Solve 5(ic-2)=2(c-1)

$$5 ( i c - 2 ) = 2 ( c - 1 )$$

Solve for c

$$c=-\frac{16}{29}-\frac{40}{29}i\approx -0.551724138-1.379310345i$$

Steps for Solving Linear Equation

Use the distributive property to multiply 5 by ic-2.

$$5ic-10=2\left(c-1\right)$$

Use the distributive property to multiply 2 by c-1.

$$5ic-10=2c-2$$

Subtract 2c from both sides.

$$5ic-10-2c=-2$$

Combine 5ic and -2c to get \left(-2+5i\right)c.

$$\left(-2+5i\right)c-10=-2$$

Add 10 to both sides.

$$\left(-2+5i\right)c=-2+10$$

Add -2 and 10 to get 8.

$$\left(-2+5i\right)c=8$$

Divide both sides by -2+5i.

$$c=\frac{8}{-2+5i}$$

Multiply both numerator and denominator of \frac{8}{-2+5i} by the complex conjugate of the denominator, -2-5i.

$$c=\frac{8\left(-2-5i\right)}{\left(-2+5i\right)\left(-2-5i\right)}$$

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$c=\frac{8\left(-2-5i\right)}{\left(-2\right)^{2}-5^{2}i^{2}}$$

By definition, i^{2} is -1. Calculate the denominator.

$$c=\frac{8\left(-2-5i\right)}{29}$$

Multiply 8 times -2-5i.

$$c=\frac{8\left(-2\right)+8\times \left(-5i\right)}{29}$$

Do the multiplications in 8\left(-2\right)+8\times \left(-5i\right).

$$c=\frac{-16-40i}{29}$$

Divide -16-40i by 29 to get -\frac{16}{29}-\frac{40}{29}i.

$$c=-\frac{16}{29}-\frac{40}{29}i$$