How would you solve: 3- qx= 14- rx?

Question

homework help. it says to solve each equation. you need to solve for "x"

Answer 1

3 - qx = 14 - rx

add rx to both sides while subtracting 3 from both sides:

rx - qx = 14 - 3

factor x out on the left side while doing subtraction on right:

x(r - q) = 11

divide both sides by (r - q):

x = 11/(r - q)

that's it! ;)

Answer 2

First, get all the x's to one side.

3 - qx + rx = 14 (I added rx to both sides).  Next, move the 3 by subtracting it to both sides.

-qx + rx = 14 - 3 (now simplify)

-qx + rx = 11 (now factor out the x)

x(-q + r) = 11 (now divide both sides by -q + r to get x alone)

x = 11/(-q + r)

Answer 3

my attempt:

x = 11/(-q-r)

Answer 4

3 - qx = 14 - rx

-qx = 11 - rx

-qx + rx = 11

x(r-q) = 11

x = 11/(r-q)

Answer 5

First get the terms with x in them on the left and the terms without x in them on the right, by adding the opposites of what needs to be moved.  It comes out

rx - qx = 14 - 3 which is 11

Then factor out the x using distributive: x(r - q) = 11

Then divide both sides by (r - q): x = 11/(r - q)