Solve {l}{S+C=1/2}{S^2+C^2=1}

$$\left. \begin{array} { l } { S + C = \frac { 1 } { 2 } } \\ { S ^ { 2 } + C ^ { 2 } = 1 } \end{array} \right.$$

Solve for S, C

$$S=\frac{\sqrt{7}+1}{4}\approx 0.911437828\text{, }C=\frac{1-\sqrt{7}}{4}\approx -0.411437828$$

$$S=\frac{1-\sqrt{7}}{4}\approx -0.411437828\text{, }C=\frac{\sqrt{7}+1}{4}\approx 0.911437828$$

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.

$$S+C=\frac{1}{2},C^{2}+S^{2}=1$$

Solve S+C=\frac{1}{2} for S by isolating S on the left hand side of the equal sign.

$$S+C=\frac{1}{2}$$

Subtract C from both sides of the equation.

$$S=-C+\frac{1}{2}$$

Substitute -C+\frac{1}{2} for S in the other equation, C^{2}+S^{2}=1.

$$C^{2}+\left(-C+\frac{1}{2}\right)^{2}=1$$

Square -C+\frac{1}{2}.

$$C^{2}+C^{2}-C+\frac{1}{4}=1$$

Add C^{2} to C^{2}.

$$2C^{2}-C+\frac{1}{4}=1$$

Subtract 1 from both sides of the equation.

$$2C^{2}-C-\frac{3}{4}=0$$

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

$$C=\frac{-\left(-1\right)±\sqrt{1-4\times 2\left(-\frac{3}{4}\right)}}{2\times 2}$$

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

$$C=\frac{-\left(-1\right)±\sqrt{1-8\left(-\frac{3}{4}\right)}}{2\times 2}$$

Multiply -8 times -\frac{3}{4}.

$$C=\frac{-\left(-1\right)±\sqrt{1+6}}{2\times 2}$$

Add 1 to 6.

$$C=\frac{-\left(-1\right)±\sqrt{7}}{2\times 2}$$

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

$$C=\frac{1±\sqrt{7}}{2\times 2}$$

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

$$C=\frac{1±\sqrt{7}}{4}$$

Now solve the equation C=\frac{1±\sqrt{7}}{4} when ± is plus. Add 1 to \sqrt{7}.

$$C=\frac{\sqrt{7}+1}{4}$$

Now solve the equation C=\frac{1±\sqrt{7}}{4} when ± is minus. Subtract \sqrt{7} from 1.

$$C=\frac{1-\sqrt{7}}{4}$$

There are two solutions for C: \frac{1+\sqrt{7}}{4} and \frac{1-\sqrt{7}}{4}. Substitute \frac{1+\sqrt{7}}{4} for C in the equation S=-C+\frac{1}{2} to find the corresponding solution for S that satisfies both equations.

$$S=-\frac{\sqrt{7}+1}{4}+\frac{1}{2}$$

Now substitute \frac{1-\sqrt{7}}{4} for C in the equation S=-C+\frac{1}{2} and solve to find the corresponding solution for S that satisfies both equations.

$$S=-\frac{1-\sqrt{7}}{4}+\frac{1}{2}$$

The system is now solved.

$$S=-\frac{\sqrt{7}+1}{4}+\frac{1}{2},C=\frac{\sqrt{7}+1}{4}\text{ or }S=-\frac{1-\sqrt{7}}{4}+\frac{1}{2},C=\frac{1-\sqrt{7}}{4}$$