Solve {l}{a+b=3}{a^2+b^2=5}

$$\left. \begin{array} { l } { a + b = 3 } \\ { a ^ { 2 } + b ^ { 2 } = 5 } \end{array} \right.$$

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.

$$a+b=3,b^{2}+a^{2}=5$$

Solve a+b=3 for a by isolating a on the left hand side of the equal sign.

$$a+b=3$$

Subtract b from both sides of the equation.

$$a=-b+3$$

Substitute -b+3 for a in the other equation, b^{2}+a^{2}=5.

$$b^{2}+\left(-b+3\right)^{2}=5$$

Square -b+3.

$$b^{2}+b^{2}-6b+9=5$$

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

$$2b^{2}-6b+9=5$$

Subtract 5 from both sides of the equation.

$$2b^{2}-6b+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\times 3\left(-1\right)\times 2 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 2\times 4}}{2\times 2}$$

Square 1\times 3\left(-1\right)\times 2.

$$b=\frac{-\left(-6\right)±\sqrt{36-4\times 2\times 4}}{2\times 2}$$

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

$$b=\frac{-\left(-6\right)±\sqrt{36-8\times 4}}{2\times 2}$$

Multiply -8 times 4.

$$b=\frac{-\left(-6\right)±\sqrt{36-32}}{2\times 2}$$

Add 36 to -32.

$$b=\frac{-\left(-6\right)±\sqrt{4}}{2\times 2}$$

Take the square root of 4.

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

The opposite of 1\times 3\left(-1\right)\times 2 is 6.

$$b=\frac{6±2}{2\times 2}$$

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

$$b=\frac{6±2}{4}$$

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

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

Divide 8 by 4.

$$b=2$$

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

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

Divide 4 by 4.

$$b=1$$

There are two solutions for b: 2 and 1. Substitute 2 for b in the equation a=-b+3 to find the corresponding solution for a that satisfies both equations.

$$a=-2+3$$

Add -2 to 3.

$$a=1$$

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

$$a=-1+3$$

Add -1 to 3.

$$a=2$$

The system is now solved.

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