Solve b{left(3/2right)}^5-8{left(3/2right)}^2-a(3/2)+8=-10

$$b { \left( \frac{ 3 }{ 2 } \right) }^{ 5 } -8 { \left( \frac{ 3 }{ 2 } \right) }^{ 2 } -a( \frac{ 3 }{ 2 } )+8 = -10$$

Solve for a

$$a=\frac{81b}{16}$$

Steps for Solving Linear Equation

Calculate \frac{3}{2} to the power of 5 and get \frac{243}{32}.

$$b\times \frac{243}{32}-8\times \left(\frac{3}{2}\right)^{2}-a\times \frac{3}{2}+8=-10$$

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

$$b\times \frac{243}{32}-8\times \frac{9}{4}-a\times \frac{3}{2}+8=-10$$

Multiply 8 and \frac{9}{4} to get 18.

$$b\times \frac{243}{32}-18-a\times \frac{3}{2}+8=-10$$

Subtract 8 from both sides.

$$b\times \frac{243}{32}-18-a\times \frac{3}{2}=-10-8$$

Subtract 8 from -10 to get -18.

$$b\times \frac{243}{32}-18-a\times \frac{3}{2}=-18$$

Multiply -1 and \frac{3}{2} to get -\frac{3}{2}.

$$b\times \frac{243}{32}-18-\frac{3}{2}a=-18$$

Subtract b\times \frac{243}{32} from both sides.

$$-18-\frac{3}{2}a=-18-b\times \frac{243}{32}$$

Add 18 to both sides.

$$-\frac{3}{2}a=-18-b\times \frac{243}{32}+18$$

The equation is in standard form.

$$-\frac{3}{2}a=-\frac{243b}{32}$$

Divide both sides of the equation by -\frac{3}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{3}{2}a}{-\frac{3}{2}}=-\frac{\frac{243b}{32}}{-\frac{3}{2}}$$

Dividing by -\frac{3}{2} undoes the multiplication by -\frac{3}{2}.

$$a=-\frac{\frac{243b}{32}}{-\frac{3}{2}}$$

Divide -\frac{243b}{32} by -\frac{3}{2} by multiplying -\frac{243b}{32} by the reciprocal of -\frac{3}{2}.

$$a=\frac{81b}{16}$$

Solve for b

$$b=\frac{16a}{81}$$

Steps for Solving Linear Equation

Calculate \frac{3}{2} to the power of 5 and get \frac{243}{32}.

$$b\times \frac{243}{32}-8\times \left(\frac{3}{2}\right)^{2}-a\times \frac{3}{2}+8=-10$$

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

$$b\times \frac{243}{32}-8\times \frac{9}{4}-a\times \frac{3}{2}+8=-10$$

Multiply 8 and \frac{9}{4} to get 18.

$$b\times \frac{243}{32}-18-a\times \frac{3}{2}+8=-10$$

Subtract 8 from both sides.

$$b\times \frac{243}{32}-18-a\times \frac{3}{2}=-10-8$$

Subtract 8 from -10 to get -18.

$$b\times \frac{243}{32}-18-a\times \frac{3}{2}=-18$$

Add a\times \frac{3}{2} to both sides.

$$b\times \frac{243}{32}-18=-18+a\times \frac{3}{2}$$

Add 18 to both sides.

$$b\times \frac{243}{32}=-18+a\times \frac{3}{2}+18$$

Add -18 and 18 to get 0.

$$b\times \frac{243}{32}=a\times \frac{3}{2}$$

The equation is in standard form.

$$\frac{243}{32}b=\frac{3a}{2}$$

Divide both sides of the equation by \frac{243}{32}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{243}{32}b}{\frac{243}{32}}=\frac{3a}{2\times \frac{243}{32}}$$

Dividing by \frac{243}{32} undoes the multiplication by \frac{243}{32}.

$$b=\frac{3a}{2\times \frac{243}{32}}$$

Divide \frac{3a}{2} by \frac{243}{32} by multiplying \frac{3a}{2} by the reciprocal of \frac{243}{32}.

$$b=\frac{16a}{81}$$