Solve {l}{a^2+b^2=13}{a+b=5}

$$\left. \begin{array} { l } { a ^ { 2 } + b ^ { 2 } = 13 } \\ { a + b = 5 } \end{array} \right.$$

Solve for a, b

$$a=3\text{, }b=2$$

$$a=2\text{, }b=3$$

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.

$$a+b=5,b^{2}+a^{2}=13$$

Solve a+b=5 for a by isolating a on the left hand side of the equal sign.

$$a+b=5$$

Subtract b from both sides of the equation.

$$a=-b+5$$

Substitute -b+5 for a in the other equation, b^{2}+a^{2}=13.

$$b^{2}+\left(-b+5\right)^{2}=13$$

Square -b+5.

$$b^{2}+b^{2}-10b+25=13$$

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

$$2b^{2}-10b+25=13$$

Subtract 13 from both sides of the equation.

$$2b^{2}-10b+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\times 5\left(-1\right)\times 2 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 2\times 12}}{2\times 2}$$

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

$$b=\frac{-\left(-10\right)±\sqrt{100-4\times 2\times 12}}{2\times 2}$$

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

$$b=\frac{-\left(-10\right)±\sqrt{100-8\times 12}}{2\times 2}$$

Multiply -8 times 12.

$$b=\frac{-\left(-10\right)±\sqrt{100-96}}{2\times 2}$$

Add 100 to -96.

$$b=\frac{-\left(-10\right)±\sqrt{4}}{2\times 2}$$

Take the square root of 4.

$$b=\frac{-\left(-10\right)±2}{2\times 2}$$

The opposite of 1\times 5\left(-1\right)\times 2 is 10.

$$b=\frac{10±2}{2\times 2}$$

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

$$b=\frac{10±2}{4}$$

Now solve the equation b=\frac{10±2}{4} when ± is plus. Add 10 to 2.

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

Divide 12 by 4.

$$b=3$$

Now solve the equation b=\frac{10±2}{4} when ± is minus. Subtract 2 from 10.

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

Divide 8 by 4.

$$b=2$$

There are two solutions for b: 3 and 2. Substitute 3 for b in the equation a=-b+5 to find the corresponding solution for a that satisfies both equations.

$$a=-3+5$$

Add -3 to 5.

$$a=2$$

Now substitute 2 for b in the equation a=-b+5 and solve to find the corresponding solution for a that satisfies both equations.

$$a=-2+5$$

Add -2 to 5.

$$a=3$$

The system is now solved.

$$a=2,b=3\text{ or }a=3,b=2$$