Multiplying or dividing an equation with variables

1. Nov 14, 2015

Mr Davis 97

I need to get something cleared up. Let's say we have the equation $x = 1$. For this equation, the solution set is {1}. However, if I do the valid operation of multiplying both sides by x, we get $x^2 = x$. Now the solution set is {0, 1}. Additionally, if I take the original equation and divide by x, I get $\displaystyle 1 = \frac{1}{x}$. The solution set is still {1}, but now the domain of x values has changed, such that I can no longer try x = 0. My question is, how can we ever be sure of the solution set or the domain if they can change with doing operations with variables to both sides? An additional example is that we start with $x^2 = x$. We think to divide both sides by x, since there is a common factor. However, doesn't this get rid of the solution x = 0? How would we ever know that $x^2 = x$ was our original equation, with the solution set {0, 1}?

Last edited: Nov 14, 2015
2. Nov 14, 2015

Staff: Mentor

That's easy. You simply have to distinguish the cases. You aren't allowed to divide by 0 even if you hide it by saying x. And if you hide 0 in x and multiply, then your equation gets trivial. You have to assure that your operation is allowed! Same goes for < operations when multiplying with negative numbers. If you hide them in x, you have to calculate two cases: x > 0 and x < 0. Anything else is just wrong.

3. Nov 14, 2015

jbriggs444

If the operation that you carry out on the equation is reversible then the solution set will not change. For instance, you could add x to both sides of the equation. This is reversible because you can subtract x and recover the original equation. But if you multiply by x, you cannot always divide by x -- because x could be zero. That manipulation is not reversible.

If you perform the same manipulation (reversible or not) to both sides of an equation, the new equation will hold true whenever the old one would have held true. That means that all solutions to the original equation will also solve the modified equation. That, in turn means that the old solution set will be a subset of the new solution set. It may or may not be a proper subset.

If the manipulation is reversible then applying the reverse manipulation to the new equation will demonstrate that the solution set to the new equation is a subset of the solution set to the original equation. If the two solution sets are subsets of each other then those sets are equal. QED.

4. Nov 14, 2015

Staff: Mentor

No, the solution set is still {1}. The solution set did not include 0 when you started, and it still doesn't.

5. Nov 14, 2015

Mr Davis 97

That helps.

So imagine I have x = 2t - 1, which we will say models the linear displacement of a ball. At t = 0, the ball is at x = -1.
Now imagine that I want to find when displacement is 0, so I use 2t - 1 = 0. If I were to go from here, I would simply get t = 0.5. However, if I multiply both sides by t, I get 2t2 - t = 0, in which case we have two solutions, 0 and 0.5.
So what's going on here? Multiplying by t was a valid operation, and I end up with two solutions. In fact, now t = 0 does not agree with the original function, yet the operation was algebraically valid. What I am getting at is that doing a reversible operation seems to change the solutions and the function in arbitrary ways that don't reflect the original goal of finding when x = 0. It's all confusing me...

6. Nov 14, 2015

Staff: Mentor

No, it's not valid since you can't reverse it. You hided a possible 0 in t. Therefore you changed the set of solutions. Graphically you turned your straight into a parabola which happens to intersect at your original solution at t = ½.

7. Nov 14, 2015

Staff: Mentor

When you multiply both sides of an equation by a variable, there is the potential of increasing the size of the solution set.

Consider x - 1 = 0, with {1} as the solution set.
Now multiply both sides by x -3, getting (x - 1)(x - 3) = 0. The solution set is now {1, 3}.

When you multiply both sides of an equation by a nonzero constant, that's a reversible action, so there's no change in the solution set.

8. Nov 15, 2015

Mr Davis 97

So basically, by multiplying both sides by x, you are doing an irreversible operation since x could potentially be zero? So in the end, for the second equation, would we just look back at the original equation and consider 3 an extraneous solution?

9. Nov 15, 2015

Staff: Mentor

No, because again you multiplied with 0. This time disguised as a 3 in (x - 3).
You may multiply with x-3 if and only if x - 3 ≠ 0. Otherwise you change the task and the solution set.

10. Nov 15, 2015

Staff: Mentor

Yes.
I guess, although I'm not sure that you would want to dignify it in this way.

Extraneous solutions often come up in the context of equations with radicals, where they can creep in in ways that aren't obvious. Here's an example:
$\sqrt{x + 4} = x - 2$
If you square both sides (which is NOT a reversible operation), you get
$x + 4 = x^2 - 4x + 4$
$\Rightarrow x^2 - 5x = 0$
$\Rightarrow x(x - 5) = 0$
$\Rightarrow x = 5 \text{or} x = 0$
By substitution in the original equation, we see that x = 0 is extraneous, leaving only x = 5 as the solution of the original equation.

11. Nov 15, 2015

Mr Davis 97

So what about when we divide an equation like $2x^2 = x$ by $x$? Is that a reversible or irreversible operation? By doing that we eliminate 0 as a potential solution it seems...

12. Nov 15, 2015

Staff: Mentor

You pretty much never want to divide both sides of an equation by a variable, for exactly this reason - you lose the solution x = 0. Of course, if x = 0 is still a solution of the reduced equation, then you haven't lost it.
For example, if you divide both sides of $2x^3 = x^2$ by x, you are left with $2x^2 = x$, which has x = 0 and x = 1/2 as solutions. I wouldn't actually do this, though -- I'm just giving a cooked-up example.

With regard to your earlier question about multiplying both sides of an equation by a variable, the only place I can think of where this is appropriate is in converting polar-form equations to rectangular form.

A typical example:
Convert $r = \cos \theta$ to rectangular form.
Solution: Multiply both sides by r,
$r^2 = r\cos \theta$
$x^2 + y^2 = x$
$x^2 - x + y^2 = 0$
$x^2 - x + 1/4 + y^2 = 1/4$
$(x - 1/2)^2 + y^2 = 1/4$, a circle whose center is at (1/2, 0) with radius = 1/2

In this example, multiplying by r is valid, as the original equation already has r = 0 (when $\theta = \pi/2 + k\pi$. So multiplying by a variable did not change the solution set in this example.