Solve {l}{a=2b-3}{a^2=b^2+9}

$$\left\{ \begin{array} { l } { a = 2 b - 3 } \\ { a ^ { 2 } = b ^ { 2 } + 9 } \end{array} \right.$$

Solve for a, b

$$a=5\text{, }b=4$$

$$a=-3\text{, }b=0$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Subtract 2b from both sides.

$$a-2b=-3$$

Consider the second equation. Subtract b^{2} from both sides.

$$a^{2}-b^{2}=9$$

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.

$$a-2b=-3,-b^{2}+a^{2}=9$$

Solve a-2b=-3 for a by isolating a on the left hand side of the equal sign.

$$a-2b=-3$$

Subtract -2b from both sides of the equation.

$$a=2b-3$$

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

$$-b^{2}+\left(2b-3\right)^{2}=9$$

Square 2b-3.

$$-b^{2}+4b^{2}-12b+9=9$$

Add -b^{2} to 4b^{2}.

$$3b^{2}-12b+9=9$$

Subtract 9 from both sides of the equation.

$$3b^{2}-12b=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1+1\times 2^{2} for a, 1\left(-3\right)\times 2\times 2 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}}}{2\times 3}$$

Take the square root of \left(-12\right)^{2}.

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

The opposite of 1\left(-3\right)\times 2\times 2 is 12.

$$b=\frac{12±12}{2\times 3}$$

Multiply 2 times -1+1\times 2^{2}.

$$b=\frac{12±12}{6}$$

Now solve the equation b=\frac{12±12}{6} when ± is plus. Add 12 to 12.

$$b=\frac{24}{6}$$

Divide 24 by 6.

$$b=4$$

Now solve the equation b=\frac{12±12}{6} when ± is minus. Subtract 12 from 12.

$$b=\frac{0}{6}$$

Divide 0 by 6.

$$b=0$$

There are two solutions for b: 4 and 0. Substitute 4 for b in the equation a=2b-3 to find the corresponding solution for a that satisfies both equations.

$$a=2\times 4-3$$

Multiply 2 times 4.

$$a=8-3$$

Add 2\times 4 to -3.

$$a=5$$

Now substitute 0 for b in the equation a=2b-3 and solve to find the corresponding solution for a that satisfies both equations.

$$a=-3$$

The system is now solved.

$$a=5,b=4\text{ or }a=-3,b=0$$