Solve {l}{b^2+c^2=169}{3b+4c=0}

$$\left. \begin{array} { l } { b ^ { 2 } + c ^ { 2 } = 169 } \\ { 3 b + 4 c = 0 } \end{array} \right.$$

Solve for b, c

$$b=-\frac{52}{5}=-10.4\text{, }c=\frac{39}{5}=7.8$$

$$b=\frac{52}{5}=10.4\text{, }c=-\frac{39}{5}=-7.8$$

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.

$$3b+4c=0,c^{2}+b^{2}=169$$

Solve 3b+4c=0 for b by isolating b on the left hand side of the equal sign.

$$3b+4c=0$$

Subtract 4c from both sides of the equation.

$$3b=-4c$$

Divide both sides by 3.

$$b=-\frac{4}{3}c$$

Substitute -\frac{4}{3}c for b in the other equation, c^{2}+b^{2}=169.

$$c^{2}+\left(-\frac{4}{3}c\right)^{2}=169$$

Square -\frac{4}{3}c.

$$c^{2}+\frac{16}{9}c^{2}=169$$

Add c^{2} to \frac{16}{9}c^{2}.

$$\frac{25}{9}c^{2}=169$$

Subtract 169 from both sides of the equation.

$$\frac{25}{9}c^{2}-169=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-\frac{4}{3}\right)^{2} for a, 1\times 0\left(-\frac{4}{3}\right)\times 2 for b, and -169 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$c=\frac{0±\sqrt{0^{2}-4\times \frac{25}{9}\left(-169\right)}}{2\times \frac{25}{9}}$$

Square 1\times 0\left(-\frac{4}{3}\right)\times 2.

$$c=\frac{0±\sqrt{-4\times \frac{25}{9}\left(-169\right)}}{2\times \frac{25}{9}}$$

Multiply -4 times 1+1\left(-\frac{4}{3}\right)^{2}.

$$c=\frac{0±\sqrt{-\frac{100}{9}\left(-169\right)}}{2\times \frac{25}{9}}$$

Multiply -\frac{100}{9} times -169.

$$c=\frac{0±\sqrt{\frac{16900}{9}}}{2\times \frac{25}{9}}$$

Take the square root of \frac{16900}{9}.

$$c=\frac{0±\frac{130}{3}}{2\times \frac{25}{9}}$$

Multiply 2 times 1+1\left(-\frac{4}{3}\right)^{2}.

$$c=\frac{0±\frac{130}{3}}{\frac{50}{9}}$$

Now solve the equation c=\frac{0±\frac{130}{3}}{\frac{50}{9}} when ± is plus.

$$c=\frac{39}{5}$$

Now solve the equation c=\frac{0±\frac{130}{3}}{\frac{50}{9}} when ± is minus.

$$c=-\frac{39}{5}$$

There are two solutions for c: \frac{39}{5} and -\frac{39}{5}. Substitute \frac{39}{5} for c in the equation b=-\frac{4}{3}c to find the corresponding solution for b that satisfies both equations.

$$b=-\frac{4}{3}\times \frac{39}{5}$$

Multiply -\frac{4}{3} times \frac{39}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$b=-\frac{52}{5}$$

Now substitute -\frac{39}{5} for c in the equation b=-\frac{4}{3}c and solve to find the corresponding solution for b that satisfies both equations.

$$b=-\frac{4}{3}\left(-\frac{39}{5}\right)$$

Multiply -\frac{4}{3} times -\frac{39}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$b=\frac{52}{5}$$

The system is now solved.

$$b=-\frac{52}{5},c=\frac{39}{5}\text{ or }b=\frac{52}{5},c=-\frac{39}{5}$$