Solve {l}{70/100x+98/100y=58(x+y)/100}{170/100x-frac{45}{100}y=37}

$$\left\{ \begin{array} { l } { \frac { 70 } { 100 } x + \frac { 98 } { 100 } y = \frac { 58 ( x + y ) } { 100 } } \\ { \frac { 170 } { 100 } x - \frac { 45 } { 100 } y = 37 } \end{array} \right.$$

Solve for x, y

$$x = \frac{7400}{367} = 20\frac{60}{367} \approx 20.163487738$$

$$y = -\frac{2220}{367} = -6\frac{18}{367} \approx -6.049046322$$

Steps Using Substitution

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

$$70x+98y=58\left(x+y\right)$$

Use the distributive property to multiply 58 by x+y.

$$70x+98y=58x+58y$$

Subtract 58x from both sides.

$$70x+98y-58x=58y$$

Combine 70x and -58x to get 12x.

$$12x+98y=58y$$

Subtract 58y from both sides.

$$12x+98y-58y=0$$

Combine 98y and -58y to get 40y.

$$12x+40y=0$$

Consider the second equation. Reduce the fraction \frac{170}{100} to lowest terms by extracting and canceling out 10.

$$\frac{17}{10}x-\frac{45}{100}y=37$$

Reduce the fraction \frac{45}{100} to lowest terms by extracting and canceling out 5.

$$\frac{17}{10}x-\frac{9}{20}y=37$$

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.

$$12x+40y=0,\frac{17}{10}x-\frac{9}{20}y=37$$

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

$$12x+40y=0$$

Subtract 40y from both sides of the equation.

$$12x=-40y$$

Divide both sides by 12.

$$x=\frac{1}{12}\left(-40\right)y$$

Multiply \frac{1}{12} times -40y.

$$x=-\frac{10}{3}y$$

Substitute -\frac{10y}{3} for x in the other equation, \frac{17}{10}x-\frac{9}{20}y=37.

$$\frac{17}{10}\left(-\frac{10}{3}\right)y-\frac{9}{20}y=37$$

Multiply \frac{17}{10} times -\frac{10y}{3}.

$$-\frac{17}{3}y-\frac{9}{20}y=37$$

Add -\frac{17y}{3} to -\frac{9y}{20}.

$$-\frac{367}{60}y=37$$

Divide both sides of the equation by -\frac{367}{60}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=-\frac{2220}{367}$$

Substitute -\frac{2220}{367} for y in x=-\frac{10}{3}y. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{10}{3}\left(-\frac{2220}{367}\right)$$

Multiply -\frac{10}{3} times -\frac{2220}{367} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{7400}{367}$$

The system is now solved.

$$x=\frac{7400}{367},y=-\frac{2220}{367}$$

Steps Using Matrices

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

$$70x+98y=58\left(x+y\right)$$

Use the distributive property to multiply 58 by x+y.

$$70x+98y=58x+58y$$

Subtract 58x from both sides.

$$70x+98y-58x=58y$$

Combine 70x and -58x to get 12x.

$$12x+98y=58y$$

Subtract 58y from both sides.

$$12x+98y-58y=0$$

Combine 98y and -58y to get 40y.

$$12x+40y=0$$

Consider the second equation. Reduce the fraction \frac{170}{100} to lowest terms by extracting and canceling out 10.

$$\frac{17}{10}x-\frac{45}{100}y=37$$

Reduce the fraction \frac{45}{100} to lowest terms by extracting and canceling out 5.

$$\frac{17}{10}x-\frac{9}{20}y=37$$

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

$$12x+40y=0,\frac{17}{10}x-\frac{9}{20}y=37$$

Write the equations in matrix form.

$$\left(\begin{matrix}12&40\\\frac{17}{10}&-\frac{9}{20}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\37\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}12&40\\\frac{17}{10}&-\frac{9}{20}\end{matrix}\right).

