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Mathematics 14 Online
OpenStudy (anonymous):

1/5x +3=-4

OpenStudy (debbieg):

Is it... \(\large \dfrac{1}{5}\cdot x+3=-4\) or..... \(\large \dfrac{1}{5x}+3=-4\)

OpenStudy (debbieg):

And, what do you think should be your first step?

OpenStudy (anonymous):

It's one fifth x.. It in the middle of the 1 and 5

OpenStudy (anonymous):

@DebbieG

OpenStudy (debbieg):

I don't know what "it in the middle of the 1 and 5" means, lol... I'm assuming you mean that it's the 1st of the two I posted above, yes?

OpenStudy (debbieg):

And what do you think is the first step?

OpenStudy (anonymous):

Yes the 1st one sorry lol. And multiply?

OpenStudy (debbieg):

You COULD multiply first (we will have to eventually) but I don't think that's the best thing to do FIRST.

OpenStudy (debbieg):

You can isolate the TERM that involves the x, if you do WHAT first?

OpenStudy (anonymous):

Multiply 5 on both sides

OpenStudy (debbieg):

Well, again, you COULD do that first but I'm suggesting that you do something else first.

OpenStudy (debbieg):

What will give you the \(\large \dfrac{1}{5}\cdot x\) all alone on the left hand side?

OpenStudy (anonymous):

5/1x1/5

OpenStudy (debbieg):

ok, let's just go with that then. :) Let's multiply both sides by 5, now what do you get for your equation? What does it become?

OpenStudy (anonymous):

X+3=-20

OpenStudy (debbieg):

Careful. You fell for the TRAP! :) you must DISTRIBUTE that 5 on the left hand side when you multiply!

OpenStudy (debbieg):

\(\large 5(\dfrac{1}{5}\cdot x+3)=-4\cdot 5\)

OpenStudy (debbieg):

\(\large 5(\dfrac{1}{5}\cdot x)+5\cdot3=-4\cdot 5\) Right?

OpenStudy (debbieg):

So, what do we get?

OpenStudy (debbieg):

@cheetahcatz96 I see you're offline now so maybe having technical problems, and I have to leave shortly. So fix the distribution error as I said above. Then you only have one step left to solve for x, just move the constant term that is on the left side to the right (by subtracting it) and you should find x! OF COURSE, the beauty of algebra - you can always check your answer by plugging it back into the equation that you started with, and make sure that it's true. :)

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