Solve {l}{5x+8y=120}{40x+500y=81}

$$\left. \begin{array} { l } { 5 x + 8 y = 120 } \\ { 40 x + 500 y = 81 } \end{array} \right.$$

Solve for x, y

$$x = \frac{14838}{545} = 27\frac{123}{545} \approx 27.225688073$$

$$y = -\frac{879}{436} = -2\frac{7}{436} \approx -2.016055046$$

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.

$$5x+8y=120,40x+500y=81$$

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

$$5x+8y=120$$

Subtract 8y from both sides of the equation.

$$5x=-8y+120$$

Divide both sides by 5.

$$x=\frac{1}{5}\left(-8y+120\right)$$

Multiply \frac{1}{5} times -8y+120.

$$x=-\frac{8}{5}y+24$$

Substitute -\frac{8y}{5}+24 for x in the other equation, 40x+500y=81.

$$40\left(-\frac{8}{5}y+24\right)+500y=81$$

Multiply 40 times -\frac{8y}{5}+24.

$$-64y+960+500y=81$$

Add -64y to 500y.

$$436y+960=81$$

Subtract 960 from both sides of the equation.

$$436y=-879$$

Divide both sides by 436.

$$y=-\frac{879}{436}$$

Substitute -\frac{879}{436} for y in x=-\frac{8}{5}y+24. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{8}{5}\left(-\frac{879}{436}\right)+24$$

Multiply -\frac{8}{5} times -\frac{879}{436} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{1758}{545}+24$$

Add 24 to \frac{1758}{545}.

$$x=\frac{14838}{545}$$

The system is now solved.

$$x=\frac{14838}{545},y=-\frac{879}{436}$$

Steps Using Matrices

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

$$5x+8y=120,40x+500y=81$$

Write the equations in matrix form.

$$\left(\begin{matrix}5&8\\40&500\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}120\\81\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}5&8\\40&500\end{matrix}\right).

$$inverse(\left(\begin{matrix}5&8\\40&500\end{matrix}\right))\left(\begin{matrix}5&8\\40&500\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&8\\40&500\end{matrix}\right))\left(\begin{matrix}120\\81\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&8\\40&500\end{matrix}\right))\left(\begin{matrix}120\\81\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&8\\40&500\end{matrix}\right))\left(\begin{matrix}120\\81\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{500}{5\times 500-8\times 40}&-\frac{8}{5\times 500-8\times 40}\\-\frac{40}{5\times 500-8\times 40}&\frac{5}{5\times 500-8\times 40}\end{matrix}\right)\left(\begin{matrix}120\\81\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{25}{109}&-\frac{2}{545}\\-\frac{2}{109}&\frac{1}{436}\end{matrix}\right)\left(\begin{matrix}120\\81\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{25}{109}\times 120-\frac{2}{545}\times 81\\-\frac{2}{109}\times 120+\frac{1}{436}\times 81\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{14838}{545}\\-\frac{879}{436}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{14838}{545},y=-\frac{879}{436}$$

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.

$$5x+8y=120,40x+500y=81$$

To make 5x 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 5.

$$40\times 5x+40\times 8y=40\times 120,5\times 40x+5\times 500y=5\times 81$$

Simplify.

$$200x+320y=4800,200x+2500y=405$$

Subtract 200x+2500y=405 from 200x+320y=4800 by subtracting like terms on each side of the equal sign.

$$200x-200x+320y-2500y=4800-405$$

Add 200x to -200x. Terms 200x and -200x cancel out, leaving an equation with only one variable that can be solved.

$$320y-2500y=4800-405$$

Add 320y to -2500y.

$$-2180y=4800-405$$

Add 4800 to -405.

$$-2180y=4395$$

Divide both sides by -2180.

$$y=-\frac{879}{436}$$

Substitute -\frac{879}{436} for y in 40x+500y=81. Because the resulting equation contains only one variable, you can solve for x directly.

$$40x+500\left(-\frac{879}{436}\right)=81$$

Multiply 500 times -\frac{879}{436}.

$$40x-\frac{109875}{109}=81$$

Add \frac{109875}{109} to both sides of the equation.

$$40x=\frac{118704}{109}$$

Divide both sides by 40.

$$x=\frac{14838}{545}$$

The system is now solved.

$$x=\frac{14838}{545},y=-\frac{879}{436}$$