Solve {l}{6x-18y=-85}{24x-5y=5}

$$\left. \begin{array} { l } { 6 x - 18 y = - 85 } \\ { 24 x - 5 y = 5 } \end{array} \right.$$

Solve for x, y

$$x = \frac{515}{402} = 1\frac{113}{402} \approx 1.281094527$$

$$y = \frac{345}{67} = 5\frac{10}{67} \approx 5.149253731$$

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.

$$6x-18y=-85,24x-5y=5$$

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

$$6x-18y=-85$$

Add 18y to both sides of the equation.

$$6x=18y-85$$

Divide both sides by 6.

$$x=\frac{1}{6}\left(18y-85\right)$$

Multiply \frac{1}{6} times 18y-85.

$$x=3y-\frac{85}{6}$$

Substitute 3y-\frac{85}{6} for x in the other equation, 24x-5y=5.

$$24\left(3y-\frac{85}{6}\right)-5y=5$$

Multiply 24 times 3y-\frac{85}{6}.

$$72y-340-5y=5$$

Add 72y to -5y.

$$67y-340=5$$

Add 340 to both sides of the equation.

$$67y=345$$

Divide both sides by 67.

$$y=\frac{345}{67}$$

Substitute \frac{345}{67} for y in x=3y-\frac{85}{6}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=3\times \frac{345}{67}-\frac{85}{6}$$

Multiply 3 times \frac{345}{67}.

$$x=\frac{1035}{67}-\frac{85}{6}$$

Add -\frac{85}{6} to \frac{1035}{67} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{515}{402}$$

The system is now solved.

$$x=\frac{515}{402},y=\frac{345}{67}$$

Steps Using Matrices

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

$$6x-18y=-85,24x-5y=5$$

Write the equations in matrix form.

$$\left(\begin{matrix}6&-18\\24&-5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-85\\5\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}6&-18\\24&-5\end{matrix}\right).

$$inverse(\left(\begin{matrix}6&-18\\24&-5\end{matrix}\right))\left(\begin{matrix}6&-18\\24&-5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}6&-18\\24&-5\end{matrix}\right))\left(\begin{matrix}-85\\5\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}6&-18\\24&-5\end{matrix}\right))\left(\begin{matrix}-85\\5\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}6&-18\\24&-5\end{matrix}\right))\left(\begin{matrix}-85\\5\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{5}{6\left(-5\right)-\left(-18\times 24\right)}&-\frac{-18}{6\left(-5\right)-\left(-18\times 24\right)}\\-\frac{24}{6\left(-5\right)-\left(-18\times 24\right)}&\frac{6}{6\left(-5\right)-\left(-18\times 24\right)}\end{matrix}\right)\left(\begin{matrix}-85\\5\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{5}{402}&\frac{3}{67}\\-\frac{4}{67}&\frac{1}{67}\end{matrix}\right)\left(\begin{matrix}-85\\5\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{5}{402}\left(-85\right)+\frac{3}{67}\times 5\\-\frac{4}{67}\left(-85\right)+\frac{1}{67}\times 5\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{515}{402}\\\frac{345}{67}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{515}{402},y=\frac{345}{67}$$

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.

$$6x-18y=-85,24x-5y=5$$

To make 6x and 24x equal, multiply all terms on each side of the first equation by 24 and all terms on each side of the second by 6.

$$24\times 6x+24\left(-18\right)y=24\left(-85\right),6\times 24x+6\left(-5\right)y=6\times 5$$

Simplify.

$$144x-432y=-2040,144x-30y=30$$

Subtract 144x-30y=30 from 144x-432y=-2040 by subtracting like terms on each side of the equal sign.

$$144x-144x-432y+30y=-2040-30$$

Add 144x to -144x. Terms 144x and -144x cancel out, leaving an equation with only one variable that can be solved.

$$-432y+30y=-2040-30$$

Add -432y to 30y.

$$-402y=-2040-30$$

Add -2040 to -30.

$$-402y=-2070$$

Divide both sides by -402.

$$y=\frac{345}{67}$$

Substitute \frac{345}{67} for y in 24x-5y=5. Because the resulting equation contains only one variable, you can solve for x directly.

$$24x-5\times \frac{345}{67}=5$$

Multiply -5 times \frac{345}{67}.

$$24x-\frac{1725}{67}=5$$

Add \frac{1725}{67} to both sides of the equation.

$$24x=\frac{2060}{67}$$

Divide both sides by 24.

$$x=\frac{515}{402}$$

The system is now solved.

$$x=\frac{515}{402},y=\frac{345}{67}$$