Solve {l}{a_{1}{(r^2+r)}=15}{r=3/2}{a=a_{1}}{text{Solvefor}btext{where}}{b=a}

$$\left. \begin{array} { l } { a_{1} {(r ^ {2} + r)} = 15 }\\ { r = \frac{3}{2} }\\ { a = a_{1} }\\ { \text{Solve for } b \text{ where} } \\ { b = a } \end{array} \right.$$

Solve for a_1, r, a, b

$$b=4$$

Solution Steps

Consider the first equation. Insert the known values of variables into the equation.

$$a_{1}\left(\left(\frac{3}{2}\right)^{2}+\frac{3}{2}\right)=15$$

Calculate \frac{3}{2} to the power of 2 and get \frac{9}{4}.

$$a_{1}\left(\frac{9}{4}+\frac{3}{2}\right)=15$$

Add \frac{9}{4} and \frac{3}{2} to get \frac{15}{4}.

$$a_{1}\times \frac{15}{4}=15$$

Multiply both sides by \frac{4}{15}, the reciprocal of \frac{15}{4}.

$$a_{1}=15\times \frac{4}{15}$$

Multiply 15 and \frac{4}{15} to get 4.

$$a_{1}=4$$

Consider the third equation. Insert the known values of variables into the equation.

$$a=4$$

Consider the fourth equation. Insert the known values of variables into the equation.

$$b=4$$

The system is now solved.

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