Solve {l}{x/30+2=y/30-2+1/2}{x/30-2=frac{y}{30+2}+(1.5-0.5)}

$$\left. \begin{array} { l } { \frac { x } { 30 + 2 } = \frac { y } { 30 - 2 } + \frac { 1 } { 2 } } \\ { \frac { x } { 30 - 2 } = \frac { y } { 30 + 2 } + ( 1.5 - 0.5 ) } \end{array} \right.$$

Solve for x, y

$$x=67.2$$

$$y=44.8$$

Steps Using Substitution

Consider the first equation. Add 30 and 2 to get 32.

$$\frac{x}{32}=\frac{y}{30-2}+\frac{1}{2}$$

Subtract 2 from 30 to get 28.

$$\frac{x}{32}=\frac{y}{28}+\frac{1}{2}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 28 and 2 is 28. Multiply \frac{1}{2} times \frac{14}{14}.

$$\frac{x}{32}=\frac{y}{28}+\frac{14}{28}$$

Since \frac{y}{28} and \frac{14}{28} have the same denominator, add them by adding their numerators.

$$\frac{x}{32}=\frac{y+14}{28}$$

Divide each term of y+14 by 28 to get \frac{1}{28}y+\frac{1}{2}.

$$\frac{x}{32}=\frac{1}{28}y+\frac{1}{2}$$

Subtract \frac{1}{28}y from both sides.

$$\frac{x}{32}-\frac{1}{28}y=\frac{1}{2}$$

Multiply both sides of the equation by 224, the least common multiple of 32,28,2.

$$7x-8y=112$$

Consider the second equation. Subtract 2 from 30 to get 28.

$$\frac{x}{28}=\frac{y}{30+2}+1.5-0.5$$

Add 30 and 2 to get 32.

$$\frac{x}{28}=\frac{y}{32}+1.5-0.5$$

Subtract 0.5 from 1.5 to get 1.

$$\frac{x}{28}=\frac{y}{32}+1$$

Subtract \frac{y}{32} from both sides.

$$\frac{x}{28}-\frac{y}{32}=1$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 28 and 32 is 224. Multiply \frac{x}{28} times \frac{8}{8}. Multiply \frac{y}{32} times \frac{7}{7}.

$$\frac{8x}{224}-\frac{7y}{224}=1$$

Since \frac{8x}{224} and \frac{7y}{224} have the same denominator, subtract them by subtracting their numerators.

$$\frac{8x-7y}{224}=1$$

Divide each term of 8x-7y by 224 to get \frac{1}{28}x-\frac{1}{32}y.

$$\frac{1}{28}x-\frac{1}{32}y=1$$

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.

$$7x-8y=112,\frac{1}{28}x-\frac{1}{32}y=1$$

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

$$7x-8y=112$$

Add 8y to both sides of the equation.

$$7x=8y+112$$

Divide both sides by 7.

$$x=\frac{1}{7}\left(8y+112\right)$$

Multiply \frac{1}{7} times 112+8y.

$$x=\frac{8}{7}y+16$$

Substitute 16+\frac{8y}{7} for x in the other equation, \frac{1}{28}x-\frac{1}{32}y=1.

$$\frac{1}{28}\left(\frac{8}{7}y+16\right)-\frac{1}{32}y=1$$

Multiply \frac{1}{28} times 16+\frac{8y}{7}.

$$\frac{2}{49}y+\frac{4}{7}-\frac{1}{32}y=1$$

Add \frac{2y}{49} to -\frac{y}{32}.

$$\frac{15}{1568}y+\frac{4}{7}=1$$

Subtract \frac{4}{7} from both sides of the equation.

$$\frac{15}{1568}y=\frac{3}{7}$$

Divide both sides of the equation by \frac{15}{1568}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=\frac{224}{5}$$

Substitute \frac{224}{5} for y in x=\frac{8}{7}y+16. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{8}{7}\times \frac{224}{5}+16$$

Multiply \frac{8}{7} times \frac{224}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{256}{5}+16$$

Add 16 to \frac{256}{5}.

$$x=\frac{336}{5}$$

The system is now solved.

