MHB Solving An Equation With Rational Terms

[Simon T]

Hey guys, I am not to good in Math and I am having issues solving this equation.

2x-9/3x + 8 = 3/7x How do you solve this equation? The answer is supposed to be x = 0.3956?

 

[MarkFL]

Re: equation help

Okay, what you mean is:

(2x - 9)/(3x) + 8 = 3/(7x)

or, using $\LaTeX$:

$$\frac{2x-9}{3x}+8=\frac{3}{7x}$$

The first thing we could do is multiply through by the LCD...can you state what the LCD is in this problem?

 

[Simon T]

Re: equation help

it doesn't say what the LCD is, we have to figure that out. and yes, the second part of your post is correct

 

[MarkFL]

Re: equation help

it doesn't say what the LCD is, we have to figure that out. and yes, the second part of your post is correct

Yes, we need to come up with that on our own...can you find it? :D

 

[Simon T]

Re: equation help

would it be 21?

 

[MarkFL]

Re: equation help

would it be 21?

It would be 21 if the denominators were 3 and 7, but there is also $x$ as a factor in both, so the LCD is:

$21x$

So, what do you get when you multiply through by $21x$?

 

[Simon T]

Re: equation help

do you multiply both dominators by 21?

 

[MarkFL]

Re: equation help

do you multiply both dominators by 21?

No, you multiply everything by $21x$:

$$\frac{2x-9}{3x}\cdot21x+8\cdot21x=\frac{3}{7x}\cdot21x$$

After dividing out common factors, we then have:

$$(2x-9)7+8\cdot21x=3\cdot3$$

Now it is a matter of distributing, collecting like terms, and solving for $x$...what do you get?

 

[Simon T]

Re: equation help

where did you get 7 + 8 from?

 

[MarkFL]

Re: equation help

where did you get 7 + 8 from?

$$\frac{2x-9}{3x}\cdot21x+8\cdot21x=\frac{3}{7x}\cdot21x$$

$$\frac{2x-9}{3x}(21x)+8\cdot21x=\frac{3}{7x}(21x)$$

$$\frac{2x-9}{\cancel{3x}}(\cancel{3x}\cdot7)+8\cdot21x=\frac{3}{\cancel{7x}}(\cancel{7x}\cdot3)$$

$$(2x-9)(7)+8\cdot21x=3(3)$$

$$7(2x-9)+8\cdot21x=3\cdot3$$

 

[Simon T]

Re: equation help

where did you get the 7? I am very confused :/

 

[MarkFL]

Re: equation help

where did you get the 7? I am very confused :/

$$\frac{2x-9}{3x}\cdot21x=(2x-9)\frac{21x}{3x}=(2x-9)\frac{3x\cdot7}{3x}=(2x-9)\frac{\cancel{3x}\cdot7}{\cancel{3x}}=(2x-9)7=7(2x-9)$$