Solve {l}{h{(x)}=4x+3}{i=h{(-1)}}{text{Solvefor}jtext{where}}{j=i}

$$\left. \begin{array} { l } { h {(x)} = 4 x + 3 }\\ { i = h {(-1)} }\\ { \text{Solve for } j \text{ where} } \\ { j = i } \end{array} \right.$$

Solve for h, x, j

$$j=i$$

Solution Steps

Consider the second equation. Divide both sides by -1.

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

Divide i by -1 to get -i.

$$-i=h$$

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

$$h=-i$$

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

$$-ix=4x+3$$

Subtract 4x from both sides.

$$-ix-4x=3$$

Combine -ix and -4x to get \left(-4-i\right)x.

$$\left(-4-i\right)x=3$$

Divide both sides by -4-i.

$$x=\frac{3}{-4-i}$$

Multiply both numerator and denominator of \frac{3}{-4-i} by the complex conjugate of the denominator, -4+i.

$$x=\frac{3\left(-4+i\right)}{\left(-4-i\right)\left(-4+i\right)}$$

Do the multiplications in \frac{3\left(-4+i\right)}{\left(-4-i\right)\left(-4+i\right)}.

$$x=\frac{-12+3i}{17}$$

Divide -12+3i by 17 to get -\frac{12}{17}+\frac{3}{17}i.

$$x=-\frac{12}{17}+\frac{3}{17}i$$

The system is now solved.

$$h=-i x=-\frac{12}{17}+\frac{3}{17}i j=i$$