Solve {l}{a^2+b^2=5}{a-b=6}

$$\left. \begin{array} { l } { a ^ { 2 } + b ^ { 2 } = 5 } \\ { a - b = 6 } \end{array} \right.$$

Solve for a, b

$$a=-\frac{\sqrt{26}i}{2}+3\approx 3-2.549509757i\text{, }b=-\frac{\sqrt{26}i}{2}-3\approx -3-2.549509757i$$

$$a=\frac{\sqrt{26}i}{2}+3\approx 3+2.549509757i\text{, }b=\frac{\sqrt{26}i}{2}-3\approx -3+2.549509757i$$

Steps Using Substitution and Quadratic Formula

Solve a-b=6 for a by isolating a on the left hand side of the equal sign.

$$a-b=6$$

Subtract -b from both sides of the equation.

$$a=b+6$$

Substitute b+6 for a in the other equation, b^{2}+a^{2}=5.

$$b^{2}+\left(b+6\right)^{2}=5$$

Square b+6.

$$b^{2}+b^{2}+12b+36=5$$

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

$$2b^{2}+12b+36=5$$

Subtract 5 from both sides of the equation.

$$2b^{2}+12b+31=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times 1^{2} for a, 1\times 6\times 1\times 2 for b, and 31 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-12±\sqrt{12^{2}-4\times 2\times 31}}{2\times 2}$$

Square 1\times 6\times 1\times 2.

$$b=\frac{-12±\sqrt{144-4\times 2\times 31}}{2\times 2}$$

Multiply -4 times 1+1\times 1^{2}.

$$b=\frac{-12±\sqrt{144-8\times 31}}{2\times 2}$$

Multiply -8 times 31.

$$b=\frac{-12±\sqrt{144-248}}{2\times 2}$$

Add 144 to -248.

$$b=\frac{-12±\sqrt{-104}}{2\times 2}$$

Take the square root of -104.

$$b=\frac{-12±2\sqrt{26}i}{2\times 2}$$

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

$$b=\frac{-12±2\sqrt{26}i}{4}$$

Now solve the equation b=\frac{-12±2\sqrt{26}i}{4} when ± is plus. Add -12 to 2i\sqrt{26}.

$$b=\frac{-12+2\sqrt{26}i}{4}$$

Divide -12+2i\sqrt{26} by 4.

$$b=\frac{\sqrt{26}i}{2}-3$$

Now solve the equation b=\frac{-12±2\sqrt{26}i}{4} when ± is minus. Subtract 2i\sqrt{26} from -12.

$$b=\frac{-2\sqrt{26}i-12}{4}$$

Divide -12-2i\sqrt{26} by 4.

$$b=-\frac{\sqrt{26}i}{2}-3$$

There are two solutions for b: -3+\frac{i\sqrt{26}}{2} and -3-\frac{i\sqrt{26}}{2}. Substitute -3+\frac{i\sqrt{26}}{2} for b in the equation a=b+6 to find the corresponding solution for a that satisfies both equations.

$$a=\frac{\sqrt{26}i}{2}-3+6$$

Add 1\left(-3+\frac{i\sqrt{26}}{2}\right) to 6.

$$a=\frac{\sqrt{26}i}{2}+3$$

Now substitute -3-\frac{i\sqrt{26}}{2} for b in the equation a=b+6 and solve to find the corresponding solution for a that satisfies both equations.

$$a=-\frac{\sqrt{26}i}{2}-3+6$$

Add 1\left(-3-\frac{i\sqrt{26}}{2}\right) to 6.

$$a=-\frac{\sqrt{26}i}{2}+3$$

The system is now solved.

$$a=\frac{\sqrt{26}i}{2}+3,b=\frac{\sqrt{26}i}{2}-3\text{ or }a=-\frac{\sqrt{26}i}{2}+3,b=-\frac{\sqrt{26}i}{2}-3$$