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

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

Solve for a, b

$$a=0\text{, }b=1$$

$$a=1\text{, }b=0$$

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.

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

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}=1.

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

Square -b+1.

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

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

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

Subtract 1 from both sides of the equation.

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

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 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

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

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

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

$$b=\frac{2±2}{2\times 2}$$

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

$$b=\frac{2±2}{4}$$

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

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

Divide 4 by 4.

$$b=1$$

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

$$b=\frac{0}{4}$$

Divide 0 by 4.

$$b=0$$

There are two solutions for b: 1 and 0. Substitute 1 for b in the equation a=-b+1 to find the corresponding solution for a that satisfies both equations.

$$a=-1+1$$

Add -1 to 1.

$$a=0$$

Now substitute 0 for b in the equation a=-b+1 and solve to find the corresponding solution for a that satisfies both equations.

$$a=1$$

The system is now solved.

$$a=0,b=1\text{ or }a=1,b=0$$