### Instruction: Solve $2z + i = iz + 3.$  Express your answer in the form $x + yi.$.### Response: I want to find the value of $z$ that makes this equation true.
$z$ is a complex number, so it has a real part and an imaginary part.
I can write $z = x + yi$ for some real numbers $x$ and $y$.
Then I can substitute $z = x + yi$ into the equation and simplify it.
$2z + i = iz + 3$
$2(x + yi) + i = i(x + yi) + 3$
$2x + 2yi + i = ix - y + 3$
To make this equation true, the real parts and the imaginary parts must be equal.
So I can equate the real parts and the imaginary parts separately.
For the real parts, I have $2x = -y + 3$
For the imaginary parts, I have $2y + 1 = x$
I can solve this system of equations by substitution or elimination.
I'll use substitution.
I'll solve for $x$ in the second equation and plug it into the first equation.
$x = 2y + 1$
$2(2y + 1) = -y + 3$
$4y + 2 = -y + 3$
$5y = 1$
$y = \frac{1}{5}$
Now I can use this value of $y$ to find $x$.
$x = 2y + 1$
$x = 2(\frac{1}{5}) + 1$
$x = \frac{2}{5} + 1$
$x = \frac{7}{5}$
So I have $x = \frac{7}{5}$ and $y = \frac{1}{5}$.
That means $z = x + yi$ is