Solve {l}{a^2+b^2=100}{a+b=20}

$$\left\{ \begin{array} { l } { a ^ { 2 } + b ^ { 2 } = 100 } \\ { a + b = 20 } \end{array} \right.$$

Solve for a, b

$$a=10+5\sqrt{2}i\approx 10+7.071067812i\text{, }b=-5\sqrt{2}i+10\approx 10-7.071067812i$$

$$a=-5\sqrt{2}i+10\approx 10-7.071067812i\text{, }b=10+5\sqrt{2}i\approx 10+7.071067812i$$

Steps Using Substitution and Quadratic Formula

Solve a+b=20 for a by isolating a on the left hand side of the equal sign.

$$a+b=20$$

Subtract b from both sides of the equation.

$$a=-b+20$$

Substitute -b+20 for a in the other equation, b^{2}+a^{2}=100.

$$b^{2}+\left(-b+20\right)^{2}=100$$

Square -b+20.

$$b^{2}+b^{2}-40b+400=100$$

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

$$2b^{2}-40b+400=100$$

Subtract 100 from both sides of the equation.

$$2b^{2}-40b+300=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\times 20\left(-1\right)\times 2 for b, and 300 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\times 2\times 300}}{2\times 2}$$

Square 1\times 20\left(-1\right)\times 2.

$$b=\frac{-\left(-40\right)±\sqrt{1600-4\times 2\times 300}}{2\times 2}$$

Multiply -4 times 1+1\left(-1\right)^{2}.

$$b=\frac{-\left(-40\right)±\sqrt{1600-8\times 300}}{2\times 2}$$

Multiply -8 times 300.

$$b=\frac{-\left(-40\right)±\sqrt{1600-2400}}{2\times 2}$$

Add 1600 to -2400.

$$b=\frac{-\left(-40\right)±\sqrt{-800}}{2\times 2}$$

Take the square root of -800.

$$b=\frac{-\left(-40\right)±20\sqrt{2}i}{2\times 2}$$

The opposite of 1\times 20\left(-1\right)\times 2 is 40.

$$b=\frac{40±20\sqrt{2}i}{2\times 2}$$

Multiply 2 times 1+1\left(-1\right)^{2}.

$$b=\frac{40±20\sqrt{2}i}{4}$$

Now solve the equation b=\frac{40±20\sqrt{2}i}{4} when ± is plus. Add 40 to 20i\sqrt{2}.

$$b=\frac{40+20\sqrt{2}i}{4}$$

Divide 40+20i\sqrt{2} by 4.

$$b=10+5\sqrt{2}i$$

Now solve the equation b=\frac{40±20\sqrt{2}i}{4} when ± is minus. Subtract 20i\sqrt{2} from 40.

$$b=\frac{-20\sqrt{2}i+40}{4}$$

Divide 40-20i\sqrt{2} by 4.

$$b=-5\sqrt{2}i+10$$

There are two solutions for b: 10+5i\sqrt{2} and 10-5i\sqrt{2}. Substitute 10+5i\sqrt{2} for b in the equation a=-b+20 to find the corresponding solution for a that satisfies both equations.

$$a=-\left(10+5\sqrt{2}i\right)+20$$

Now substitute 10-5i\sqrt{2} for b in the equation a=-b+20 and solve to find the corresponding solution for a that satisfies both equations.

$$a=-\left(-5\sqrt{2}i+10\right)+20$$

The system is now solved.

$$a=-\left(10+5\sqrt{2}i\right)+20,b=10+5\sqrt{2}i\text{ or }a=-\left(-5\sqrt{2}i+10\right)+20,b=-5\sqrt{2}i+10$$