Solve 17x+23y=40text{stit}23x+17y=80

$$17 x + 23 y = 40 \text { stit } 23 x + 17 y = 80$$

Solve for x, y

$$x = \frac{29}{6} = 4\frac{5}{6} \approx 4.833333333$$

$$y = -\frac{11}{6} = -1\frac{5}{6} \approx -1.833333333$$

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.

$$17x+23y=40,23x+17y=80$$

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

$$17x+23y=40$$

Subtract 23y from both sides of the equation.

$$17x=-23y+40$$

Divide both sides by 17.

$$x=\frac{1}{17}\left(-23y+40\right)$$

Multiply \frac{1}{17} times -23y+40.

$$x=-\frac{23}{17}y+\frac{40}{17}$$

Substitute \frac{-23y+40}{17} for x in the other equation, 23x+17y=80.

$$23\left(-\frac{23}{17}y+\frac{40}{17}\right)+17y=80$$

Multiply 23 times \frac{-23y+40}{17}.

$$-\frac{529}{17}y+\frac{920}{17}+17y=80$$

Add -\frac{529y}{17} to 17y.

$$-\frac{240}{17}y+\frac{920}{17}=80$$

Subtract \frac{920}{17} from both sides of the equation.

$$-\frac{240}{17}y=\frac{440}{17}$$

Divide both sides of the equation by -\frac{240}{17}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=-\frac{11}{6}$$

Substitute -\frac{11}{6} for y in x=-\frac{23}{17}y+\frac{40}{17}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{23}{17}\left(-\frac{11}{6}\right)+\frac{40}{17}$$

Multiply -\frac{23}{17} times -\frac{11}{6} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{253}{102}+\frac{40}{17}$$

Add \frac{40}{17} to \frac{253}{102} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{29}{6}$$

The system is now solved.

$$x=\frac{29}{6},y=-\frac{11}{6}$$

Steps Using Matrices

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

$$17x+23y=40,23x+17y=80$$

Write the equations in matrix form.

$$\left(\begin{matrix}17&23\\23&17\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}40\\80\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}17&23\\23&17\end{matrix}\right).

$$inverse(\left(\begin{matrix}17&23\\23&17\end{matrix}\right))\left(\begin{matrix}17&23\\23&17\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}17&23\\23&17\end{matrix}\right))\left(\begin{matrix}40\\80\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}17&23\\23&17\end{matrix}\right))\left(\begin{matrix}40\\80\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}17&23\\23&17\end{matrix}\right))\left(\begin{matrix}40\\80\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{17}{17\times 17-23\times 23}&-\frac{23}{17\times 17-23\times 23}\\-\frac{23}{17\times 17-23\times 23}&\frac{17}{17\times 17-23\times 23}\end{matrix}\right)\left(\begin{matrix}40\\80\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{17}{240}&\frac{23}{240}\\\frac{23}{240}&-\frac{17}{240}\end{matrix}\right)\left(\begin{matrix}40\\80\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{17}{240}\times 40+\frac{23}{240}\times 80\\\frac{23}{240}\times 40-\frac{17}{240}\times 80\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{29}{6}\\-\frac{11}{6}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{29}{6},y=-\frac{11}{6}$$

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.

$$17x+23y=40,23x+17y=80$$

To make 17x and 23x equal, multiply all terms on each side of the first equation by 23 and all terms on each side of the second by 17.

$$23\times 17x+23\times 23y=23\times 40,17\times 23x+17\times 17y=17\times 80$$

Simplify.

$$391x+529y=920,391x+289y=1360$$

Subtract 391x+289y=1360 from 391x+529y=920 by subtracting like terms on each side of the equal sign.

$$391x-391x+529y-289y=920-1360$$

Add 391x to -391x. Terms 391x and -391x cancel out, leaving an equation with only one variable that can be solved.

$$529y-289y=920-1360$$

Add 529y to -289y.

$$240y=920-1360$$

Add 920 to -1360.

$$240y=-440$$

Divide both sides by 240.

$$y=-\frac{11}{6}$$

Substitute -\frac{11}{6} for y in 23x+17y=80. Because the resulting equation contains only one variable, you can solve for x directly.

$$23x+17\left(-\frac{11}{6}\right)=80$$

Multiply 17 times -\frac{11}{6}.

$$23x-\frac{187}{6}=80$$

Add \frac{187}{6} to both sides of the equation.

$$23x=\frac{667}{6}$$

Divide both sides by 23.

$$x=\frac{29}{6}$$

The system is now solved.

$$x=\frac{29}{6},y=-\frac{11}{6}$$