Solve {l}{50x-40y=-50}{50y-40x=136}

$$\left. \begin{array} { l } { 50 x - 40 y = - 50 } \\ { 50 y - 40 x = 136 } \end{array} \right.$$

Solve for x, y

$$x = \frac{49}{15} = 3\frac{4}{15} \approx 3.266666667$$

$$y = \frac{16}{3} = 5\frac{1}{3} \approx 5.333333333$$

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.

$$50x-40y=-50,-40x+50y=136$$

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

$$50x-40y=-50$$

Add 40y to both sides of the equation.

$$50x=40y-50$$

Divide both sides by 50.

$$x=\frac{1}{50}\left(40y-50\right)$$

Multiply \frac{1}{50} times 40y-50.

$$x=\frac{4}{5}y-1$$

Substitute \frac{4y}{5}-1 for x in the other equation, -40x+50y=136.

$$-40\left(\frac{4}{5}y-1\right)+50y=136$$

Multiply -40 times \frac{4y}{5}-1.

$$-32y+40+50y=136$$

Add -32y to 50y.

$$18y+40=136$$

Subtract 40 from both sides of the equation.

$$18y=96$$

Divide both sides by 18.

$$y=\frac{16}{3}$$

Substitute \frac{16}{3} for y in x=\frac{4}{5}y-1. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{4}{5}\times \frac{16}{3}-1$$

Multiply \frac{4}{5} times \frac{16}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{64}{15}-1$$

Add -1 to \frac{64}{15}.

$$x=\frac{49}{15}$$

The system is now solved.

$$x=\frac{49}{15},y=\frac{16}{3}$$

Steps Using Matrices

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

$$50x-40y=-50,-40x+50y=136$$

Write the equations in matrix form.

$$\left(\begin{matrix}50&-40\\-40&50\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-50\\136\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}50&-40\\-40&50\end{matrix}\right).

$$inverse(\left(\begin{matrix}50&-40\\-40&50\end{matrix}\right))\left(\begin{matrix}50&-40\\-40&50\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}50&-40\\-40&50\end{matrix}\right))\left(\begin{matrix}-50\\136\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}50&-40\\-40&50\end{matrix}\right))\left(\begin{matrix}-50\\136\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}50&-40\\-40&50\end{matrix}\right))\left(\begin{matrix}-50\\136\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{50}{50\times 50-\left(-40\left(-40\right)\right)}&-\frac{-40}{50\times 50-\left(-40\left(-40\right)\right)}\\-\frac{-40}{50\times 50-\left(-40\left(-40\right)\right)}&\frac{50}{50\times 50-\left(-40\left(-40\right)\right)}\end{matrix}\right)\left(\begin{matrix}-50\\136\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{18}&\frac{2}{45}\\\frac{2}{45}&\frac{1}{18}\end{matrix}\right)\left(\begin{matrix}-50\\136\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{18}\left(-50\right)+\frac{2}{45}\times 136\\\frac{2}{45}\left(-50\right)+\frac{1}{18}\times 136\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{49}{15}\\\frac{16}{3}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{49}{15},y=\frac{16}{3}$$

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.

$$50x-40y=-50,-40x+50y=136$$

To make 50x and -40x equal, multiply all terms on each side of the first equation by -40 and all terms on each side of the second by 50.

$$-40\times 50x-40\left(-40\right)y=-40\left(-50\right),50\left(-40\right)x+50\times 50y=50\times 136$$

Simplify.

$$-2000x+1600y=2000,-2000x+2500y=6800$$

Subtract -2000x+2500y=6800 from -2000x+1600y=2000 by subtracting like terms on each side of the equal sign.

$$-2000x+2000x+1600y-2500y=2000-6800$$

Add -2000x to 2000x. Terms -2000x and 2000x cancel out, leaving an equation with only one variable that can be solved.

$$1600y-2500y=2000-6800$$

Add 1600y to -2500y.

$$-900y=2000-6800$$

Add 2000 to -6800.

$$-900y=-4800$$

Divide both sides by -900.

$$y=\frac{16}{3}$$

Substitute \frac{16}{3} for y in -40x+50y=136. Because the resulting equation contains only one variable, you can solve for x directly.

$$-40x+50\times \frac{16}{3}=136$$

Multiply 50 times \frac{16}{3}.

$$-40x+\frac{800}{3}=136$$

Subtract \frac{800}{3} from both sides of the equation.

$$-40x=-\frac{392}{3}$$

Divide both sides by -40.

$$x=\frac{49}{15}$$

The system is now solved.

$$x=\frac{49}{15},y=\frac{16}{3}$$