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Mathematics 11 Online
OpenStudy (anonymous):

(2/5) + (3/5x) = (x + 5/10) The answer is NOT x = 2, as I originally thought.

OpenStudy (anonymous):

LCD is 10x. 4x/10x + 6/10x = x^2 + 5x/10x 4x + 6 = x^2 + 5x 0 = x^2 + x - 6 Where to from there? Factoring?

OpenStudy (anonymous):

That would give me 0 = (x - 2)(x + 3), correct? And the solutions would be x = 2, x = -3. Now I would need to plug them back in.

OpenStudy (anonymous):

2/5 + 3/10 = 7/10

OpenStudy (anonymous):

2/5 x 2/2 to create a common denominator...4/10 + 3/10 = 7/10. This is true. Now we check -3.

OpenStudy (anonymous):

2/5 + 3/-15 = 2/10...need -15 as the LCD

OpenStudy (anonymous):

Wouldn't x= -1/4 ?

OpenStudy (anonymous):

I don't have any fractions in my answers...and the LCD wouldn't be -15. Hm.

OpenStudy (anonymous):

@absurdism Come back, please? I'm stuck. :)

OpenStudy (anonymous):

-3 pretty much has to work when you plug it back in...but I'm trying to find the LCD so I can check it.

OpenStudy (anonymous):

(I say that it has to work, because my options are no solutions, 2, -3, or 2 and -3, and I know that 2 works but 2 is not the only answer, thus no solutions can't be the answer, so I have to make sure -3 works).

OpenStudy (anonymous):

sorry, is the equation 2/5 + 3/(5x) = x + 5/10? plus if you have solved the equation by factoring you needn't do anything else

OpenStudy (anonymous):

Yes, that is the equation. But I have to check my work. :)

OpenStudy (anonymous):

To make sure it's not an extraneous solution.

OpenStudy (anonymous):

it must be 2 and -3 (x-2)(x+3) = x^2 + 3x - 2x - 6 = x^2 + x - 6

OpenStudy (anonymous):

that's as much work check as you need

OpenStudy (anonymous):

Okay. Thank you!

OpenStudy (anonymous):

no worries

OpenStudy (anonymous):

@absurdism Can you help with a slightly harder one? Same topic.

OpenStudy (anonymous):

mhm

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