Solve {l}{a=3a+1}{b={(a+1/a)}^2}{c=b}{text{Solvefor}dtext{where}}{d=c}

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

Solve for a, b, c, d

$$d = \frac{25}{4} = 6\frac{1}{4} = 6.25$$

Solution Steps

Consider the first equation. Subtract 3a from both sides.

$$a-3a=1$$

Combine a and -3a to get -2a.

$$-2a=1$$

Divide both sides by -2.

$$a=-\frac{1}{2}$$

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

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

Divide 1 by -\frac{1}{2} by multiplying 1 by the reciprocal of -\frac{1}{2}.

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

Multiply 1 and -2 to get -2.

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

Subtract 2 from -\frac{1}{2} to get -\frac{5}{2}.

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

Calculate -\frac{5}{2} to the power of 2 and get \frac{25}{4}.

$$b=\frac{25}{4}$$

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

$$c=\frac{25}{4}$$

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

$$d=\frac{25}{4}$$

The system is now solved.

$$a=-\frac{1}{2} b=\frac{25}{4} c=\frac{25}{4} d=\frac{25}{4}$$