$$x=\frac{336}{5},y=\frac{224}{5}$$

Steps Using Matrices

Consider the first equation. Add 30 and 2 to get 32.

$$\frac{x}{32}=\frac{y}{30-2}+\frac{1}{2}$$

Subtract 2 from 30 to get 28.

$$\frac{x}{32}=\frac{y}{28}+\frac{1}{2}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 28 and 2 is 28. Multiply \frac{1}{2} times \frac{14}{14}.

$$\frac{x}{32}=\frac{y}{28}+\frac{14}{28}$$

Since \frac{y}{28} and \frac{14}{28} have the same denominator, add them by adding their numerators.

$$\frac{x}{32}=\frac{y+14}{28}$$

Divide each term of y+14 by 28 to get \frac{1}{28}y+\frac{1}{2}.

$$\frac{x}{32}=\frac{1}{28}y+\frac{1}{2}$$

Subtract \frac{1}{28}y from both sides.

$$\frac{x}{32}-\frac{1}{28}y=\frac{1}{2}$$

Multiply both sides of the equation by 224, the least common multiple of 32,28,2.

$$7x-8y=112$$

Consider the second equation. Subtract 2 from 30 to get 28.

$$\frac{x}{28}=\frac{y}{30+2}+1.5-0.5$$

Add 30 and 2 to get 32.

$$\frac{x}{28}=\frac{y}{32}+1.5-0.5$$

Subtract 0.5 from 1.5 to get 1.

$$\frac{x}{28}=\frac{y}{32}+1$$

Subtract \frac{y}{32} from both sides.

$$\frac{x}{28}-\frac{y}{32}=1$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 28 and 32 is 224. Multiply \frac{x}{28} times \frac{8}{8}. Multiply \frac{y}{32} times \frac{7}{7}.

$$\frac{8x}{224}-\frac{7y}{224}=1$$

Since \frac{8x}{224} and \frac{7y}{224} have the same denominator, subtract them by subtracting their numerators.

$$\frac{8x-7y}{224}=1$$

Divide each term of 8x-7y by 224 to get \frac{1}{28}x-\frac{1}{32}y.

$$\frac{1}{28}x-\frac{1}{32}y=1$$

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

$$7x-8y=112,\frac{1}{28}x-\frac{1}{32}y=1$$

Write the equations in matrix form.

$$\left(\begin{matrix}7&-8\\\frac{1}{28}&-\frac{1}{32}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}112\\1\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}7&-8\\\frac{1}{28}&-\frac{1}{32}\end{matrix}\right).

$$inverse(\left(\begin{matrix}7&-8\\\frac{1}{28}&-\frac{1}{32}\end{matrix}\right))\left(\begin{matrix}7&-8\\\frac{1}{28}&-\frac{1}{32}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}7&-8\\\frac{1}{28}&-\frac{1}{32}\end{matrix}\right))\left(\begin{matrix}112\\1\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}7&-8\\\frac{1}{28}&-\frac{1}{32}\end{matrix}\right))\left(\begin{matrix}112\\1\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}7&-8\\\frac{1}{28}&-\frac{1}{32}\end{matrix}\right))\left(\begin{matrix}112\\1\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{\frac{1}{32}}{7\left(-\frac{1}{32}\right)-\left(-8\times \frac{1}{28}\right)}&-\frac{-8}{7\left(-\frac{1}{32}\right)-\left(-8\times \frac{1}{28}\right)}\\-\frac{\frac{1}{28}}{7\left(-\frac{1}{32}\right)-\left(-8\times \frac{1}{28}\right)}&\frac{7}{7\left(-\frac{1}{32}\right)-\left(-8\times \frac{1}{28}\right)}\end{matrix}\right)\left(\begin{matrix}112\\1\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{7}{15}&\frac{1792}{15}\\-\frac{8}{15}&\frac{1568}{15}\end{matrix}\right)\left(\begin{matrix}112\\1\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{7}{15}\times 112+\frac{1792}{15}\\-\frac{8}{15}\times 112+\frac{1568}{15}\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{336}{5}\\\frac{224}{5}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{336}{5},y=\frac{224}{5}$$

Steps Using Elimination

Consider the first equation. Add 30 and 2 to get 32.

