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Mathematics 21 Online
OpenStudy (anonymous):

Solve |5x + 2| = 13

OpenStudy (anonymous):

seeing how we have \[\left| x \right|\] it will never be a negative

OpenStudy (anonymous):

Your next step is to subtract 2 from each side, so that you can get x alone and solve for x.

OpenStudy (anonymous):

You then have 5x = 11. Solve from there by dividing by 5 on each side.

OpenStudy (anonymous):

???

OpenStudy (anonymous):

Okay, Jerry, the straight lines around 5x+2 just mean that your answer won't be negative, as Nick said. From there, focus on the equation 5x+2 = 13 To solve this equation, you need to find what x is equal to. To do this, you have to go step by step. Your first step is to subtract 2 from each side. You have to subtract from each side to make everything even. 5x+2=13 -2 -2 5x=11 Do you see what I've done so far?

OpenStudy (anonymous):

yes

OpenStudy (mertsj):

This equation means: 5x+2 is exactly 13 units from 0. What are the numbers that are exactly 13 units from 0?

OpenStudy (mertsj):

|dw:1366936148493:dw|

OpenStudy (mertsj):

Let's allow the asker to answer the question.

OpenStudy (anonymous):

Okay, Jerry. Next we have to get JUST x, not 5x To do this, we have to divide. 5x=11 -- -- 5 5 Tell me what you get for 11 divided by 5.

OpenStudy (anonymous):

Mertsj is explaining a visual way to look at this to help with future problems. Good luck!

OpenStudy (mertsj):

@Cookieyumm I guess the asker no longer cares about this problem.

OpenStudy (anonymous):

Maybe he understood from a previous explanation. Nice try explaining it, though, I see where you were going (:

OpenStudy (anonymous):

these are the answers{3, -2.2} {13, -13} {2.2, -2.2} {2.2, -3}

OpenStudy (mertsj):

We are not going to tell you the answers. We will help YOU discover the answers.

OpenStudy (mertsj):

This equation means: 5x+2 is exactly 13 units from 0. What are the numbers that are exactly 13 units from 0?

OpenStudy (mertsj):

Can you answer that question?

OpenStudy (netlopes1):

If you want, there is another form to understand this question, please look this "attach"

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