Question

4x-9y+13=1/5(25-5y)
Rewrite the equation in terms of y
Can anyone help me out plzz ?

Answer 1

Start by expanding the bracket on the right hand side.
Then take all the terms with y's to the left, and everything else to the right.
Let me know what you get :)

Answer 2

I'll do a similar example:

\[2x + 3y -7 = \frac{2}3(6+9y)\]First thing I like to do is get rid of any fractions by multiplying the entire equation by the denominator. You might have to do this repeatedly if you have fractions with different denominators, but here there is only one, so we multiply everything by its denominator, 3:
\[3*2x + 3*3y - 3*7 = 3*\frac{2}{3}(6+9y)\]\[6x+9y-21=2(6+9y)\]With the fraction cleaned up, let's expand the right side by the distributive property:\[6x+9y-21=2*6+2*9y\]\[6x+9y-21=12+18y\]Now, because the directions are to rewrite the equation in terms of \(y\), that means we want to get all of the \(y\) terms collected together on the right, with just \(x\) on the left. We'll subtract any terms containing \(y\) on the left side from both sides of the equation:\[6x+9y-21-9y = 12 + 18y - 9y\]\[6x-21=12+9y\]Now we do the same with any numbers:\[6x-21-(-21) = 12 +9y -(-21)\]subtracting a negative number is like adding the corresponding positive number, so that could be written as\[6x-21+21 = 12 + 9y + 21\]\[6x=33+9y\]Now we finally have \(6x\) in terms of \(y\), but we want \(x\), so we divide both sides by 6 and reduce any fractions:\[\frac{6x}6 = \frac{33}{6}+\frac{9y}6\]\[x=\frac{11}3 + \frac{3y}2\]

The key thing to remember is that anything you do to one side of the equation you need to do to the other side as well. Think of it as a balance (or a teetertotter on a playground): if you add or remove something on one side, the equation will stay in balance as long as you do the same to the other side.