}{2}=2$$

Add \frac{5y}{4} to \frac{y}{2}.

$$\frac{7}{4}y-\frac{43}{4}+\frac{1}{2}=2$$

Add -\frac{43}{4} to \frac{1}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$\frac{7}{4}y-\frac{41}{4}=2$$

Add \frac{41}{4} to both sides of the equation.

$$\frac{7}{4}y=\frac{49}{4}$$

Divide both sides of the equation by \frac{7}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=7$$

Substitute 7 for y in x=5y-42. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=5\times 7-42$$

Multiply 5 times 7.

$$x=35-42$$

Add -42 to 35.

$$x=-7$$

The system is now solved.

$$x=-7,y=7$$

Steps Using Matrices

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

$$\frac{1}{5}\left(x+2\right)-y=-8,\frac{1}{4}\left(x-1\right)+\frac{1}{2}\left(y+1\right)=2$$

Simplify the first equation to put it in standard form.

$$\frac{1}{5}\left(x+2\right)-y=-8$$

Multiply \frac{1}{5} times x+2.

$$\frac{1}{5}x+\frac{2}{5}-y=-8$$

Subtract \frac{2}{5} from both sides of the equation.

$$\frac{1}{5}x-y=-\frac{42}{5}$$

Simplify the second equation to put it in standard form.

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

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

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

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

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

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.

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

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

$$\frac{1}{4}x+\frac{1}{2}y=\frac{7}{4}$$

Write the equations in matrix form.

$$\left(\begin{matrix}\frac{1}{5}&-1\\\frac{1}{4}&\frac{1}{2}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{42}{5}\\\frac{7}{4}\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}\frac{1}{5}&-1\\\frac{1}{4}&\frac{1}{2}\end{matrix}\right).

$$inverse(\left(\begin{matrix}\frac{1}{5}&-1\\\frac{1}{4}&\frac{1}{2}\end{matrix}\right))\left(\begin{matrix}\frac{1}{5}&-1\\\frac{1}{4}&\frac{1}{2}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{5}&-1\\\frac{1}{4}&\frac{1}{2}\end{matrix}\right))\left(\begin{matrix}-\frac{42}{5}\\\frac{7}{4}\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{5}&-1\\\frac{1}{4}&\frac{1}{2}\end{matrix}\right))\left(\begin{matrix}-\frac{42}{5}\\\frac{7}{4}\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{5}&-1\\\frac{1}{4}&\frac{1}{2}\end{matrix}\right))\left(\begin{matrix}-\frac{42}{5}\\\frac{7}{4}\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{\frac{1}{2}}{\frac{1}{5}\times \frac{1}{2}-\left(-\frac{1}{4}\right)}&-\frac{-1}{\frac{1}{5}\times \frac{1}{2}-\left(-\frac{1}{4}\right)}\\-\frac{\frac{1}{4}}{\frac{1}{5}\times \frac{1}{2}-\left(-\frac{1}{4}\right)}&\frac{\frac{1}{5}}{\frac{1}{5}\times \frac{1}{2}-\left(-\frac{1}{4}\right)}\end{matrix}\right)\left(\begin{matrix}-\frac{42}{5}\\\frac{7}{4}\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{10}{7}&\frac{20}{7}\\-\frac{5}{7}&\frac{4}{7}\end{matrix}\right)\left(\begin{matrix}-\frac{42}{5}\\\frac{7}{4}\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{10}{7}\left(-\frac{42}{5}\right)+\frac{20}{7}\times \frac{7}{4}\\-\frac{5}{7}\left(-\frac{42}{5}\right)+\frac{4}{7}\times \frac{7}{4}\end{matrix}\right)$$

Do the arithmetic.

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

Extract the matrix elements x and y.

$$x=-7,y=7$$