Solve {l}{g{(x)}=-3x+4}{h=g{(10)}}{text{Solvefor}itext{where}}{i=h}

$$\left. \begin{array} { l } { g {(x)} = -3 x + 4 }\\ { h = g {(10)} }\\ { \text{Solve for } i \text{ where} } \\ { i = h } \end{array} \right.$$

Solve for g, x, h

$$x=\frac{1200}{901}-\frac{40}{901}i\approx 1.331853496-0.044395117i$$

$$g=\frac{1}{10}i=0.1i$$

$$h=i$$

Solution Steps

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

$$h=i$$

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

$$i=g\times 10$$

Divide both sides by 10.

$$\frac{i}{10}=g$$

Divide i by 10 to get \frac{1}{10}i.

$$\frac{1}{10}i=g$$

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

$$g=\frac{1}{10}i$$

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

$$\frac{1}{10}ix=-3x+4$$

Add 3x to both sides.

$$\frac{1}{10}ix+3x=4$$

Combine \frac{1}{10}ix and 3x to get \left(3+\frac{1}{10}i\right)x.

$$\left(3+\frac{1}{10}i\right)x=4$$

Divide both sides by 3+\frac{1}{10}i.

$$x=\frac{4}{3+\frac{1}{10}i}$$

Multiply both numerator and denominator of \frac{4}{3+\frac{1}{10}i} by the complex conjugate of the denominator, 3-\frac{1}{10}i.

$$x=\frac{4\left(3-\frac{1}{10}i\right)}{\left(3+\frac{1}{10}i\right)\left(3-\frac{1}{10}i\right)}$$

Do the multiplications in \frac{4\left(3-\frac{1}{10}i\right)}{\left(3+\frac{1}{10}i\right)\left(3-\frac{1}{10}i\right)}.

$$x=\frac{12-\frac{2}{5}i}{\frac{901}{100}}$$

Divide 12-\frac{2}{5}i by \frac{901}{100} to get \frac{1200}{901}-\frac{40}{901}i.

$$x=\frac{1200}{901}-\frac{40}{901}i$$

The system is now solved.

$$g=\frac{1}{10}i x=\frac{1200}{901}-\frac{40}{901}i h=i$$