WILL MEDAL AND FAN, JUST NEED HELP REVIEWING

- anonymous

WILL MEDAL AND FAN, JUST NEED HELP REVIEWING

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- schrodinger

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- anonymous

I just need someone to help me work out the answer!

- anonymous

Solve for x: 3(x + 1) = −2(x − 1) − 4.

- anonymous

answer choices:
1
−1
−5
−25

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## More answers

- anonymous

I know you do the parenthesis first, right?

- mathstudent55

Yes.
Distribute the 3 on the left side, and distribute the -2 on the right side.
That gets rid of the parentheses on both sides.
What do you get?

- anonymous

-2x +1 = 3x-5?? I duck at this stuff

- anonymous

lol I meant suck

- mathstudent55

Let's do it one step at a time.
We start with the original equation:
|dw:1433462580564:dw|

- mathstudent55

Now we distribute the 3 on the left side. To distribute a number by a quantity in parentheses, we multiply the number outside the parentheses by each term inside the parentheses.
Let's do just the left side first.

- anonymous

Oh, the distributive property!

- mathstudent55

|dw:1433462685387:dw|

- anonymous

-2x -2 x 1 -4 ??

- mathstudent55

Yes, we are using the distributive property of multiplication over addition on the left side.
You see, 3(x + 1) turns into 3 * x + 3 * 1.
We will simplify that in the next step.

- mathstudent55

Now let's do the distributive property on the right side.
|dw:1433462795731:dw|

- mathstudent55

You need to be careful with the signs on the left side.

- mathstudent55

Now we simplify both sides.

- mathstudent55

|dw:1433462877297:dw|

- anonymous

3x+3= -2x +2-4 ??

- mathstudent55

Exactly.

- anonymous

now, we have to solve it!?

- mathstudent55

Now we collect like terms on the right side.
2 and -4 are like terms. They both have no variable. We can combine them together.
|dw:1433462994203:dw|

- anonymous

ok I am following you

- mathstudent55

Since we are solving for x, we want x by itself on the left side and the numbers by themselves on the right side.
Let's deal with the terms with x first.
We see -2x on the right side. Since the term -2x and the term -2 (both on the right side) are being added together, we add 2x to both sides to move the -2x to the left side.
|dw:1433463153571:dw|

- mathstudent55

Great.

- anonymous

then you add 3 to both sides? Or do you get rid of x?

- anonymous

I mean subtract

- mathstudent55

Subtract is correct.
Now we have all terms with x on the left side.
Now we need to get all terms that are just numbers on the right side.
We see on the left side that 3 is being added to 5x.
We want that 3 on the right side, so we subtract 3 from both sides.

- mathstudent55

|dw:1433463277247:dw|

- anonymous

5x = -5

- mathstudent55

Exactly.

- anonymous

divide both sides by positive 5

- anonymous

to get x= -1 !!??

- mathstudent55

We still want the x by itself, but there is a 5 with the x.
The 5 is multiplying the x.
We do the opposite operation to multiplication which is division.
We divide both sides by 5.

- mathstudent55

|dw:1433463388210:dw|

- anonymous

Thank you so much for your help! Wish me luck....

- mathstudent55

One final thing.
With many problems, you answer them, and you don;t really know whether the answer is correct or incorrect until your teacher grades your work.
With an equation, we can find out by ourselves if x = -1 is really correct or not.

- anonymous

you fill in the number?

- mathstudent55

By stating that x = -1, we are saying that if we plug in -1 for x in the equation, the equation will be true.
Let's check it for ourselves:
In the second line below, I replaced x with our solution, -1.
|dw:1433463548144:dw|

- mathstudent55

Now we work out the math to see if the statement above is true.
If it is true, then x = -1 is the correct solution.

- mathstudent55

|dw:1433463631425:dw|

- mathstudent55

The solution is x = -1.
We were correct.

- anonymous

YAYY! Thanks, again.. I gave you a medal and fanned you.

- mathstudent55

Good luck with your studies.
If you see me online again and have a question, don't hesitate to ask.
It's been great working with you!

- anonymous

Wait, do you mind helping with one more? I will open new question if you want?!

- mathstudent55

No problem.
Please start a new post. I'll help you there.

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