Solve {l}{a+b=1/5}{a^2+b^2=1}

$$\left. \begin{array} { l } { a + b = \frac { 1 } { 5 } } \\ { a ^ { 2 } + b ^ { 2 } = 1 } \end{array} \right.$$

Solve for a, b

$$a=\frac{4}{5}=0.8\text{, }b=-\frac{3}{5}=-0.6$$

$$a=-\frac{3}{5}=-0.6\text{, }b=\frac{4}{5}=0.8$$

Steps Using Substitution and Quadratic Formula

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.

$$a+b=\frac{1}{5},b^{2}+a^{2}=1$$

Solve a+b=\frac{1}{5} for a by isolating a on the left hand side of the equal sign.

$$a+b=\frac{1}{5}$$

Subtract b from both sides of the equation.

$$a=-b+\frac{1}{5}$$

Substitute -b+\frac{1}{5} for a in the other equation, b^{2}+a^{2}=1.

$$b^{2}+\left(-b+\frac{1}{5}\right)^{2}=1$$

Square -b+\frac{1}{5}.

$$b^{2}+b^{2}-\frac{2}{5}b+\frac{1}{25}=1$$

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

$$2b^{2}-\frac{2}{5}b+\frac{1}{25}=1$$

Subtract 1 from both sides of the equation.

$$2b^{2}-\frac{2}{5}b-\frac{24}{25}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\times \frac{1}{5}\left(-1\right)\times 2 for b, and -\frac{24}{25} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-\left(-\frac{2}{5}\right)±\sqrt{\left(-\frac{2}{5}\right)^{2}-4\times 2\left(-\frac{24}{25}\right)}}{2\times 2}$$

Square 1\times \frac{1}{5}\left(-1\right)\times 2.

$$b=\frac{-\left(-\frac{2}{5}\right)±\sqrt{\frac{4}{25}-4\times 2\left(-\frac{24}{25}\right)}}{2\times 2}$$

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

$$b=\frac{-\left(-\frac{2}{5}\right)±\sqrt{\frac{4}{25}-8\left(-\frac{24}{25}\right)}}{2\times 2}$$

Multiply -8 times -\frac{24}{25}.

$$b=\frac{-\left(-\frac{2}{5}\right)±\sqrt{\frac{4+192}{25}}}{2\times 2}$$

Add \frac{4}{25} to \frac{192}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$b=\frac{-\left(-\frac{2}{5}\right)±\sqrt{\frac{196}{25}}}{2\times 2}$$

Take the square root of \frac{196}{25}.

$$b=\frac{-\left(-\frac{2}{5}\right)±\frac{14}{5}}{2\times 2}$$

The opposite of 1\times \frac{1}{5}\left(-1\right)\times 2 is \frac{2}{5}.

$$b=\frac{\frac{2}{5}±\frac{14}{5}}{2\times 2}$$

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

$$b=\frac{\frac{2}{5}±\frac{14}{5}}{4}$$

Now solve the equation b=\frac{\frac{2}{5}±\frac{14}{5}}{4} when ± is plus. Add \frac{2}{5} to \frac{14}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$b=\frac{\frac{16}{5}}{4}$$

Divide \frac{16}{5} by 4.

$$b=\frac{4}{5}$$

Now solve the equation b=\frac{\frac{2}{5}±\frac{14}{5}}{4} when ± is minus. Subtract \frac{14}{5} from \frac{2}{5} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$b=-\frac{\frac{12}{5}}{4}$$

Divide -\frac{12}{5} by 4.

$$b=-\frac{3}{5}$$

There are two solutions for b: \frac{4}{5} and -\frac{3}{5}. Substitute \frac{4}{5} for b in the equation a=-b+\frac{1}{5} to find the corresponding solution for a that satisfies both equations.

$$a=-\frac{4}{5}+\frac{1}{5}$$

Multiply -1 times \frac{4}{5}.

$$a=\frac{-4+1}{5}$$

Add -\frac{4}{5} to \frac{1}{5}.

$$a=-\frac{3}{5}$$

Now substitute -\frac{3}{5} for b in the equation a=-b+\frac{1}{5} and solve to find the corresponding solution for a that satisfies both equations.

$$a=-\left(-\frac{3}{5}\right)+\frac{1}{5}$$

Multiply -1 times -\frac{3}{5}.

$$a=\frac{3+1}{5}$$

Add -\left(-\frac{3}{5}\right) to \frac{1}{5}.

$$a=\frac{4}{5}$$

The system is now solved.

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