Solve {l}{12a+16c=176}{10a+8c=120}

$$\left. \begin{array} { l } { 12 a + 16 c = 176 } \\ { 10 a + 8 c = 120 } \end{array} \right.$$

Solve for a, c

$$a=8$$

$$c=5$$

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.

$$12a+16c=176,10a+8c=120$$

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

$$12a+16c=176$$

Subtract 16c from both sides of the equation.

$$12a=-16c+176$$

Divide both sides by 12.

$$a=\frac{1}{12}\left(-16c+176\right)$$

Multiply \frac{1}{12} times -16c+176.

$$a=-\frac{4}{3}c+\frac{44}{3}$$

Substitute \frac{-4c+44}{3} for a in the other equation, 10a+8c=120.

$$10\left(-\frac{4}{3}c+\frac{44}{3}\right)+8c=120$$

Multiply 10 times \frac{-4c+44}{3}.

$$-\frac{40}{3}c+\frac{440}{3}+8c=120$$

Add -\frac{40c}{3} to 8c.

$$-\frac{16}{3}c+\frac{440}{3}=120$$

Subtract \frac{440}{3} from both sides of the equation.

$$-\frac{16}{3}c=-\frac{80}{3}$$

Divide both sides of the equation by -\frac{16}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$c=5$$

Substitute 5 for c in a=-\frac{4}{3}c+\frac{44}{3}. Because the resulting equation contains only one variable, you can solve for a directly.

$$a=-\frac{4}{3}\times 5+\frac{44}{3}$$

Multiply -\frac{4}{3} times 5.

$$a=\frac{-20+44}{3}$$

Add \frac{44}{3} to -\frac{20}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a=8$$

The system is now solved.

$$a=8,c=5$$

Steps Using Matrices

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

$$12a+16c=176,10a+8c=120$$

Write the equations in matrix form.

$$\left(\begin{matrix}12&16\\10&8\end{matrix}\right)\left(\begin{matrix}a\\c\end{matrix}\right)=\left(\begin{matrix}176\\120\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}12&16\\10&8\end{matrix}\right).

$$inverse(\left(\begin{matrix}12&16\\10&8\end{matrix}\right))\left(\begin{matrix}12&16\\10&8\end{matrix}\right)\left(\begin{matrix}a\\c\end{matrix}\right)=inverse(\left(\begin{matrix}12&16\\10&8\end{matrix}\right))\left(\begin{matrix}176\\120\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a\\c\end{matrix}\right)=inverse(\left(\begin{matrix}12&16\\10&8\end{matrix}\right))\left(\begin{matrix}176\\120\end{matrix}\right)$$

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

$$\left(\begin{matrix}a\\c\end{matrix}\right)=inverse(\left(\begin{matrix}12&16\\10&8\end{matrix}\right))\left(\begin{matrix}176\\120\end{matrix}\right)$$

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.

$$\left(\begin{matrix}a\\c\end{matrix}\right)=\left(\begin{matrix}\frac{8}{12\times 8-16\times 10}&-\frac{16}{12\times 8-16\times 10}\\-\frac{10}{12\times 8-16\times 10}&\frac{12}{12\times 8-16\times 10}\end{matrix}\right)\left(\begin{matrix}176\\120\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\c\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{8}&\frac{1}{4}\\\frac{5}{32}&-\frac{3}{16}\end{matrix}\right)\left(\begin{matrix}176\\120\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a\\c\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{8}\times 176+\frac{1}{4}\times 120\\\frac{5}{32}\times 176-\frac{3}{16}\times 120\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\c\end{matrix}\right)=\left(\begin{matrix}8\\5\end{matrix}\right)$$

Extract the matrix elements a and c.

$$a=8,c=5$$

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.

$$12a+16c=176,10a+8c=120$$

To make 12a and 10a equal, multiply all terms on each side of the first equation by 10 and all terms on each side of the second by 12.

$$10\times 12a+10\times 16c=10\times 176,12\times 10a+12\times 8c=12\times 120$$

Simplify.

$$120a+160c=1760,120a+96c=1440$$

Subtract 120a+96c=1440 from 120a+160c=1760 by subtracting like terms on each side of the equal sign.

$$120a-120a+160c-96c=1760-1440$$

Add 120a to -120a. Terms 120a and -120a cancel out, leaving an equation with only one variable that can be solved.

$$160c-96c=1760-1440$$

Add 160c to -96c.

$$64c=1760-1440$$

Add 1760 to -1440.

$$64c=320$$

Divide both sides by 64.

$$c=5$$

Substitute 5 for c in 10a+8c=120. Because the resulting equation contains only one variable, you can solve for a directly.

$$10a+8\times 5=120$$

Multiply 8 times 5.

$$10a+40=120$$

Subtract 40 from both sides of the equation.

$$10a=80$$

Divide both sides by 10.

$$a=8$$

The system is now solved.

$$a=8,c=5$$