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Mathematics 7 Online
OpenStudy (anonymous):

Solve the following linear equation: I will put the equation below. I think where I am going wrong is finding a least common denominator to try to get rid of the denominator.

OpenStudy (anonymous):

Ok, well let's see what you have done so far.

OpenStudy (anonymous):

OpenStudy (anonymous):

So I factored the denominators to be (2x-5)(x+0), x, and 2(x-5)

OpenStudy (anonymous):

2x^2 - 5x -> x(2x-5) 4x - 10 -> 2(2x-5)

OpenStudy (anonymous):

ohh okay, that would make a lot more sense than (x+0) haha

OpenStudy (anonymous):

=0)

OpenStudy (anonymous):

okay let me work through it again and see what i get. I'll tag you when im done or if i have questions :)

OpenStudy (anonymous):

Okay, @Cosmichaotic i'm still doing something wrong.. haha. Would the LCD be 2x(2x-5)?

OpenStudy (anonymous):

Yea, I think so!

OpenStudy (anonymous):

You gotta make each denominator = 2x(2x-5) by multiplying the numerators by the missing components of that denominator.

OpenStudy (anonymous):

If I remember correctly.

OpenStudy (anonymous):

we were told to multiply the LCD by the numerator and cancel the denominators out, so that you aren't dealing with fractions

OpenStudy (anonymous):

That's a good way to do it.

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