Solve {l}{6a+8b+b=0}{8a-2b+12=0}

$$\left. \begin{array} { l } { 6 a + 8 b + b = 0 } \\ { 8 a - 2 b + 12 = 0 } \end{array} \right.$$

Solve for a, b

$$a = -\frac{9}{7} = -1\frac{2}{7} \approx -1.285714286$$

$$b=\frac{6}{7}\approx 0.857142857$$

Steps Using Substitution

Consider the first equation. Combine 8b and b to get 9b.

$$6a+9b=0$$

Consider the second equation. Subtract 12 from both sides. Anything subtracted from zero gives its negation.

$$8a-2b=-12$$

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.

$$6a+9b=0,8a-2b=-12$$

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

$$6a+9b=0$$

Subtract 9b from both sides of the equation.

$$6a=-9b$$

Divide both sides by 6.

$$a=\frac{1}{6}\left(-9\right)b$$

Multiply \frac{1}{6} times -9b.

$$a=-\frac{3}{2}b$$

Substitute -\frac{3b}{2} for a in the other equation, 8a-2b=-12.

$$8\left(-\frac{3}{2}\right)b-2b=-12$$

Multiply 8 times -\frac{3b}{2}.

$$-12b-2b=-12$$

Add -12b to -2b.

$$-14b=-12$$

Divide both sides by -14.

$$b=\frac{6}{7}$$

Substitute \frac{6}{7} for b in a=-\frac{3}{2}b. Because the resulting equation contains only one variable, you can solve for a directly.

$$a=-\frac{3}{2}\times \frac{6}{7}$$

Multiply -\frac{3}{2} times \frac{6}{7} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$a=-\frac{9}{7}$$

The system is now solved.

$$a=-\frac{9}{7},b=\frac{6}{7}$$

Steps Using Matrices

Consider the first equation. Combine 8b and b to get 9b.

$$6a+9b=0$$

Consider the second equation. Subtract 12 from both sides. Anything subtracted from zero gives its negation.

$$8a-2b=-12$$

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

$$6a+9b=0,8a-2b=-12$$

Write the equations in matrix form.

$$\left(\begin{matrix}6&9\\8&-2\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}0\\-12\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}6&9\\8&-2\end{matrix}\right).

$$inverse(\left(\begin{matrix}6&9\\8&-2\end{matrix}\right))\left(\begin{matrix}6&9\\8&-2\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}6&9\\8&-2\end{matrix}\right))\left(\begin{matrix}0\\-12\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}6&9\\8&-2\end{matrix}\right))\left(\begin{matrix}0\\-12\end{matrix}\right)$$

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

$$\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}6&9\\8&-2\end{matrix}\right))\left(\begin{matrix}0\\-12\end{matrix}\right)$$

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.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{6\left(-2\right)-9\times 8}&-\frac{9}{6\left(-2\right)-9\times 8}\\-\frac{8}{6\left(-2\right)-9\times 8}&\frac{6}{6\left(-2\right)-9\times 8}\end{matrix}\right)\left(\begin{matrix}0\\-12\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{1}{42}&\frac{3}{28}\\\frac{2}{21}&-\frac{1}{14}\end{matrix}\right)\left(\begin{matrix}0\\-12\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{3}{28}\left(-12\right)\\-\frac{1}{14}\left(-12\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-\frac{9}{7}\\\frac{6}{7}\end{matrix}\right)$$

Extract the matrix elements a and b.

$$a=-\frac{9}{7},b=\frac{6}{7}$$

Steps Using Elimination

Consider the first equation. Combine 8b and b to get 9b.

$$6a+9b=0$$

Consider the second equation. Subtract 12 from both sides. Anything subtracted from zero gives its negation.

$$8a-2b=-12$$

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.

$$6a+9b=0,8a-2b=-12$$

To make 6a and 8a equal, multiply all terms on each side of the first equation by 8 and all terms on each side of the second by 6.

$$8\times 6a+8\times 9b=0,6\times 8a+6\left(-2\right)b=6\left(-12\right)$$

Simplify.

$$48a+72b=0,48a-12b=-72$$

Subtract 48a-12b=-72 from 48a+72b=0 by subtracting like terms on each side of the equal sign.

$$48a-48a+72b+12b=72$$

Add 48a to -48a. Terms 48a and -48a cancel out, leaving an equation with only one variable that can be solved.

$$72b+12b=72$$

Add 72b to 12b.

$$84b=72$$

Divide both sides by 84.

$$b=\frac{6}{7}$$

Substitute \frac{6}{7} for b in 8a-2b=-12. Because the resulting equation contains only one variable, you can solve for a directly.

$$8a-2\times \frac{6}{7}=-12$$

Multiply -2 times \frac{6}{7}.

$$8a-\frac{12}{7}=-12$$

Add \frac{12}{7} to both sides of the equation.

$$8a=-\frac{72}{7}$$

Divide both sides by 8.

$$a=-\frac{9}{7}$$

The system is now solved.

$$a=-\frac{9}{7},b=\frac{6}{7}$$