Solve {l}{4a-3b=2}{5a^2-3b^2=17}

$$\left\{ \begin{array} { l } { 4 a - 3 b = 2 } \\ { 5 a ^ { 2 } - 3 b ^ { 2 } = 17 } \end{array} \right.$$

Solve for a, b

$$a=11\text{, }b=14$$

$$a=5\text{, }b=6$$

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.

$$4a-3b=2,-3b^{2}+5a^{2}=17$$

Solve 4a-3b=2 for a by isolating a on the left hand side of the equal sign.

$$4a-3b=2$$

Subtract -3b from both sides of the equation.

$$4a=3b+2$$

Divide both sides by 4.

$$a=\frac{3}{4}b+\frac{1}{2}$$

Substitute \frac{3}{4}b+\frac{1}{2} for a in the other equation, -3b^{2}+5a^{2}=17.

$$-3b^{2}+5\left(\frac{3}{4}b+\frac{1}{2}\right)^{2}=17$$

Square \frac{3}{4}b+\frac{1}{2}.

$$-3b^{2}+5\left(\frac{9}{16}b^{2}+\frac{3}{4}b+\frac{1}{4}\right)=17$$

Multiply 5 times \frac{9}{16}b^{2}+\frac{3}{4}b+\frac{1}{4}.

$$-3b^{2}+\frac{45}{16}b^{2}+\frac{15}{4}b+\frac{5}{4}=17$$

Add -3b^{2} to \frac{45}{16}b^{2}.

$$-\frac{3}{16}b^{2}+\frac{15}{4}b+\frac{5}{4}=17$$

Subtract 17 from both sides of the equation.

$$-\frac{3}{16}b^{2}+\frac{15}{4}b-\frac{63}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3+5\times \left(\frac{3}{4}\right)^{2} for a, 5\times \frac{1}{2}\times \frac{3}{4}\times 2 for b, and -\frac{63}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-\frac{15}{4}±\sqrt{\left(\frac{15}{4}\right)^{2}-4\left(-\frac{3}{16}\right)\left(-\frac{63}{4}\right)}}{2\left(-\frac{3}{16}\right)}$$

Square 5\times \frac{1}{2}\times \frac{3}{4}\times 2.

$$b=\frac{-\frac{15}{4}±\sqrt{\frac{225}{16}-4\left(-\frac{3}{16}\right)\left(-\frac{63}{4}\right)}}{2\left(-\frac{3}{16}\right)}$$

Multiply -4 times -3+5\times \left(\frac{3}{4}\right)^{2}.

$$b=\frac{-\frac{15}{4}±\sqrt{\frac{225}{16}+\frac{3}{4}\left(-\frac{63}{4}\right)}}{2\left(-\frac{3}{16}\right)}$$

Multiply \frac{3}{4} times -\frac{63}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$b=\frac{-\frac{15}{4}±\sqrt{\frac{225-189}{16}}}{2\left(-\frac{3}{16}\right)}$$

Add \frac{225}{16} to -\frac{189}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$b=\frac{-\frac{15}{4}±\sqrt{\frac{9}{4}}}{2\left(-\frac{3}{16}\right)}$$

Take the square root of \frac{9}{4}.

$$b=\frac{-\frac{15}{4}±\frac{3}{2}}{2\left(-\frac{3}{16}\right)}$$

Multiply 2 times -3+5\times \left(\frac{3}{4}\right)^{2}.

$$b=\frac{-\frac{15}{4}±\frac{3}{2}}{-\frac{3}{8}}$$

Now solve the equation b=\frac{-\frac{15}{4}±\frac{3}{2}}{-\frac{3}{8}} when ± is plus. Add -\frac{15}{4} to \frac{3}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$b=-\frac{\frac{9}{4}}{-\frac{3}{8}}$$

Divide -\frac{9}{4} by -\frac{3}{8} by multiplying -\frac{9}{4} by the reciprocal of -\frac{3}{8}.

$$b=6$$

Now solve the equation b=\frac{-\frac{15}{4}±\frac{3}{2}}{-\frac{3}{8}} when ± is minus. Subtract \frac{3}{2} from -\frac{15}{4} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$b=-\frac{\frac{21}{4}}{-\frac{3}{8}}$$

Divide -\frac{21}{4} by -\frac{3}{8} by multiplying -\frac{21}{4} by the reciprocal of -\frac{3}{8}.

$$b=14$$

There are two solutions for b: 6 and 14. Substitute 6 for b in the equation a=\frac{3}{4}b+\frac{1}{2} to find the corresponding solution for a that satisfies both equations.

$$a=\frac{3}{4}\times 6+\frac{1}{2}$$

Multiply \frac{3}{4} times 6.

$$a=\frac{9+1}{2}$$

Add \frac{3}{4}\times 6 to \frac{1}{2}.

$$a=5$$

Now substitute 14 for b in the equation a=\frac{3}{4}b+\frac{1}{2} and solve to find the corresponding solution for a that satisfies both equations.

$$a=\frac{3}{4}\times 14+\frac{1}{2}$$

Multiply \frac{3}{4} times 14.

$$a=\frac{21+1}{2}$$

Add \frac{3}{4}\times 14 to \frac{1}{2}.

$$a=11$$

The system is now solved.

$$a=5,b=6\text{ or }a=11,b=14$$