Solve {l}{108/100x+110y/100=1008}{110/100x+108/100y=1028}

$$\left. \begin{array} { l } { \frac { 108 } { 100 } x + \frac { 110 y } { 100 } = 1008 } \\ { \frac { 110 } { 100 } x + \frac { 108 } { 100 } y = 1028 } \end{array} \right.$$

Solve for x, y

$$x = \frac{105400}{109} = 966\frac{106}{109} \approx 966.972477064$$

$$y = -\frac{3600}{109} = -33\frac{3}{109} \approx -33.027522936$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 100.

$$108x+110y=100800$$

Consider the second equation. Reduce the fraction \frac{110}{100} to lowest terms by extracting and canceling out 10.

$$\frac{11}{10}x+\frac{108}{100}y=1028$$

Reduce the fraction \frac{108}{100} to lowest terms by extracting and canceling out 4.

$$\frac{11}{10}x+\frac{27}{25}y=1028$$

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.

$$108x+110y=100800,\frac{11}{10}x+\frac{27}{25}y=1028$$

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

$$108x+110y=100800$$

Subtract 110y from both sides of the equation.

$$108x=-110y+100800$$

Divide both sides by 108.

$$x=\frac{1}{108}\left(-110y+100800\right)$$

Multiply \frac{1}{108} times -110y+100800.

$$x=-\frac{55}{54}y+\frac{2800}{3}$$

Substitute -\frac{55y}{54}+\frac{2800}{3} for x in the other equation, \frac{11}{10}x+\frac{27}{25}y=1028.

$$\frac{11}{10}\left(-\frac{55}{54}y+\frac{2800}{3}\right)+\frac{27}{25}y=1028$$

Multiply \frac{11}{10} times -\frac{55y}{54}+\frac{2800}{3}.

$$-\frac{121}{108}y+\frac{3080}{3}+\frac{27}{25}y=1028$$

Add -\frac{121y}{108} to \frac{27y}{25}.

$$-\frac{109}{2700}y+\frac{3080}{3}=1028$$

Subtract \frac{3080}{3} from both sides of the equation.

$$-\frac{109}{2700}y=\frac{4}{3}$$

Divide both sides of the equation by -\frac{109}{2700}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=-\frac{3600}{109}$$

Substitute -\frac{3600}{109} for y in x=-\frac{55}{54}y+\frac{2800}{3}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{55}{54}\left(-\frac{3600}{109}\right)+\frac{2800}{3}$$

Multiply -\frac{55}{54} times -\frac{3600}{109} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{11000}{327}+\frac{2800}{3}$$

Add \frac{2800}{3} to \frac{11000}{327} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{105400}{109}$$

The system is now solved.

$$x=\frac{105400}{109},y=-\frac{3600}{109}$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 100.

$$108x+110y=100800$$

Consider the second equation. Reduce the fraction \frac{110}{100} to lowest terms by extracting and canceling out 10.

$$\frac{11}{10}x+\frac{108}{100}y=1028$$

Reduce the fraction \frac{108}{100} to lowest terms by extracting and canceling out 4.

$$\frac{11}{10}x+\frac{27}{25}y=1028$$

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

$$108x+110y=100800,\frac{11}{10}x+\frac{27}{25}y=1028$$

Write the equations in matrix form.

$$\left(\begin{matrix}108&110\\\frac{11}{10}&\frac{27}{25}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}100800\\1028\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}108&110\\\frac{11}{10}&\frac{27}{25}\end{matrix}\right).

$$inverse(\left(\begin{matrix}108&110\\\frac{11}{10}&\frac{27}{25}\end{matrix}\right))\left(\begin{matrix}108&110\\\frac{11}{10}&\frac{27}{25}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}108&110\\\frac{11}{10}&\frac{27}{25}\end{matrix}\right))\left(\begin{matrix}100800\\1028\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}108&110\\\frac{11}{10}&\frac{27}{25}\end{matrix}\right))\left(\begin{matrix}100800\\1028\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}108&110\\\frac{11}{10}&\frac{27}{25}\end{matrix}\right))\left(\begin{matrix}100800\\1028\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{\frac{27}{25}}{108\times \frac{27}{25}-110\times \frac{11}{10}}&-\frac{110}{108\times \frac{27}{25}-110\times \frac{11}{10}}\\-\frac{\frac{11}{10}}{108\times \frac{27}{25}-110\times \frac{11}{10}}&\frac{108}{108\times \frac{27}{25}-110\times \frac{11}{10}}\end{matrix}\right)\left(\begin{matrix}100800\\1028\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{27}{109}&\frac{2750}{109}\\\frac{55}{218}&-\frac{2700}{109}\end{matrix}\right)\left(\begin{matrix}100800\\1028\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{27}{109}\times 100800+\frac{2750}{109}\times 1028\\\frac{55}{218}\times 100800-\frac{2700}{109}\times 1028\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{105400}{109}\\-\frac{3600}{109}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{105400}{109},y=-\frac{3600}{109}$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 100.

$$108x+110y=100800$$

Consider the second equation. Reduce the fraction \frac{110}{100} to lowest terms by extracting and canceling out 10.

$$\frac{11}{10}x+\frac{108}{100}y=1028$$

Reduce the fraction \frac{108}{100} to lowest terms by extracting and canceling out 4.

$$\frac{11}{10}x+\frac{27}{25}y=1028$$

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.

$$108x+110y=100800,\frac{11}{10}x+\frac{27}{25}y=1028$$

To make 108x and \frac{11x}{10} equal, multiply all terms on each side of the first equation by \frac{11}{10} and all terms on each side of the second by 108.

$$\frac{11}{10}\times 108x+\frac{11}{10}\times 110y=\frac{11}{10}\times 100800,108\times \frac{11}{10}x+108\times \frac{27}{25}y=108\times 1028$$

Simplify.

$$\frac{594}{5}x+121y=110880,\frac{594}{5}x+\frac{2916}{25}y=111024$$

Subtract \frac{594}{5}x+\frac{2916}{25}y=111024 from \frac{594}{5}x+121y=110880 by subtracting like terms on each side of the equal sign.

$$\frac{594}{5}x-\frac{594}{5}x+121y-\frac{2916}{25}y=110880-111024$$

Add \frac{594x}{5} to -\frac{594x}{5}. Terms \frac{594x}{5} and -\frac{594x}{5} cancel out, leaving an equation with only one variable that can be solved.

$$121y-\frac{2916}{25}y=110880-111024$$

Add 121y to -\frac{2916y}{25}.

$$\frac{109}{25}y=110880-111024$$

Add 110880 to -111024.

$$\frac{109}{25}y=-144$$

Divide both sides of the equation by \frac{109}{25}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=-\frac{3600}{109}$$

Substitute -\frac{3600}{109} for y in \frac{11}{10}x+\frac{27}{25}y=1028. Because the resulting equation contains only one variable, you can solve for x directly.

$$\frac{11}{10}x+\frac{27}{25}\left(-\frac{3600}{109}\right)=1028$$

Multiply \frac{27}{25} times -\frac{3600}{109} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$\frac{11}{10}x-\frac{3888}{109}=1028$$

Add \frac{3888}{109} to both sides of the equation.

$$\frac{11}{10}x=\frac{115940}{109}$$

Divide both sides of the equation by \frac{11}{10}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$x=\frac{105400}{109}$$

The system is now solved.

$$x=\frac{105400}{109},y=-\frac{3600}{109}$$