Solve Algebra Problem: 3x+5 - 2x-1 = 3

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You get the same answer if you subtract the two numerators in their entirety from the first numerator's denominator:\frac{5x+17}{12} = \frac{36}{12}Now, you can see that the two denominators are the same, but the two numerators are not. This is because 9x+15-4x-2 is not the same as (9x+15)-(4x-2). The key is to remember that the subtraction sign in front of the second term means "minus" and the subtraction sign between the two terms means "minus the whole second term from the first term." So, the subtraction sign between the two terms is NOT the same as the subtraction
  • #1
david18
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Hi I'm on this question where I have to find x:

3x+5 - 2x-1 = 3
4 6

and i ended up with 9x + 15 - 4x - 2 = 36
12 = 12

so i removed the 12s to leave me with 9x + 15 - 4x - 2 = 36

Now for my question:

When I am simplifying the left hand side; does it become 5x + 13 or 5x + 17 because if the equation should be written as (9x + 15) - (4x - 2) it would mean a -- which is a plus to make 5x + 17...edit: for some reason the denominators haven't come up where i wanted them to be. but its 3x + 5 all over 4 and 2x - 1 all over 6
and the second error is that both sides should have 12 as their denominators
 
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  • #2
david18 said:
Hi I'm on this question where I have to find x:

3x+5 - 2x-1 = 3
4 6

and i ended up with 9x + 15 - 4x - 2 = 36
12 = 12

so i removed the 12s to leave me with 9x + 15 - 4x - 2 = 36

Now for my question:

When I am simplifying the left hand side; does it become 5x + 13 or 5x + 17 because if the equation should be written as (9x + 15) - (4x - 2) it would mean a -- which is a plus to make 5x + 17...


edit: for some reason the denominators haven't come up where i wanted them to be. but its 3x + 5 all over 4 and 2x - 1 all over 6
and the second error is that both sides should have 12 as their denominators

I'm not really sure how the problem is supposed to read. is 4 the denominator of 3x+5 and 6 the denominator of -2x-1?
 
  • #3
I'm confused. Is this the original equation?

[tex]\frac{3x+5}{4} - \frac{2x-1}{6} = 3[/tex]
 
  • #4
[tex] \frac {3x+5} {4} - \frac {2x-1} {6}=3 [/tex]
 
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  • #5
You beat me to it :(
 
  • #6
Error with LaTeX.

So wait you have [tex] \frac {3x+5} {4} [/tex] and that is all divided by 12?
 
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  • #7
yes the 2 equations you said are correct, thanks for sorting my one out
 
  • #8
david18 said:
yes the 2 equations you said are correct, thanks for sorting my one out

So then everything is divided by 12?
 
  • #9
I think there is an error in your work. If both sides have a denominator of 12 are you referring to the left side and right side of the "=" sign. If so you would have [tex] \frac {3} {12} [/tex] not 36.
 
  • #10
the original equation eventually became
9x + 15 - 4x - 2 (all divided by 12) = 36 (all divided by 12) when i simplified it.
This is because I multiplied the denominators to make them all become 12

then i would be able to multiple either side by 12 to leave the equation:

9x + 15 - 4x - 2 = 36

But the problem is whether that becomes 5x + 13 or 5x + 17 is what I am wondering.
 
  • #11
Stevedye56 said:
I think there is an error in your work. If both sides have a denominator of 12 are you referring to the left side and right side of the "=" sign. If so you would have [tex] \frac {3} {12} [/tex] not 36.

i think my message has become a bit too confusing lol. basically the question are the ones that you 2 said. The denominator of 12 bit only comes up after some steps. and if i want to make them all have equal denominators, in this case 12, it would mean i would write 3 as 36/12
 
  • #12
Im totally confused to what is going on right now.
 
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  • #13
david18, where did the 12 come from? What were you trying to do, cross multiply? I honestly can't see how you got from your first equation to the second.
 
  • #14
actually I've got it... it becomes 5x + 17 = 36 because i used the method that most people use... which is a lot better than my 'make everything have a denominator of 12' method...

instead if you just make the left hand side have a denominator of 12 and leave the right hand side as 3...

