Solve {l}{8+1=x+6}{I=5{(1-x/-6)}^2}{a=8*I}{text{Solvefor}btext{where}}{b=a}

$$\left. \begin{array} { l } { 8 + 1 = x + 6 }\\ { I = 5 {(1 - \frac{x}{-6})} ^ {2} }\\ { a = 8 \cdot I }\\ { \text{Solve for } b \text{ where} } \\ { b = a } \end{array} \right.$$

Solve for x, I, a, b

$$b=90$$

Solution Steps

Consider the first equation. Add 8 and 1 to get 9.

$$9=x+6$$

Swap sides so that all variable terms are on the left hand side.

$$x+6=9$$

Subtract 6 from both sides.

$$x=9-6$$

Subtract 6 from 9 to get 3.

$$x=3$$

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

$$I=5\left(1-\frac{3}{-6}\right)^{2}$$

Reduce the fraction \frac{3}{-6} to lowest terms by extracting and canceling out 3.

$$I=5\left(1-\left(-\frac{1}{2}\right)\right)^{2}$$

The opposite of -\frac{1}{2} is \frac{1}{2}.

$$I=5\left(1+\frac{1}{2}\right)^{2}$$

Add 1 and \frac{1}{2} to get \frac{3}{2}.

$$I=5\times \left(\frac{3}{2}\right)^{2}$$

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

$$I=5\times \frac{9}{4}$$

Multiply 5 and \frac{9}{4} to get \frac{45}{4}.

$$I=\frac{45}{4}$$

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

$$a=8\times \frac{45}{4}$$

Multiply 8 and \frac{45}{4} to get 90.

$$a=90$$

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

$$b=90$$

The system is now solved.

$$x=3 I=\frac{45}{4} a=90 b=90$$