Solve {l}{7x+12y=120}{93x+88y=1080}

$$\left. \begin{array} { l } { 7 x + 12 y = 120 } \\ { 93 x + 88 y = 1080 } \end{array} \right.$$

Solve for x, y

$$x = \frac{24}{5} = 4\frac{4}{5} = 4.8$$

$$y = \frac{36}{5} = 7\frac{1}{5} = 7.2$$

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.

$$7x+12y=120,93x+88y=1080$$

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

$$7x+12y=120$$

Subtract 12y from both sides of the equation.

$$7x=-12y+120$$

Divide both sides by 7.

$$x=\frac{1}{7}\left(-12y+120\right)$$

Multiply \frac{1}{7} times -12y+120.

$$x=-\frac{12}{7}y+\frac{120}{7}$$

Substitute \frac{-12y+120}{7} for x in the other equation, 93x+88y=1080.

$$93\left(-\frac{12}{7}y+\frac{120}{7}\right)+88y=1080$$

Multiply 93 times \frac{-12y+120}{7}.

$$-\frac{1116}{7}y+\frac{11160}{7}+88y=1080$$

Add -\frac{1116y}{7} to 88y.

$$-\frac{500}{7}y+\frac{11160}{7}=1080$$

Subtract \frac{11160}{7} from both sides of the equation.

$$-\frac{500}{7}y=-\frac{3600}{7}$$

Divide both sides of the equation by -\frac{500}{7}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=\frac{36}{5}$$

Substitute \frac{36}{5} for y in x=-\frac{12}{7}y+\frac{120}{7}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{12}{7}\times \frac{36}{5}+\frac{120}{7}$$

Multiply -\frac{12}{7} times \frac{36}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{432}{35}+\frac{120}{7}$$

Add \frac{120}{7} to -\frac{432}{35} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{24}{5}$$

The system is now solved.

$$x=\frac{24}{5},y=\frac{36}{5}$$

Steps Using Matrices

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

$$7x+12y=120,93x+88y=1080$$

Write the equations in matrix form.

$$\left(\begin{matrix}7&12\\93&88\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}120\\1080\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}7&12\\93&88\end{matrix}\right).

$$inverse(\left(\begin{matrix}7&12\\93&88\end{matrix}\right))\left(\begin{matrix}7&12\\93&88\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}7&12\\93&88\end{matrix}\right))\left(\begin{matrix}120\\1080\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}7&12\\93&88\end{matrix}\right))\left(\begin{matrix}120\\1080\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}7&12\\93&88\end{matrix}\right))\left(\begin{matrix}120\\1080\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{88}{7\times 88-12\times 93}&-\frac{12}{7\times 88-12\times 93}\\-\frac{93}{7\times 88-12\times 93}&\frac{7}{7\times 88-12\times 93}\end{matrix}\right)\left(\begin{matrix}120\\1080\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{22}{125}&\frac{3}{125}\\\frac{93}{500}&-\frac{7}{500}\end{matrix}\right)\left(\begin{matrix}120\\1080\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{22}{125}\times 120+\frac{3}{125}\times 1080\\\frac{93}{500}\times 120-\frac{7}{500}\times 1080\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{24}{5}\\\frac{36}{5}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{24}{5},y=\frac{36}{5}$$

Steps Using Elimination

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.

$$7x+12y=120,93x+88y=1080$$

To make 7x and 93x equal, multiply all terms on each side of the first equation by 93 and all terms on each side of the second by 7.

$$93\times 7x+93\times 12y=93\times 120,7\times 93x+7\times 88y=7\times 1080$$

Simplify.

$$651x+1116y=11160,651x+616y=7560$$

Subtract 651x+616y=7560 from 651x+1116y=11160 by subtracting like terms on each side of the equal sign.

$$651x-651x+1116y-616y=11160-7560$$

Add 651x to -651x. Terms 651x and -651x cancel out, leaving an equation with only one variable that can be solved.

$$1116y-616y=11160-7560$$

Add 1116y to -616y.

$$500y=11160-7560$$

Add 11160 to -7560.

$$500y=3600$$

Divide both sides by 500.

$$y=\frac{36}{5}$$

Substitute \frac{36}{5} for y in 93x+88y=1080. Because the resulting equation contains only one variable, you can solve for x directly.

$$93x+88\times \frac{36}{5}=1080$$

Multiply 88 times \frac{36}{5}.

$$93x+\frac{3168}{5}=1080$$

Subtract \frac{3168}{5} from both sides of the equation.

$$93x=\frac{2232}{5}$$

Divide both sides by 93.

$$x=\frac{24}{5}$$

The system is now solved.

$$x=\frac{24}{5},y=\frac{36}{5}$$