Quickest way to do this calculation

[Manasan3010]

Homework Statement

Find value of $x$

Relevant Equations

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$\frac{-3x^4}{7}+x=\frac{-3x^4}{7}+10 \\x=10$

I solved the equation and got x=10, My question is how can I solve for x without brute-force method( Multiply all terms by (x-13)(x-7)(x-14)(x-6) and solving for x)

 

[phinds]

I can't see any other method.

 

[fresh_42]

\begin{align*}
\dfrac{5x-64}{x-13}+\dfrac{x-6}{x-7}&=\dfrac{4x-55}{x-14}+\dfrac{2x-11}{x-6}\\
\dfrac{5(x-13)+1}{x-13}+\dfrac{x-7+1}{x-7}&=\dfrac{4(x-14)+1}{x-14}+\dfrac{2(x-6)+1}{x-6}\\
5 + \dfrac{1}{x-13} + 1 + \dfrac{1}{x-7}&=4+\dfrac{1}{x-14}+2+\dfrac{1}{x-6}\\
\dfrac{1}{x-13} + \dfrac{1}{x-7}&=\dfrac{1}{x-14}+\dfrac{1}{x-6}\\
&\text{ etc. }
\end{align*}
looks a lot easier!

 

[SammyS]

Continuing with:
$\dfrac{1}{x-13} + \dfrac{1}{x-7}=\dfrac{1}{x-14}+\dfrac{1}{x-6}$
From there ...

$\dfrac{x-7 + x-13}{(x-13)(x-7)}=\dfrac{x-6 + x-14}{(x-14)(x-6)}$

Some simplifying gives

$\dfrac{2(x-10)}{(x-13)(x-7)}=\dfrac{2(x-10)}{(x-14)(x-6)}$

Equate the denominators & solve for $x$.

 

[fresh_42]

From
$$
\dfrac{2(x-10)}{p(x)}=\dfrac{2(x-10)}{q(x)} \quad (p\neq q \in \mathbb{R}[x])
$$
is nothing left to solve, it is already written there.

 

[Manasan3010]

From
$$
\dfrac{2(x-10)}{p(x)}=\dfrac{2(x-10)}{q(x)} \quad (p\neq q \in \mathbb{R}[x])
$$
is nothing left to solve, it is already written there.

Can you please elaborate on how the numerators of the fractions are being equalled.
Even if the numerators are equal, equation would become
2x-10 = 2x-10
x=x(How do I find the value of x from this)
I also want to know what is meant by$\quad (p\neq q \in \mathbb{R}[x])$ in words.

 

[phinds]

Some simplifying gives

$\dfrac{2(x-10)}{(x-13)(x-7)}=\dfrac{2(x-10)}{(x-14)(x-6)}$

Equate the denominators & solve for $x$.

That equation simplifies to 91 = 84 so (1) clearly you've made a mistake somewhere, and (2) it can't be used to find x since the x's all drop out and just leave 91 = 84. Not very useful.

 

[ehild]

That equation simplifies to 91 = 84 so (1) clearly you've made a mistake somewhere, and (2) it can't be used to find x since the x's all drop out and just leave 91 = 84. Not very useful.

$\dfrac{2(x-10)}{(x-13)(x-7)}=\dfrac{2(x-10)}{(x-14)(x-6)}$
Following @fresh 42's hint, if x=10 both sides of the equation are zero, independently of the denominators. They need not be equal. So the solution is x=10

 

[Manasan3010]

$\dfrac{2(x-10)}{(x-13)(x-7)}=\dfrac{2(x-10)}{(x-14)(x-6)}$
Following @fresh 42's hint, if x=10 both sides of the equation are zero, independently of the denominators. They need not be equal. So the solution is x=10

But How do I mathematically show that 10 is a possible solution for x?
In test can I say that If you plug 10 for x, Equation is veracious so 10 is a possible solution?

 

[Ibix]

Equation is veracious so 10 is a possible solution?

And, as per fresh's post #5, there are no others.

Alternatively, rewrite the last expression in post #3 as$$\frac 1{x-7}-\frac 1{x-6}=\frac 1{x-14}-\frac 1{x-13}$$and put each side over a common denominator. The result is one over a quadratic in both cases, and the result follows simply.

 

[fresh_42]

But How do I mathematically show that 10 is a possible solution for x?
In test can I say that If you plug 10 for x, Equation is veracious so 10 is a possible solution?

$\dfrac{a(x)}{p(x)}=\dfrac{a(x)}{q(x)} \Longleftrightarrow a(x)(q(x)-q(x))=0 \Longleftrightarrow a(x)=0 \text{ or }p(x)=q(x)$
As we can rule out $p=q$ we are left with $a(x)=2(x-10)=0$ and it can be seen that $x=10$ is the only way for $a(x)$ to be zero.

So all steps which are missing in post #3 can actually be done in mind.

 

[SammyS]

I have a bad habit of writing up a post, previewing and editing until most typos are fixed, then when it's ready to go, I neglect clicking to post it.

The following is part of a post I wrote up yesterday evening, shortly after @phinds (that big white dog could pass for a polar bear) posted that enlightening observation.

That equation simplifies to 91 = 84 so (1) clearly you've made a mistake somewhere, and (2) it can't be used to find x since the x's all drop out and just leave 91 = 84. Not very useful.

Oh my! Yes, of course.

... And since you point out that the denominator on the left is 7 units greater than the denominator on the right (for all values of $x$), the only solution possible is for the numerators to be zero.

 

[phinds]

... And since you point out that the denominator on the left is 7 units greater than the denominator on the right (for all values of $x$), the only solution possible is for the numerators to be zero.

Good point. That's a point of view I've never had to take before so basically rejected it. Not too bright sometimes, but I mean, come on ... I'm just a dog after all

 

[Mark44]

From
$$
\dfrac{2(x-10)}{p(x)}=\dfrac{2(x-10)}{q(x)} \quad (p\neq q \in \mathbb{R}[x])
$$
is nothing left to solve, it is already written there.

Even if the numerators are equal, equation would become
2x-10 = 2x-10

@Manasan3010, did you mean "even if the denominators are equal..."
If so, the numerators are $2(x - 10) = 2x - 20 \ne 2x - 10$

 

[FactChecker]

But How do I mathematically show that 10 is a possible solution for x?
In test can I say that If you plug 10 for x, Equation is veracious so 10 is a possible solution?

Yes. After all those manipulations of equations, it is always good to show that your answer satisfies the original equations. Then you should ask yourself if there are additional solutions. If x is not 10 and the numerators in the last equation of post #4 are equal but not zero, then the denominators must be equal for some value of x other than 10. But @phinds pointed out in post #7 that setting the denominators equal leads to 91=84, so that is not possible.

 

[Chestermiller]

In my judgment, the method recommended by @Ibix in post #10 is the clearcut winner.