Solve {l}{l^2+z^2=34}{l+z=8}

$$\left. \begin{array} { l } { l ^ { 2 } + z ^ { 2 } = 34 } \\ { l + z = 8 } \end{array} \right.$$

Solve for l, z

$$z=5\text{, }l=3$$

$$z=3\text{, }l=5$$

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.

$$l+z=8,z^{2}+l^{2}=34$$

Solve l+z=8 for l by isolating l on the left hand side of the equal sign.

$$l+z=8$$

Subtract z from both sides of the equation.

$$l=-z+8$$

Substitute -z+8 for l in the other equation, z^{2}+l^{2}=34.

$$z^{2}+\left(-z+8\right)^{2}=34$$

Square -z+8.

$$z^{2}+z^{2}-16z+64=34$$

Add z^{2} to z^{2}.

$$2z^{2}-16z+64=34$$

Subtract 34 from both sides of the equation.

$$2z^{2}-16z+30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\times 8\left(-1\right)\times 2 for b, and 30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$z=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 2\times 30}}{2\times 2}$$

Square 1\times 8\left(-1\right)\times 2.

$$z=\frac{-\left(-16\right)±\sqrt{256-4\times 2\times 30}}{2\times 2}$$

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

$$z=\frac{-\left(-16\right)±\sqrt{256-8\times 30}}{2\times 2}$$

Multiply -8 times 30.

$$z=\frac{-\left(-16\right)±\sqrt{256-240}}{2\times 2}$$

Add 256 to -240.

$$z=\frac{-\left(-16\right)±\sqrt{16}}{2\times 2}$$

Take the square root of 16.

$$z=\frac{-\left(-16\right)±4}{2\times 2}$$

The opposite of 1\times 8\left(-1\right)\times 2 is 16.

$$z=\frac{16±4}{2\times 2}$$

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

$$z=\frac{16±4}{4}$$

Now solve the equation z=\frac{16±4}{4} when ± is plus. Add 16 to 4.

$$z=\frac{20}{4}$$

Divide 20 by 4.

$$z=5$$

Now solve the equation z=\frac{16±4}{4} when ± is minus. Subtract 4 from 16.

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

Divide 12 by 4.

$$z=3$$

There are two solutions for z: 5 and 3. Substitute 5 for z in the equation l=-z+8 to find the corresponding solution for l that satisfies both equations.

$$l=-5+8$$

Add -5 to 8.

$$l=3$$

Now substitute 3 for z in the equation l=-z+8 and solve to find the corresponding solution for l that satisfies both equations.

$$l=-3+8$$

Add -3 to 8.

$$l=5$$

The system is now solved.

$$l=3,z=5\text{ or }l=5,z=3$$