Solve {l}{3a+4b=0}{a^2+b^2=1}

$$\left. \begin{array} { l } { 3 a + 4 b = 0 } \\ { a ^ { 2 } + b ^ { 2 } = 1 } \end{array} \right.$$

Solve for a, b

$$a=-\frac{4}{5}=-0.8\text{, }b=\frac{3}{5}=0.6$$

$$a=\frac{4}{5}=0.8\text{, }b=-\frac{3}{5}=-0.6$$

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.

$$3a+4b=0,b^{2}+a^{2}=1$$

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

$$3a+4b=0$$

Subtract 4b from both sides of the equation.

$$3a=-4b$$

Divide both sides by 3.

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

Substitute -\frac{4}{3}b for a in the other equation, b^{2}+a^{2}=1.

$$b^{2}+\left(-\frac{4}{3}b\right)^{2}=1$$

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

$$b^{2}+\frac{16}{9}b^{2}=1$$

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

$$\frac{25}{9}b^{2}=1$$

Subtract 1 from both sides of the equation.

$$\frac{25}{9}b^{2}-1=0$$

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 -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{0±\sqrt{0^{2}-4\times \frac{25}{9}\left(-1\right)}}{2\times \frac{25}{9}}$$

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

$$b=\frac{0±\sqrt{-4\times \frac{25}{9}\left(-1\right)}}{2\times \frac{25}{9}}$$

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

$$b=\frac{0±\sqrt{-\frac{100}{9}\left(-1\right)}}{2\times \frac{25}{9}}$$

Multiply -\frac{100}{9} times -1.

$$b=\frac{0±\sqrt{\frac{100}{9}}}{2\times \frac{25}{9}}$$

Take the square root of \frac{100}{9}.

$$b=\frac{0±\frac{10}{3}}{2\times \frac{25}{9}}$$

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

$$b=\frac{0±\frac{10}{3}}{\frac{50}{9}}$$

Now solve the equation b=\frac{0±\frac{10}{3}}{\frac{50}{9}} when ± is plus.

$$b=\frac{3}{5}$$

Now solve the equation b=\frac{0±\frac{10}{3}}{\frac{50}{9}} when ± is minus.

$$b=-\frac{3}{5}$$

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

$$a=-\frac{4}{3}\times \frac{3}{5}$$

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

$$a=-\frac{4}{5}$$

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

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

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

$$a=\frac{4}{5}$$

The system is now solved.

$$a=-\frac{4}{5},b=\frac{3}{5}\text{ or }a=\frac{4}{5},b=-\frac{3}{5}$$