Solve {l}{z+b=7}{z^2+b^2=25}

$$\left. \begin{array} { l } { z + b = 7 } \\ { z ^ { 2 } + b ^ { 2 } = 25 } \end{array} \right.$$

Solve for z, b

$$z=4\text{, }b=3$$

$$z=3\text{, }b=4$$

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.

$$z+b=7,b^{2}+z^{2}=25$$

Solve z+b=7 for z by isolating z on the left hand side of the equal sign.

$$z+b=7$$

Subtract b from both sides of the equation.

$$z=-b+7$$

Substitute -b+7 for z in the other equation, b^{2}+z^{2}=25.

$$b^{2}+\left(-b+7\right)^{2}=25$$

Square -b+7.

$$b^{2}+b^{2}-14b+49=25$$

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

$$2b^{2}-14b+49=25$$

Subtract 25 from both sides of the equation.

$$2b^{2}-14b+24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\times 7\left(-1\right)\times 2 for b, and 24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 2\times 24}}{2\times 2}$$

Square 1\times 7\left(-1\right)\times 2.

$$b=\frac{-\left(-14\right)±\sqrt{196-4\times 2\times 24}}{2\times 2}$$

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

$$b=\frac{-\left(-14\right)±\sqrt{196-8\times 24}}{2\times 2}$$

Multiply -8 times 24.

$$b=\frac{-\left(-14\right)±\sqrt{196-192}}{2\times 2}$$

Add 196 to -192.

$$b=\frac{-\left(-14\right)±\sqrt{4}}{2\times 2}$$

Take the square root of 4.

$$b=\frac{-\left(-14\right)±2}{2\times 2}$$

The opposite of 1\times 7\left(-1\right)\times 2 is 14.

$$b=\frac{14±2}{2\times 2}$$

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

$$b=\frac{14±2}{4}$$

Now solve the equation b=\frac{14±2}{4} when ± is plus. Add 14 to 2.

$$b=\frac{16}{4}$$

Divide 16 by 4.

$$b=4$$

Now solve the equation b=\frac{14±2}{4} when ± is minus. Subtract 2 from 14.

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

Divide 12 by 4.

$$b=3$$

There are two solutions for b: 4 and 3. Substitute 4 for b in the equation z=-b+7 to find the corresponding solution for z that satisfies both equations.

$$z=-4+7$$

Add -4 to 7.

$$z=3$$

Now substitute 3 for b in the equation z=-b+7 and solve to find the corresponding solution for z that satisfies both equations.

$$z=-3+7$$

Add -3 to 7.

$$z=4$$

The system is now solved.

$$z=3,b=4\text{ or }z=4,b=3$$