Solve {l}{8x+8y+7y=1}{12x+22y=1}

$$\left\{ \begin{array} { l } { 8 x + 8 y + 7 y = 1 } \\ { 12 x + 22 y = 1 } \end{array} \right.$$

Solve for x, y

$$x = -\frac{7}{4} = -1\frac{3}{4} = -1.75$$

$$y=1$$

Steps Using Substitution

Consider the first equation. Combine 8y and 7y to get 15y.

$$8x+15y=1$$

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.

$$8x+15y=1,12x+22y=1$$

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

$$8x+15y=1$$

Subtract 15y from both sides of the equation.

$$8x=-15y+1$$

Divide both sides by 8.

$$x=\frac{1}{8}\left(-15y+1\right)$$

Multiply \frac{1}{8} times -15y+1.

$$x=-\frac{15}{8}y+\frac{1}{8}$$

Substitute \frac{-15y+1}{8} for x in the other equation, 12x+22y=1.

$$12\left(-\frac{15}{8}y+\frac{1}{8}\right)+22y=1$$

Multiply 12 times \frac{-15y+1}{8}.

$$-\frac{45}{2}y+\frac{3}{2}+22y=1$$

Add -\frac{45y}{2} to 22y.

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

Subtract \frac{3}{2} from both sides of the equation.

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

Multiply both sides by -2.

$$y=1$$

Substitute 1 for y in x=-\frac{15}{8}y+\frac{1}{8}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{-15+1}{8}$$

Add \frac{1}{8} to -\frac{15}{8} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{7}{4}$$

The system is now solved.

$$x=-\frac{7}{4},y=1$$

Steps Using Matrices

Consider the first equation. Combine 8y and 7y to get 15y.

$$8x+15y=1$$

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

$$8x+15y=1,12x+22y=1$$

Write the equations in matrix form.

$$\left(\begin{matrix}8&15\\12&22\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1\\1\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}8&15\\12&22\end{matrix}\right).

$$inverse(\left(\begin{matrix}8&15\\12&22\end{matrix}\right))\left(\begin{matrix}8&15\\12&22\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&15\\12&22\end{matrix}\right))\left(\begin{matrix}1\\1\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&15\\12&22\end{matrix}\right))\left(\begin{matrix}1\\1\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&15\\12&22\end{matrix}\right))\left(\begin{matrix}1\\1\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{22}{8\times 22-15\times 12}&-\frac{15}{8\times 22-15\times 12}\\-\frac{12}{8\times 22-15\times 12}&\frac{8}{8\times 22-15\times 12}\end{matrix}\right)\left(\begin{matrix}1\\1\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{11}{2}&\frac{15}{4}\\3&-2\end{matrix}\right)\left(\begin{matrix}1\\1\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{11}{2}+\frac{15}{4}\\3-2\end{matrix}\right)$$

Do the arithmetic.

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

Extract the matrix elements x and y.

$$x=-\frac{7}{4},y=1$$

Steps Using Elimination

Consider the first equation. Combine 8y and 7y to get 15y.

$$8x+15y=1$$

In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.

$$8x+15y=1,12x+22y=1$$

To make 8x and 12x equal, multiply all terms on each side of the first equation by 12 and all terms on each side of the second by 8.

$$12\times 8x+12\times 15y=12,8\times 12x+8\times 22y=8$$

Simplify.

$$96x+180y=12,96x+176y=8$$

Subtract 96x+176y=8 from 96x+180y=12 by subtracting like terms on each side of the equal sign.

$$96x-96x+180y-176y=12-8$$

Add 96x to -96x. Terms 96x and -96x cancel out, leaving an equation with only one variable that can be solved.

$$180y-176y=12-8$$

Add 180y to -176y.

$$4y=12-8$$

Add 12 to -8.

$$4y=4$$

Divide both sides by 4.

$$y=1$$

Substitute 1 for y in 12x+22y=1. Because the resulting equation contains only one variable, you can solve for x directly.

$$12x+22=1$$

Subtract 22 from both sides of the equation.

$$12x=-21$$

Divide both sides by 12.

$$x=-\frac{7}{4}$$

The system is now solved.

$$x=-\frac{7}{4},y=1$$