... means the left hand side becomes:

3(3x+5) - 2(2x-1) = 3
______________
...12

(ignore dots)
which means its -2 x -1 which is +2. Yeah the method i used was bad so I'll avoid it makes myself confused.

does it make sense to everyone now?
 
  • #15
hage567 said:
david18, where did the 12 come from?

That's what I'm wondering also.
 
  • #16
hage567 said:
I'm confused. Is this the original equation?

[tex]\frac{3x+5}{4} - \frac{2x-1}{6} = 3[/tex]

This is the correct question. The Lowest common denominator is 12.
 
  • #17
Ok. Then you would have [tex] \frac {9x+15} {12} - \frac {4x-2} {12} = 3 [/tex]

Therefore

[tex] \frac {5x +17} {12} = 3 [/tex]
 
  • #18
OK, I see what you were doing.
 
  • #19
hage567 said:
OK, I see what you were doing.

This is right, correct?
 
  • #20
Yes, I come up with the same answer. Sorry.
 
  • #21
david18 said:
actually I've got it... it becomes 5x + 17 = 36 because i used the method that most people use... which is a lot better than my 'make everything have a denominator of 12' method...

instead if you just make the left hand side have a denominator of 12 and leave the right hand side as 3...

... means the left hand side becomes:

3(3x+5) - 2(2x-1) = 3
______________
...12

(ignore dots)
which means its -2 x -1 which is +2. Yeah the method i used was bad so I'll avoid it makes myself confused.

does it make sense to everyone now?

I don't understand what everybody is getting confused about! There is nothing wrong whatsoever with the method that you first tried to use, david18. Here's what you started with:

[tex]\frac{3x+5}{4} - \frac{2x-1}{6} = 3[/tex]

Then you put everything over a common denominator of 12:

[tex]\frac{9x+15}{12} - \frac{4x-2}{12} = \frac{36}{12}[/tex]

So far, you're correct! Now, to answer the question in your original post, when you combine the two terms on the left hand side, you have to be careful! You are subtracting the whole second term from the first term. As a result, you must subtract the second numerator in its entirety from the first. To make sure you get this right, use parentheses:

[tex]\frac{9x+15 - (4x-2)}{12} = \frac{36}{12}[/tex]

Now, when you remove the parentheses, be sure to get the signs right:

[tex]\frac{9x + 15 - 4x + 2}{12} = \frac{36}{12}[/tex]

[tex]\frac{5x +17}{12} = \frac{36}{12}[/tex]

Now, of course, you can multiply both sides by 12, which is equivalent to just removing the denominators:

[tex] 5x +17 = 36 [/tex]

Can you see that this method is not different from the "second method" that you used? It doesn't matter whether you express the right hand side as 3 or 36/12. Mathematically, those are just two ways of writing the same number, and you'll still end up with 5x + 17 = 36 in the end. The two methods are not distinct from each other. They're the same.
 
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  • #22
cepheid said:
I don't understand what everybody is getting confused about!

What was so confusing was that it wasn't in latex and the spaces that were used to set up the fractions didn't come out. Also the method wasn't very clear.
 

FAQ: Solve Algebra Problem: 3x+5 - 2x-1 = 3

What is the problem asking for?

The problem is asking for the value of x that makes the equation 3x+5 - 2x-1 = 3 true.

What is the first step in solving this problem?

The first step is to combine like terms on both sides of the equation. In this case, we can combine the 3x and -2x terms to get x on the left side and the 5 and -1 terms to get 4 on the right side.

What is the next step after combining like terms?

The next step is to isolate the x term by getting rid of any constants on the same side as the x term. In this case, we can subtract 4 from both sides to get x = -1.

How can we check if our solution is correct?

We can plug in the value we found for x (-1) into the original equation and see if it makes the equation true. In this case, we have 3(-1)+5 - 2(-1)-1 = 3, which simplifies to 3-1+2-1 = 3 and is true, so our solution is correct.

Are there any other solutions to this problem?

No, there is only one solution for this problem. In general, linear equations with one variable have either one solution, no solutions, or infinitely many solutions.

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