Solve {l}{3a+2b=6}{15a+3b=-33}

$$\left. \begin{array} { l } { 3 a + 2 b = 6 } \\ { 15 a + 3 b = - 33 } \end{array} \right.$$

Solve for a, b

$$a=-4$$

$$b=9$$

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.

$$3a+2b=6,15a+3b=-33$$

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

$$3a+2b=6$$

Subtract 2b from both sides of the equation.

$$3a=-2b+6$$

Divide both sides by 3.

$$a=\frac{1}{3}\left(-2b+6\right)$$

Multiply \frac{1}{3} times -2b+6.

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

Substitute -\frac{2b}{3}+2 for a in the other equation, 15a+3b=-33.

$$15\left(-\frac{2}{3}b+2\right)+3b=-33$$

Multiply 15 times -\frac{2b}{3}+2.

$$-10b+30+3b=-33$$

Add -10b to 3b.

$$-7b+30=-33$$

Subtract 30 from both sides of the equation.

$$-7b=-63$$

Divide both sides by -7.

$$b=9$$

Substitute 9 for b in a=-\frac{2}{3}b+2. Because the resulting equation contains only one variable, you can solve for a directly.

$$a=-\frac{2}{3}\times 9+2$$

Multiply -\frac{2}{3} times 9.

$$a=-6+2$$

Add 2 to -6.

$$a=-4$$

The system is now solved.

$$a=-4,b=9$$

Steps Using Matrices

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

$$3a+2b=6,15a+3b=-33$$

Write the equations in matrix form.

$$\left(\begin{matrix}3&2\\15&3\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}6\\-33\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}3&2\\15&3\end{matrix}\right).

$$inverse(\left(\begin{matrix}3&2\\15&3\end{matrix}\right))\left(\begin{matrix}3&2\\15&3\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}3&2\\15&3\end{matrix}\right))\left(\begin{matrix}6\\-33\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}3&2\\15&3\end{matrix}\right))\left(\begin{matrix}6\\-33\end{matrix}\right)$$

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

$$\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}3&2\\15&3\end{matrix}\right))\left(\begin{matrix}6\\-33\end{matrix}\right)$$

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.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{3}{3\times 3-2\times 15}&-\frac{2}{3\times 3-2\times 15}\\-\frac{15}{3\times 3-2\times 15}&\frac{3}{3\times 3-2\times 15}\end{matrix}\right)\left(\begin{matrix}6\\-33\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{7}&\frac{2}{21}\\\frac{5}{7}&-\frac{1}{7}\end{matrix}\right)\left(\begin{matrix}6\\-33\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{7}\times 6+\frac{2}{21}\left(-33\right)\\\frac{5}{7}\times 6-\frac{1}{7}\left(-33\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-4\\9\end{matrix}\right)$$

Extract the matrix elements a and b.

$$a=-4,b=9$$

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.

$$3a+2b=6,15a+3b=-33$$

To make 3a and 15a equal, multiply all terms on each side of the first equation by 15 and all terms on each side of the second by 3.

$$15\times 3a+15\times 2b=15\times 6,3\times 15a+3\times 3b=3\left(-33\right)$$

Simplify.

$$45a+30b=90,45a+9b=-99$$

Subtract 45a+9b=-99 from 45a+30b=90 by subtracting like terms on each side of the equal sign.

$$45a-45a+30b-9b=90+99$$

Add 45a to -45a. Terms 45a and -45a cancel out, leaving an equation with only one variable that can be solved.

$$30b-9b=90+99$$

Add 30b to -9b.

$$21b=90+99$$

Add 90 to 99.

$$21b=189$$

Divide both sides by 21.

$$b=9$$

Substitute 9 for b in 15a+3b=-33. Because the resulting equation contains only one variable, you can solve for a directly.

$$15a+3\times 9=-33$$

Multiply 3 times 9.

$$15a+27=-33$$

Subtract 27 from both sides of the equation.

$$15a=-60$$

Divide both sides by 15.

$$a=-4$$

The system is now solved.

$$a=-4,b=9$$