Solve {l}{h{(x)}=3x+1}{i=h{(-2)}}{text{Solvefor}jtext{where}}{j=i}

$$\left. \begin{array} { l } { h {(x)} = 3 x + 1 }\\ { i = h {(-2)} }\\ { \text{Solve for } j \text{ where} } \\ { j = i } \end{array} \right.$$

Solve for h, x, j

$$j=i$$

Solution Steps

Consider the second equation. Divide both sides by -2.

$$\frac{i}{-2}=h$$

Divide i by -2 to get -\frac{1}{2}i.

$$-\frac{1}{2}i=h$$

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

$$h=-\frac{1}{2}i$$

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

$$-\frac{1}{2}ix=3x+1$$

Subtract 3x from both sides.

$$-\frac{1}{2}ix-3x=1$$

Combine -\frac{1}{2}ix and -3x to get \left(-3-\frac{1}{2}i\right)x.

$$\left(-3-\frac{1}{2}i\right)x=1$$

Divide both sides by -3-\frac{1}{2}i.

$$x=\frac{1}{-3-\frac{1}{2}i}$$

Multiply both numerator and denominator of \frac{1}{-3-\frac{1}{2}i} by the complex conjugate of the denominator, -3+\frac{1}{2}i.

$$x=\frac{1\left(-3+\frac{1}{2}i\right)}{\left(-3-\frac{1}{2}i\right)\left(-3+\frac{1}{2}i\right)}$$

Do the multiplications in \frac{1\left(-3+\frac{1}{2}i\right)}{\left(-3-\frac{1}{2}i\right)\left(-3+\frac{1}{2}i\right)}.

$$x=\frac{-3+\frac{1}{2}i}{\frac{37}{4}}$$

Divide -3+\frac{1}{2}i by \frac{37}{4} to get -\frac{12}{37}+\frac{2}{37}i.

$$x=-\frac{12}{37}+\frac{2}{37}i$$

The system is now solved.

$$h=-\frac{1}{2}i x=-\frac{12}{37}+\frac{2}{37}i j=i$$