Solve {l}{a^2+b^2=0.5}{a-1=-b}

$$\left. \begin{array} { l } { a ^ { 2 } + b ^ { 2 } = 0.5 } \\ { a - 1 = - b } \end{array} \right.$$

Solve for a, b

$$a=0.5$$

$$b=0.5$$

Steps Using Substitution and Quadratic Formula

Consider the second equation. Add b to both sides.

$$a-1+b=0$$

Add 1 to both sides. Anything plus zero gives itself.

$$a+b=1$$

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.

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

Solve a+b=1 for a by isolating a on the left hand side of the equal sign.

$$a+b=1$$

Subtract b from both sides of the equation.

$$a=-b+1$$

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

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

Square -b+1.

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

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

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

Subtract \frac{1}{2} from both sides of the equation.

$$2b^{2}-2b+\frac{1}{2}=0$$

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

$$b=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 2\times \frac{1}{2}}}{2\times 2}$$

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

$$b=\frac{-\left(-2\right)±\sqrt{4-4\times 2\times \frac{1}{2}}}{2\times 2}$$

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

$$b=\frac{-\left(-2\right)±\sqrt{4-8\times \frac{1}{2}}}{2\times 2}$$

Multiply -8 times \frac{1}{2}.

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

Add 4 to -4.

$$b=\frac{-\left(-2\right)±\sqrt{0}}{2\times 2}$$

Take the square root of 0.

$$b=-\frac{-2}{2\times 2}$$

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

$$b=\frac{2}{2\times 2}$$

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

$$b=\frac{2}{4}$$

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

$$b=\frac{1}{2}$$

There are two solutions for b: \frac{1}{2} and \frac{1}{2}. Substitute \frac{1}{2} for b in the equation a=-b+1 to find the corresponding solution for a that satisfies both equations.

$$a=-\frac{1}{2}+1$$

Add -\frac{1}{2} to 1.

$$a=\frac{1}{2}$$

The system is now solved.

$$a=\frac{1}{2},b=\frac{1}{2}\text{ or }a=\frac{1}{2},b=\frac{1}{2}$$