a - b/ac = z solving for a
I multiplied both sides by the lcd of ac giving me a - b - acz
why can't I add b to both sides in the next step
It looks like there might be some confusion with the formatting of the initial equation and the steps you've described. It's crucial to ensure that the operations follow the correct order of precedence, such as performing division before addition or subtraction.
Assuming the original equation is:
\[ a - \frac{b}{ac} = z \]
To clear the fraction, you would indeed multiply every term by the least common denominator (LCD), which is \(ac\). When multiplying by \(ac\), be sure to distribute it correctly to each term:
\[ ac(a) - ac\left(\frac{b}{ac}\right) = ac(z) \]
Doing so would result in:
\[ a(ac) - b = acz \]
Which simplifies to:
\[ a^2c - b = acz \]
Now if you want to solve for \(a\), you cannot simply add \(b\) to both sides at this stage since the term \(a^2c\) contains an \(a\) that you're trying to solve for. You've got a quadratic equation in terms of \(a\).
To continue solving for \(a\), the equation should be rearranged to resemble a standard quadratic form. In this case:
\[ a^2c - acz = b \]
Bring all terms to one side to have zero on the other side:
\[ a^2c - acz - b = 0 \]
Now you could either factor this quadratic equation or use the quadratic formula to find the value(s) of \(a\). If factoring is difficult or not possible, using the quadratic formula can give you the solutions directly:
\[ a = \frac{-(-cz) \pm \sqrt{(-cz)^2 - 4(c)(-b)}}{2c} \]
You could simplify that to:
\[ a = \frac{cz \pm \sqrt{c^2z^2 + 4bc}}{2c} \]
This will give you the solution(s) for \(a\) in the original equation, assuming there are real solutions to the quadratic equation. If the discriminant (\(c^2z^2 + 4bc\)) is negative, then the solutions for \(a\) will be complex or imaginary.