Solve {l}{8(x-9)+6(x+y)-96=144}{x+y=40}

$$\left\{ \begin{array} { l } { 8 ( x - 9 ) + 6 ( x + y ) - 96 = 144 } \\ { x + y = 40 } \end{array} \right.$$

Solve for x, y

$$x=9$$

$$y=31$$

Steps Using Substitution

Consider the first equation. Use the distributive property to multiply 8 by x-9.

$$8x-72+6\left(x+y\right)-96=144$$

Use the distributive property to multiply 6 by x+y.

$$8x-72+6x+6y-96=144$$

Combine 8x and 6x to get 14x.

$$14x-72+6y-96=144$$

Subtract 96 from -72 to get -168.

$$14x-168+6y=144$$

Add 168 to both sides.

$$14x+6y=144+168$$

Add 144 and 168 to get 312.

$$14x+6y=312$$

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.

$$14x+6y=312,x+y=40$$

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

$$14x+6y=312$$

Subtract 6y from both sides of the equation.

$$14x=-6y+312$$

Divide both sides by 14.

$$x=\frac{1}{14}\left(-6y+312\right)$$

Multiply \frac{1}{14} times -6y+312.

$$x=-\frac{3}{7}y+\frac{156}{7}$$

Substitute \frac{-3y+156}{7} for x in the other equation, x+y=40.

$$-\frac{3}{7}y+\frac{156}{7}+y=40$$

Add -\frac{3y}{7} to y.

$$\frac{4}{7}y+\frac{156}{7}=40$$

Subtract \frac{156}{7} from both sides of the equation.

$$\frac{4}{7}y=\frac{124}{7}$$

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

$$y=31$$

Substitute 31 for y in x=-\frac{3}{7}y+\frac{156}{7}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{3}{7}\times 31+\frac{156}{7}$$

Multiply -\frac{3}{7} times 31.

$$x=\frac{-93+156}{7}$$

Add \frac{156}{7} to -\frac{93}{7} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=9$$

The system is now solved.

$$x=9,y=31$$

Steps Using Matrices

Consider the first equation. Use the distributive property to multiply 8 by x-9.

$$8x-72+6\left(x+y\right)-96=144$$

Use the distributive property to multiply 6 by x+y.

$$8x-72+6x+6y-96=144$$

Combine 8x and 6x to get 14x.

$$14x-72+6y-96=144$$

Subtract 96 from -72 to get -168.

$$14x-168+6y=144$$

Add 168 to both sides.

$$14x+6y=144+168$$

Add 144 and 168 to get 312.

$$14x+6y=312$$

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

$$14x+6y=312,x+y=40$$

Write the equations in matrix form.

$$\left(\begin{matrix}14&6\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}312\\40\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}14&6\\1&1\end{matrix}\right).

$$inverse(\left(\begin{matrix}14&6\\1&1\end{matrix}\right))\left(\begin{matrix}14&6\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}14&6\\1&1\end{matrix}\right))\left(\begin{matrix}312\\40\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}14&6\\1&1\end{matrix}\right))\left(\begin{matrix}312\\40\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}14&6\\1&1\end{matrix}\right))\left(\begin{matrix}312\\40\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{14-6}&-\frac{6}{14-6}\\-\frac{1}{14-6}&\frac{14}{14-6}\end{matrix}\right)\left(\begin{matrix}312\\40\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{8}&-\frac{3}{4}\\-\frac{1}{8}&\frac{7}{4}\end{matrix}\right)\left(\begin{matrix}312\\40\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{8}\times 312-\frac{3}{4}\times 40\\-\frac{1}{8}\times 312+\frac{7}{4}\times 40\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}9\\31\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=9,y=31$$

Steps Using Elimination

Consider the first equation. Use the distributive property to multiply 8 by x-9.

$$8x-72+6\left(x+y\right)-96=144$$

Use the distributive property to multiply 6 by x+y.

$$8x-72+6x+6y-96=144$$

Combine 8x and 6x to get 14x.

$$14x-72+6y-96=144$$

Subtract 96 from -72 to get -168.

$$14x-168+6y=144$$

Add 168 to both sides.

$$14x+6y=144+168$$

Add 144 and 168 to get 312.

$$14x+6y=312$$

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.

$$14x+6y=312,x+y=40$$

To make 14x and x equal, multiply all terms on each side of the first equation by 1 and all terms on each side of the second by 14.

$$14x+6y=312,14x+14y=14\times 40$$

Simplify.

$$14x+6y=312,14x+14y=560$$

Subtract 14x+14y=560 from 14x+6y=312 by subtracting like terms on each side of the equal sign.

$$14x-14x+6y-14y=312-560$$

Add 14x to -14x. Terms 14x and -14x cancel out, leaving an equation with only one variable that can be solved.

$$6y-14y=312-560$$

Add 6y to -14y.

$$-8y=312-560$$

Add 312 to -560.

$$-8y=-248$$

Divide both sides by -8.

$$y=31$$

Substitute 31 for y in x+y=40. Because the resulting equation contains only one variable, you can solve for x directly.

$$x+31=40$$

Subtract 31 from both sides of the equation.

$$x=9$$

The system is now solved.

$$x=9,y=31$$