Question

solve for x, b(5px-3c)+a(qx-4)

Answer 1

Distribute.
Where is your equal sign?

Answer 2

^I was gonna say the same thing

Answer 3

This is an expression not an equation so we can't solve, just simplify

Answer 4

@Luigi0210 @gypsy1274 b(5px-3c)=a(qx-4)

Answer 5

Then yea distirbute first

Answer 6

*distribute

Answer 7

Much better. That shift button can be awfully shifty...

Answer 8

ok so then you get b5px-3bc+aqx-4a

Answer 9

@gypsy1274 's got you

Answer 10

There you go with that shifty shift key again....
5bpx-3bc=aqx-4a
Now use addition and subtraction to get the x terms on the left and the non x terms on the right.

Answer 11

@Luigi0210 don't go too far...Your advice is always appreciated.

Answer 12

okk so you get \[\frac{ aqx-4a+3bc }{ 5bp}\] ?

Answer 13

wait thats not right because theres another x hmm

Answer 14

One step at a time.
5bpx-3bc=aqx-4a
Pick one term to work with first. Then decide what you need to do to get it where you want it.

Answer 15

so what do i do first? im honestly so confused

Answer 16

You started here:
\(b(5px-3c)=a(qx-4)\)
Then distributed to get:
\(5bpx-3bc=aqx-4a\)
The next step is to get your x terms to the left side.

Answer 17

so divide by aqx?

Answer 18

\(aqx\) is not multiplied by anything, so don't divide it.
I would subtract it from both sides.

Answer 19

so the answer is \[\frac{ 4a+3bc }{ 5bp-aq }\]

Answer 20

One of us dropped a negative sign on the floor, re-checking now...

Answer 21

I'm pretty sure it was you....look around the desk and floor, I'm sure you will find it. :-)
the \(4a\) was negative and should remain negative.

Answer 22

op i found it! it rolled under my bed not sure how it got there hmm haha thanks :)

Answer 23

I have that problem too. Those negatives are very slippery little suckers....