Solve {l}{a^2+b^2=25^2}{a/b=9/32}

$$\left\{ \begin{array} { l } { a ^ { 2 } + b ^ { 2 } = 25 ^ { 2 } } \\ { \frac { a } { b } = \frac { 9 } { 32 } } \end{array} \right.$$

Solve for a, b

$$a=-\frac{45\sqrt{1105}}{221}\approx -6.768639423\text{, }b=-\frac{160\sqrt{1105}}{221}\approx -24.066273504$$

$$a=\frac{45\sqrt{1105}}{221}\approx 6.768639423\text{, }b=\frac{160\sqrt{1105}}{221}\approx 24.066273504$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Calculate 25 to the power of 2 and get 625.

$$a^{2}+b^{2}=625$$

Consider the second equation. Variable b cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 32b, the least common multiple of b,32.

$$32a=9b$$

Divide both sides by 32.

$$a=\frac{9}{32}b$$

Substitute \frac{9}{32}b for a in the other equation, b^{2}+a^{2}=625.

$$b^{2}+\left(\frac{9}{32}b\right)^{2}=625$$

Square \frac{9}{32}b.

$$b^{2}+\frac{81}{1024}b^{2}=625$$

Add b^{2} to \frac{81}{1024}b^{2}.

$$\frac{1105}{1024}b^{2}=625$$

Subtract 625 from both sides of the equation.

$$\frac{1105}{1024}b^{2}-625=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times \left(\frac{9}{32}\right)^{2} for a, 1\times 0\times \frac{9}{32}\times 2 for b, and -625 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{0±\sqrt{0^{2}-4\times \frac{1105}{1024}\left(-625\right)}}{2\times \frac{1105}{1024}}$$

Square 1\times 0\times \frac{9}{32}\times 2.

$$b=\frac{0±\sqrt{-4\times \frac{1105}{1024}\left(-625\right)}}{2\times \frac{1105}{1024}}$$

Multiply -4 times 1+1\times \left(\frac{9}{32}\right)^{2}.

$$b=\frac{0±\sqrt{-\frac{1105}{256}\left(-625\right)}}{2\times \frac{1105}{1024}}$$

Multiply -\frac{1105}{256} times -625.

$$b=\frac{0±\sqrt{\frac{690625}{256}}}{2\times \frac{1105}{1024}}$$

Take the square root of \frac{690625}{256}.

$$b=\frac{0±\frac{25\sqrt{1105}}{16}}{2\times \frac{1105}{1024}}$$

Multiply 2 times 1+1\times \left(\frac{9}{32}\right)^{2}.

$$b=\frac{0±\frac{25\sqrt{1105}}{16}}{\frac{1105}{512}}$$

Now solve the equation b=\frac{0±\frac{25\sqrt{1105}}{16}}{\frac{1105}{512}} when ± is plus.

$$b=\frac{160\sqrt{1105}}{221}$$

Now solve the equation b=\frac{0±\frac{25\sqrt{1105}}{16}}{\frac{1105}{512}} when ± is minus.

$$b=-\frac{160\sqrt{1105}}{221}$$

There are two solutions for b: \frac{160\sqrt{1105}}{221} and -\frac{160\sqrt{1105}}{221}. Substitute \frac{160\sqrt{1105}}{221} for b in the equation a=\frac{9}{32}b to find the corresponding solution for a that satisfies both equations.

$$a=\frac{9}{32}\times \frac{160\sqrt{1105}}{221}$$

Multiply \frac{9}{32} times \frac{160\sqrt{1105}}{221}.

$$a=\frac{9\times \frac{160\sqrt{1105}}{221}}{32}$$

Now substitute -\frac{160\sqrt{1105}}{221} for b in the equation a=\frac{9}{32}b and solve to find the corresponding solution for a that satisfies both equations.

$$a=\frac{9}{32}\left(-\frac{160\sqrt{1105}}{221}\right)$$

Multiply \frac{9}{32} times -\frac{160\sqrt{1105}}{221}.

$$a=\frac{9\left(-\frac{160\sqrt{1105}}{221}\right)}{32}$$

The system is now solved.

$$a=\frac{9\times \frac{160\sqrt{1105}}{221}}{32},b=\frac{160\sqrt{1105}}{221}\text{ or }a=\frac{9\left(-\frac{160\sqrt{1105}}{221}\right)}{32},b=-\frac{160\sqrt{1105}}{221}$$