Question

5x(3x+2)-6 < 4X(2X-5) +12

Answer 1

Hello

Answer 2

Do you have any guesses on how to start this problem?

Answer 3

I multipy 5x 15+10-6< 4x 12x-20+12, I am not good at all with algebra.

Answer 4

You were indeed correct to start off by distributing. But we have a couple minor errors with your distribution, however.

Answer 5

When you multiply 5x times 3x, what do you get?

Answer 6

15

Answer 7

Actually, you get 15x^2. The x's multiply with each other. But the 5*3 was spot on.

Answer 8

okay, thanks!

Answer 9

Do you see how that works?

Answer 10

Do you see how that works?

Answer 11

Alright, now let's try to multiply the 5x and the +2 together.

Answer 12

5x + 2 -7

Answer 13

5x(3x+2)-6

Hmm.. Not quite. Maybe you don't know where I am talking about. In the above phrase, will you take the 5x and multiply it with the 2 in the parenthesis?

Answer 14

Yes, 5x2+10

Answer 15

Mm.. Not quite. Remember, the x is a variable, not the same as a multiplication symbol.

5x * 2 = 10x

Answer 16

Oh, and by the way, * is the computer symbol for multiplication. And that's the reason why, we don't want to get confused about whether it's a letter x or an operation so we use *.

Answer 17

Okay I understand

Answer 18

Great. So if you want to double (or multiply by two) your five X's, you now have 10 X's. Is that clear?

Answer 19

Yes!

Answer 20

So on the left side we went from:
5x(3x+2)-6

To:
15x^2 + 10x - 6

Following? Now we'll want to do distribute on the right side:
4X(2X-5) +12

Answer 21

Okay I am following!

Answer 22

Cool, can you distribute on the right side then?
4X(2X-5) +12

Answer 23

yes 20x^4 +10

Answer 24

Hmm.. Let's try this together.

4X(2X)

Can you do this for me? Remember what I said above about multiply two Xs together.

Answer 25

8x^2 +12

Answer 26

Excellent, very very good. Now you left out one last part for the right side:
4X(2X-5) +12

Can you do the 4x * -5?

Answer 27

4x8-5+3

Answer 28

I'm afraid not.

4x * -5 = -20x

Pretend that an X is an item. Don't even bother calling it a variable. Now, I have 4 Xs. If I have 4 Xs and I multiply my Xs by negative five, how many Xs do I have?

Answer 29

would it be 1x

Answer 30

Hmm.. Let's try something a little different.

If I have 4 pennies and you said that you could multiply my pennies by five, by the end of it, how many pennies would I have?

Answer 31

20

Answer 32

Exactly. Now, pretend that my pennies were really just little X's all along. Not a letter, not a variable, a real X, something I could hold in my hands. I'll ask you the same question.

If I have 4 X's in my hand and you said you could multiply my X's by 5, how many X's would I have?

Answer 33

20x's

Answer 34

Great, does that make it a little more clear? That is how you should approach these problems when multiplying X's. Just pretend they are just labels. :)

Answer 35

Yes it does, thank you!

Answer 36

So we had this:
5x(3x+2)-6 < 4X(2X-5) +12

And turned it into this:
15x^2 + 10x - 6 < 20x^2 - 20x + 12

When we solve algebraically, we'll want everything that has that nasty X on the same side of the inequality. Do you know how you might accomplish that?

Answer 37

15x2+10x-6 < 20x2-20x+12
-15x+6 20x-12

Answer 38

Afraid not.

We'll want to do:
15x^2 + 10x - 6 < 20x^2 - 20x + 12
+6 -20x^2 +20x +6
To get:
15x^2 +10x - 20x^2 + 20x < 18

Answer 39

From there we can combine like terms:

The x^2s can combine.
15x^2 - 20x^2 = -5x^2

The X's can combine.
10x + 20x = 30x

So we end up with:
-5x^2 + 30x < 18.

Answer 40

You subtracted 2 on both sides to get the answer 18 and combined like terms 15x 20x and 5x and added 10+20 = 30.

Answer 41

I added six to both sides to get +18 on the right side. The rest is correct :)

Answer 42

Now, are you familiar with the quadratic equation?

Answer 43

Yes I understand somewhat, however I will continue to practice and study for a better understanding. Thank you so much for your help!

Answer 44

It's no problem. I just wish I could be a little more clear, it's hard to do that with just text.

Anyways, we'll use the quadratic equation, first we have to make the equation equal to 0:
-5x^2 + 30x < 18
-18 -18

-5x^2 + 30x -18 < 0

And then plug in the A, B, and C values into the quadratic equation:
-b plus/minus sqrt of b^2 - 4AC all divided by 2A

- 30 plus/minus sqrt 30^2 - 4*-5*-18, all divided by 2(-5) = 16.905 and 133.09. Those are your values of X and those are the answers. Two of them.

Answer 45

Oh wait. I lie.

Answer 46

Okay

Answer 47

I screwed up my math somewhere.

The system is true when:

-4.819 < X < 0.533

Answer 48

And that's the answer.

Answer 49

Okay thank you!!!! I am going to print out this information and study it and I will practice other problems. Thanks again!!!!

Answer 50

No problem Sorry I couldn't be more clear.