$$inverse(\left(\begin{matrix}12&40\\\frac{17}{10}&-\frac{9}{20}\end{matrix}\right))\left(\begin{matrix}12&40\\\frac{17}{10}&-\frac{9}{20}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}12&40\\\frac{17}{10}&-\frac{9}{20}\end{matrix}\right))\left(\begin{matrix}0\\37\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}12&40\\\frac{17}{10}&-\frac{9}{20}\end{matrix}\right))\left(\begin{matrix}0\\37\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}12&40\\\frac{17}{10}&-\frac{9}{20}\end{matrix}\right))\left(\begin{matrix}0\\37\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{\frac{9}{20}}{12\left(-\frac{9}{20}\right)-40\times \frac{17}{10}}&-\frac{40}{12\left(-\frac{9}{20}\right)-40\times \frac{17}{10}}\\-\frac{\frac{17}{10}}{12\left(-\frac{9}{20}\right)-40\times \frac{17}{10}}&\frac{12}{12\left(-\frac{9}{20}\right)-40\times \frac{17}{10}}\end{matrix}\right)\left(\begin{matrix}0\\37\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{9}{1468}&\frac{200}{367}\\\frac{17}{734}&-\frac{60}{367}\end{matrix}\right)\left(\begin{matrix}0\\37\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{200}{367}\times 37\\-\frac{60}{367}\times 37\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{7400}{367}\\-\frac{2220}{367}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{7400}{367},y=-\frac{2220}{367}$$

Steps Using Elimination

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

$$70x+98y=58\left(x+y\right)$$

Use the distributive property to multiply 58 by x+y.

$$70x+98y=58x+58y$$

Subtract 58x from both sides.

$$70x+98y-58x=58y$$

Combine 70x and -58x to get 12x.

$$12x+98y=58y$$

Subtract 58y from both sides.

$$12x+98y-58y=0$$

Combine 98y and -58y to get 40y.

$$12x+40y=0$$

Consider the second equation. Reduce the fraction \frac{170}{100} to lowest terms by extracting and canceling out 10.

$$\frac{17}{10}x-\frac{45}{100}y=37$$

Reduce the fraction \frac{45}{100} to lowest terms by extracting and canceling out 5.

$$\frac{17}{10}x-\frac{9}{20}y=37$$

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.

$$12x+40y=0,\frac{17}{10}x-\frac{9}{20}y=37$$

To make 12x and \frac{17x}{10} equal, multiply all terms on each side of the first equation by \frac{17}{10} and all terms on each side of the second by 12.

$$\frac{17}{10}\times 12x+\frac{17}{10}\times 40y=0,12\times \frac{17}{10}x+12\left(-\frac{9}{20}\right)y=12\times 37$$

Simplify.

$$\frac{102}{5}x+68y=0,\frac{102}{5}x-\frac{27}{5}y=444$$

Subtract \frac{102}{5}x-\frac{27}{5}y=444 from \frac{102}{5}x+68y=0 by subtracting like terms on each side of the equal sign.

$$\frac{102}{5}x-\frac{102}{5}x+68y+\frac{27}{5}y=-444$$

Add \frac{102x}{5} to -\frac{102x}{5}. Terms \frac{102x}{5} and -\frac{102x}{5} cancel out, leaving an equation with only one variable that can be solved.

$$68y+\frac{27}{5}y=-444$$

Add 68y to \frac{27y}{5}.

$$\frac{367}{5}y=-444$$

Divide both sides of the equation by \frac{367}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=-\frac{2220}{367}$$

Substitute -\frac{2220}{367} for y in \frac{17}{10}x-\frac{9}{20}y=37. Because the resulting equation contains only one variable, you can solve for x directly.

$$\frac{17}{10}x-\frac{9}{20}\left(-\frac{2220}{367}\right)=37$$

Multiply -\frac{9}{20} times -\frac{2220}{367} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$\frac{17}{10}x+\frac{999}{367}=37$$

Subtract \frac{999}{367} from both sides of the equation.

$$\frac{17}{10}x=\frac{12580}{367}$$

Divide both sides of the equation by \frac{17}{10}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$x=\frac{7400}{367}$$

The system is now solved.

$$x=\frac{7400}{367},y=-\frac{2220}{367}$$