Solve {l}{3A-13c=-255}{31A-6c=-180}

$$\left. \begin{array} { l } { 3 A - 13 c = - 255 } \\ { 31 A - 6 c = - 180 } \end{array} \right.$$

Solve for A, c

$$A = -\frac{162}{77} = -2\frac{8}{77} \approx -2.103896104$$

$$c = \frac{1473}{77} = 19\frac{10}{77} \approx 19.12987013$$

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.

$$3A-13c=-255,31A-6c=-180$$

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

$$3A-13c=-255$$

Add 13c to both sides of the equation.

$$3A=13c-255$$

Divide both sides by 3.

$$A=\frac{1}{3}\left(13c-255\right)$$

Multiply \frac{1}{3} times 13c-255.

$$A=\frac{13}{3}c-85$$

Substitute \frac{13c}{3}-85 for A in the other equation, 31A-6c=-180.

$$31\left(\frac{13}{3}c-85\right)-6c=-180$$

Multiply 31 times \frac{13c}{3}-85.

$$\frac{403}{3}c-2635-6c=-180$$

Add \frac{403c}{3} to -6c.

$$\frac{385}{3}c-2635=-180$$

Add 2635 to both sides of the equation.

$$\frac{385}{3}c=2455$$

Divide both sides of the equation by \frac{385}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$c=\frac{1473}{77}$$

Substitute \frac{1473}{77} for c in A=\frac{13}{3}c-85. Because the resulting equation contains only one variable, you can solve for A directly.

$$A=\frac{13}{3}\times \frac{1473}{77}-85$$

Multiply \frac{13}{3} times \frac{1473}{77} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$A=\frac{6383}{77}-85$$

Add -85 to \frac{6383}{77}.

$$A=-\frac{162}{77}$$

The system is now solved.

$$A=-\frac{162}{77},c=\frac{1473}{77}$$

Steps Using Matrices

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

$$3A-13c=-255,31A-6c=-180$$

Write the equations in matrix form.

$$\left(\begin{matrix}3&-13\\31&-6\end{matrix}\right)\left(\begin{matrix}A\\c\end{matrix}\right)=\left(\begin{matrix}-255\\-180\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}3&-13\\31&-6\end{matrix}\right).

$$inverse(\left(\begin{matrix}3&-13\\31&-6\end{matrix}\right))\left(\begin{matrix}3&-13\\31&-6\end{matrix}\right)\left(\begin{matrix}A\\c\end{matrix}\right)=inverse(\left(\begin{matrix}3&-13\\31&-6\end{matrix}\right))\left(\begin{matrix}-255\\-180\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}A\\c\end{matrix}\right)=inverse(\left(\begin{matrix}3&-13\\31&-6\end{matrix}\right))\left(\begin{matrix}-255\\-180\end{matrix}\right)$$

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

$$\left(\begin{matrix}A\\c\end{matrix}\right)=inverse(\left(\begin{matrix}3&-13\\31&-6\end{matrix}\right))\left(\begin{matrix}-255\\-180\end{matrix}\right)$$

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.

$$\left(\begin{matrix}A\\c\end{matrix}\right)=\left(\begin{matrix}-\frac{6}{3\left(-6\right)-\left(-13\times 31\right)}&-\frac{-13}{3\left(-6\right)-\left(-13\times 31\right)}\\-\frac{31}{3\left(-6\right)-\left(-13\times 31\right)}&\frac{3}{3\left(-6\right)-\left(-13\times 31\right)}\end{matrix}\right)\left(\begin{matrix}-255\\-180\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}A\\c\end{matrix}\right)=\left(\begin{matrix}-\frac{6}{385}&\frac{13}{385}\\-\frac{31}{385}&\frac{3}{385}\end{matrix}\right)\left(\begin{matrix}-255\\-180\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}A\\c\end{matrix}\right)=\left(\begin{matrix}-\frac{6}{385}\left(-255\right)+\frac{13}{385}\left(-180\right)\\-\frac{31}{385}\left(-255\right)+\frac{3}{385}\left(-180\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}A\\c\end{matrix}\right)=\left(\begin{matrix}-\frac{162}{77}\\\frac{1473}{77}\end{matrix}\right)$$

Extract the matrix elements A and c.

$$A=-\frac{162}{77},c=\frac{1473}{77}$$

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.

$$3A-13c=-255,31A-6c=-180$$

To make 3A and 31A equal, multiply all terms on each side of the first equation by 31 and all terms on each side of the second by 3.

$$31\times 3A+31\left(-13\right)c=31\left(-255\right),3\times 31A+3\left(-6\right)c=3\left(-180\right)$$

Simplify.

$$93A-403c=-7905,93A-18c=-540$$

Subtract 93A-18c=-540 from 93A-403c=-7905 by subtracting like terms on each side of the equal sign.

$$93A-93A-403c+18c=-7905+540$$

Add 93A to -93A. Terms 93A and -93A cancel out, leaving an equation with only one variable that can be solved.

$$-403c+18c=-7905+540$$

Add -403c to 18c.

$$-385c=-7905+540$$

Add -7905 to 540.

$$-385c=-7365$$

Divide both sides by -385.

$$c=\frac{1473}{77}$$

Substitute \frac{1473}{77} for c in 31A-6c=-180. Because the resulting equation contains only one variable, you can solve for A directly.

$$31A-6\times \frac{1473}{77}=-180$$

Multiply -6 times \frac{1473}{77}.

$$31A-\frac{8838}{77}=-180$$

Add \frac{8838}{77} to both sides of the equation.

$$31A=-\frac{5022}{77}$$

Divide both sides by 31.

$$A=-\frac{162}{77}$$

The system is now solved.

$$A=-\frac{162}{77},c=\frac{1473}{77}$$