Solve {l}{9a-3b=6}{12a=9b+3}

$$\left. \begin{array} { l } { 9 a - 3 b = 6 } \\ { 12 a = 9 b + 3 } \end{array} \right.$$

Solve for a, b

$$a=1$$

$$b=1$$

Steps Using Substitution

Consider the second equation. Subtract 9b from both sides.

$$12a-9b=3$$

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.

$$9a-3b=6,12a-9b=3$$

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

$$9a-3b=6$$

Add 3b to both sides of the equation.

$$9a=3b+6$$

Divide both sides by 9.

$$a=\frac{1}{9}\left(3b+6\right)$$

Multiply \frac{1}{9} times 6+3b.

$$a=\frac{1}{3}b+\frac{2}{3}$$

Substitute \frac{2+b}{3} for a in the other equation, 12a-9b=3.

$$12\left(\frac{1}{3}b+\frac{2}{3}\right)-9b=3$$

Multiply 12 times \frac{2+b}{3}.

$$4b+8-9b=3$$

Add 4b to -9b.

$$-5b+8=3$$

Subtract 8 from both sides of the equation.

$$-5b=-5$$

Divide both sides by -5.

$$b=1$$

Substitute 1 for b in a=\frac{1}{3}b+\frac{2}{3}. Because the resulting equation contains only one variable, you can solve for a directly.

$$a=\frac{1+2}{3}$$

Add \frac{2}{3} to \frac{1}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a=1$$

The system is now solved.

$$a=1,b=1$$

Steps Using Matrices

Consider the second equation. Subtract 9b from both sides.

$$12a-9b=3$$

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

$$9a-3b=6,12a-9b=3$$

Write the equations in matrix form.

$$\left(\begin{matrix}9&-3\\12&-9\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}6\\3\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}9&-3\\12&-9\end{matrix}\right).

$$inverse(\left(\begin{matrix}9&-3\\12&-9\end{matrix}\right))\left(\begin{matrix}9&-3\\12&-9\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}9&-3\\12&-9\end{matrix}\right))\left(\begin{matrix}6\\3\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}9&-3\\12&-9\end{matrix}\right))\left(\begin{matrix}6\\3\end{matrix}\right)$$

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

$$\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}9&-3\\12&-9\end{matrix}\right))\left(\begin{matrix}6\\3\end{matrix}\right)$$

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.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-\frac{9}{9\left(-9\right)-\left(-3\times 12\right)}&-\frac{-3}{9\left(-9\right)-\left(-3\times 12\right)}\\-\frac{12}{9\left(-9\right)-\left(-3\times 12\right)}&\frac{9}{9\left(-9\right)-\left(-3\times 12\right)}\end{matrix}\right)\left(\begin{matrix}6\\3\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{1}{5}&-\frac{1}{15}\\\frac{4}{15}&-\frac{1}{5}\end{matrix}\right)\left(\begin{matrix}6\\3\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{1}{5}\times 6-\frac{1}{15}\times 3\\\frac{4}{15}\times 6-\frac{1}{5}\times 3\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}1\\1\end{matrix}\right)$$

Extract the matrix elements a and b.

$$a=1,b=1$$

Steps Using Elimination

Consider the second equation. Subtract 9b from both sides.

$$12a-9b=3$$

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.

$$9a-3b=6,12a-9b=3$$

To make 9a and 12a equal, multiply all terms on each side of the first equation by 12 and all terms on each side of the second by 9.

$$12\times 9a+12\left(-3\right)b=12\times 6,9\times 12a+9\left(-9\right)b=9\times 3$$

Simplify.

$$108a-36b=72,108a-81b=27$$

Subtract 108a-81b=27 from 108a-36b=72 by subtracting like terms on each side of the equal sign.

$$108a-108a-36b+81b=72-27$$

Add 108a to -108a. Terms 108a and -108a cancel out, leaving an equation with only one variable that can be solved.

$$-36b+81b=72-27$$

Add -36b to 81b.

$$45b=72-27$$

Add 72 to -27.

$$45b=45$$

Divide both sides by 45.

$$b=1$$

Substitute 1 for b in 12a-9b=3. Because the resulting equation contains only one variable, you can solve for a directly.

$$12a-9=3$$

Add 9 to both sides of the equation.

$$12a=12$$

Divide both sides by 12.

$$a=1$$

The system is now solved.

$$a=1,b=1$$