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

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

Solve for a, b

$$a=2\text{, }b=-1$$

$$a=-1\text{, }b=2$$

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.

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

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}=5.

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

Square -b+1.

$$b^{2}+b^{2}-2b+1=5$$

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

$$2b^{2}-2b+1=5$$

Subtract 5 from both sides of the equation.

$$2b^{2}-2b-4=0$$

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

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

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

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

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

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

Multiply -8 times -4.

$$b=\frac{-\left(-2\right)±\sqrt{4+32}}{2\times 2}$$

Add 4 to 32.

$$b=\frac{-\left(-2\right)±\sqrt{36}}{2\times 2}$$

Take the square root of 36.

$$b=\frac{-\left(-2\right)±6}{2\times 2}$$

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

$$b=\frac{2±6}{2\times 2}$$

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

$$b=\frac{2±6}{4}$$

Now solve the equation b=\frac{2±6}{4} when ± is plus. Add 2 to 6.

$$b=\frac{8}{4}$$

Divide 8 by 4.

$$b=2$$

Now solve the equation b=\frac{2±6}{4} when ± is minus. Subtract 6 from 2.

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

Divide -4 by 4.

$$b=-1$$

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

$$a=-2+1$$

Add -2 to 1.

$$a=-1$$

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

$$a=-\left(-1\right)+1$$

Multiply -1 times -1.

$$a=1+1$$

Add -\left(-1\right) to 1.

$$a=2$$

The system is now solved.

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