Rational Equations using LCD. Same problem, still stuck.

[Guest]

I found where I had errored in the first line and did it again, but now I'm stuck again.


x-3...........3x
---- + 4 = ----
x-4............x

x(x-3) + 4x(x-4) = 3x(x-4)

x^2 - 3x + 4x^2 - 16x = 3x^2 - 12x

5x^2 - 19x = 3x^2 -12x

2x^2 - 19x +12x

2x^2 - 7x = 0

That is as far as I got. I don't know if that is right, but I did fix the mistake I had made in my last post. If I did it right, I don't know how to reduce it to just the x value.

 

[wjm11]

x-3...........3x
---- + 4 = ----
x-4............x

x(x-3) + 4x(x-4) = 3x(x-4)

x^2 - 3x + 4x^2 - 16x = 3x^2 - 12x

5x^2 - 19x = 3x^2 -12x

2x^2 - 19x +12x

2x^2 - 7x = 0

What you have done is fine – although it would have been much simpler if you had cancelled the x’s in (3x/x) = 3. The next step in your approach is to factor out an x from both terms:

2x^2 - 7x = 0
x(2x – 7) = 0

Now, by virtue of the Zero Product rule, you can set each factor equal to zero and solve for x. You’ll get two solutions for x:

X = 0
X = 7/2

However, you must examine the original problem statement to determine the domain (allowable values) for x. X may not be either 4 or 0, as this would result in denominators equal to zero. Therefore, x = 0 is not a solution to this problem. The solution is x = 7/2.

 

[soroban]

Hello, am3ity!

I see nothing wrong with your work.
. . But why did they give us $\displaystyle \frac{3x}{x}$ ? It's as silly as having $\displaystyle \frac{100}{50}$, isn't it?

You got: .$\displaystyle 2x^2\,-\,7x\:=\:0$ . . . It's a quadratic . . . factor it!

. . And we get: .$\displaystyle x(2x\,-\,7)\:=\:0$

. . And get two roots: .$\displaystyle x\,=\,0,\,\frac{7}{2}$

But $\displaystyle x=0$ is not allowed.

. . Therefore, $\displaystyle x=\frac{7}{2}$ is the only solution.


Edit: Too fast for me, wjm11 !

 

[wjm11]

Happy Thanksgiving, Soroban. I continue to admire your work!