Solve {l}{1z+5=4z+3c}{iz-4z=-3+3}

$$\left. \begin{array} { l } { 1 z + 5 = 4 z + 3 c } \\ { i z - 4 z = - 3 + 3 } \end{array} \right.$$

Solve for z, c

$$z=0$$

$$c = \frac{5}{3} = 1\frac{2}{3} \approx 1.666666667$$

Solution Steps

Consider the second equation. Combine iz and -4z to get \left(-4+i\right)z.

$$\left(-4+i\right)z=-3+3$$

Add -3 and 3 to get 0.

$$\left(-4+i\right)z=0$$

Divide both sides by -4+i. Zero divided by any non-zero number gives zero.

$$z=0$$

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

$$1\times 0+5=4\times 0+3c$$

Multiply 1 and 0 to get 0.

$$0+5=4\times 0+3c$$

Add 0 and 5 to get 5.

$$5=4\times 0+3c$$

Multiply 4 and 0 to get 0.

$$5=0+3c$$

Anything plus zero gives itself.

$$5=3c$$

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

$$3c=5$$

Divide both sides by 3.

$$c=\frac{5}{3}$$

The system is now solved.

$$z=0 c=\frac{5}{3}$$