Solve i2(1+y)-3y=-1

$$i 2 ( 1 + y ) - 3 y = - 1$$

Solve for y

$$y=-\frac{1}{13}+\frac{8}{13}i\approx -0.076923077+0.615384615i$$

Steps for Solving Linear Equation

Use the distributive property to multiply 2i by 1+y.

$$2i+2iy-3y=-1$$

Combine 2iy and -3y to get \left(-3+2i\right)y.

$$2i+\left(-3+2i\right)y=-1$$

Subtract 2i from both sides.

$$\left(-3+2i\right)y=-1-2i$$

Divide both sides by -3+2i.

$$y=\frac{-1-2i}{-3+2i}$$

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

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

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

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

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

$$y=\frac{\left(-1-2i\right)\left(-3-2i\right)}{13}$$

Multiply complex numbers -1-2i and -3-2i like you multiply binomials.

$$y=\frac{-\left(-3\right)-\left(-2i\right)-2i\left(-3\right)-2\left(-2\right)i^{2}}{13}$$

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

$$y=\frac{-\left(-3\right)-\left(-2i\right)-2i\left(-3\right)-2\left(-2\right)\left(-1\right)}{13}$$

Do the multiplications in -\left(-3\right)-\left(-2i\right)-2i\left(-3\right)-2\left(-2\right)\left(-1\right).

$$y=\frac{3+2i+6i-4}{13}$$

Combine the real and imaginary parts in 3+2i+6i-4.

$$y=\frac{3-4+\left(2+6\right)i}{13}$$

Do the additions in 3-4+\left(2+6\right)i.

$$y=\frac{-1+8i}{13}$$

Divide -1+8i by 13 to get -\frac{1}{13}+\frac{8}{13}i.

$$y=-\frac{1}{13}+\frac{8}{13}i$$