$$\frac{x}{32}=\frac{y}{30-2}+\frac{1}{2}$$

Subtract 2 from 30 to get 28.

$$\frac{x}{32}=\frac{y}{28}+\frac{1}{2}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 28 and 2 is 28. Multiply \frac{1}{2} times \frac{14}{14}.

$$\frac{x}{32}=\frac{y}{28}+\frac{14}{28}$$

Since \frac{y}{28} and \frac{14}{28} have the same denominator, add them by adding their numerators.

$$\frac{x}{32}=\frac{y+14}{28}$$

Divide each term of y+14 by 28 to get \frac{1}{28}y+\frac{1}{2}.

$$\frac{x}{32}=\frac{1}{28}y+\frac{1}{2}$$

Subtract \frac{1}{28}y from both sides.

$$\frac{x}{32}-\frac{1}{28}y=\frac{1}{2}$$

Multiply both sides of the equation by 224, the least common multiple of 32,28,2.

$$7x-8y=112$$

Consider the second equation. Subtract 2 from 30 to get 28.

$$\frac{x}{28}=\frac{y}{30+2}+1.5-0.5$$

Add 30 and 2 to get 32.

$$\frac{x}{28}=\frac{y}{32}+1.5-0.5$$

Subtract 0.5 from 1.5 to get 1.

$$\frac{x}{28}=\frac{y}{32}+1$$

Subtract \frac{y}{32} from both sides.

$$\frac{x}{28}-\frac{y}{32}=1$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 28 and 32 is 224. Multiply \frac{x}{28} times \frac{8}{8}. Multiply \frac{y}{32} times \frac{7}{7}.

$$\frac{8x}{224}-\frac{7y}{224}=1$$

Since \frac{8x}{224} and \frac{7y}{224} have the same denominator, subtract them by subtracting their numerators.

$$\frac{8x-7y}{224}=1$$

Divide each term of 8x-7y by 224 to get \frac{1}{28}x-\frac{1}{32}y.

$$\frac{1}{28}x-\frac{1}{32}y=1$$

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.

$$7x-8y=112,\frac{1}{28}x-\frac{1}{32}y=1$$

To make 7x and \frac{x}{28} equal, multiply all terms on each side of the first equation by \frac{1}{28} and all terms on each side of the second by 7.

$$\frac{1}{28}\times 7x+\frac{1}{28}\left(-8\right)y=\frac{1}{28}\times 112,7\times \frac{1}{28}x+7\left(-\frac{1}{32}\right)y=7$$

Simplify.

$$\frac{1}{4}x-\frac{2}{7}y=4,\frac{1}{4}x-\frac{7}{32}y=7$$

Subtract \frac{1}{4}x-\frac{7}{32}y=7 from \frac{1}{4}x-\frac{2}{7}y=4 by subtracting like terms on each side of the equal sign.

$$\frac{1}{4}x-\frac{1}{4}x-\frac{2}{7}y+\frac{7}{32}y=4-7$$

Add \frac{x}{4} to -\frac{x}{4}. Terms \frac{x}{4} and -\frac{x}{4} cancel out, leaving an equation with only one variable that can be solved.

$$-\frac{2}{7}y+\frac{7}{32}y=4-7$$

Add -\frac{2y}{7} to \frac{7y}{32}.

$$-\frac{15}{224}y=4-7$$

Add 4 to -7.

$$-\frac{15}{224}y=-3$$

Divide both sides of the equation by -\frac{15}{224}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=\frac{224}{5}$$

Substitute \frac{224}{5} for y in \frac{1}{28}x-\frac{1}{32}y=1. Because the resulting equation contains only one variable, you can solve for x directly.

$$\frac{1}{28}x-\frac{1}{32}\times \frac{224}{5}=1$$

Multiply -\frac{1}{32} times \frac{224}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$\frac{1}{28}x-\frac{7}{5}=1$$

Add \frac{7}{5} to both sides of the equation.

$$\frac{1}{28}x=\frac{12}{5}$$

Multiply both sides by 28.

$$x=\frac{336}{5}$$

The system is now solved.

$$x=\frac{336}{5},y=\frac{224}{5}$$