# B Dividing an equation by a variable

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1. Nov 2, 2016

### Kajan thana

Hello guys,

I am confused about when we can multiply and divide by the variable. I know the if zero is hidden as a factor then we can not divide or multiply.

eg. x/2x+1=0 are we allowed to divide both side 2x+1. Although zero is hidden value of x, there are no other ways to solve this equation.

Another eg. cos(x)((-4sin^2(x)-1)=0

if we pluck x=0 into the equation, we will see it will not satisfy the equation so this shows x does not take the value of zero, does this mean that we can divide the equation by the variable by the variable..

and also are allowed divide zero by the variable?

Last edited: Nov 2, 2016
2. Nov 2, 2016

### Ssnow

Hi, given an equation $a=b$, by the equivalence principle you can multiply $x$ both sides and the result doesn't change $x\cdot a=x\cdot b$ (also when $x=0$ you have trivially that $0=0$). In principle you can think that dividing will be the same thing, this is true but only when the division makes sense...

3. Nov 2, 2016

### Staff: Mentor

Do you mean $\frac x 2 x +1=0$ (what a computer would assume), $\frac x {2x} + 1 =0$ or $\frac x {2x+1} = 0$ (what I assume)? Brackets are important. 2x+1 is in a denominator. You know that it cannot be zero for all values of x where the equation makes sense.

You can multiply both sides by 0, but then the equation gets trivial, as 0=0 is always true and "[something] => [something true]" does not have any information.

Yes.

4. Nov 2, 2016

### Kajan thana

This equation.

5. Nov 2, 2016

### Kajan thana

For the equation above we know x being zero does not satisfy the equation , then if we divide it by cos(x), are we not losing a set of solution.

6. Nov 2, 2016

### pwsnafu

???
Which one? "x2x+1" would be equal to $2x^2+1$.

7. Nov 2, 2016

### Staff: Mentor

It doesn't work if you copy it like that.
x=0 and cos(x)=0 are completely different things.

8. Nov 2, 2016

### Kajan thana

And also will we loose any values if divide the 0 by the variable?

9. Nov 2, 2016

### Staff: Mentor

Possibly. It's usually not a good idea to divide both sides of an equation by a variable.
Here's a very simple example:
$x^2 = x$
If you divide both sides by x, you get x = 1, but the original equation has two solutions: x = 1 and x = 0.
For this equation, a better strategy is to bring all variables to one side, and then factor.
$x^2 - x = 0 \\ \Rightarrow x(x - 1) = 0 \\ \Rightarrow x = 1 \text{ or } x = 0$

If you do divide by a variable, keep in mind that you are possibly losing the solution x = 0.

The counterpart to dividing both sides of an equation by a variable is multiplying both sides of an equation by a variable. Here's an example of an equation in polar coordinates that shows this technique:
Convert the following equation in polar form to rectangular form:
$r = \cos(\theta)$
I'm going to mulitply both sides by r, noting that r = 0 is already a solution of this equation (when $\theta$ is any odd multiple of $\pi/2$).
$r^2 = r\cos(\theta) \\ \Rightarrow x^2 + y^2 = x \\ \Rightarrow x^2 - x + y^2 = 0 \\ \Rightarrow x^2 - x + 1/4 + y^2 = 1/4 \\ \Rightarrow (x - 1/2)^2 + y^2 = (1/2)^2$
In the second line, I converted $r^2$ to its rectangular form of $x^2 + y^2$, and converted $r\cos(\theta)$ to x.
In the fourth line above, I completed the square in the x terms, and in the fifth line, I factored the expression in x.
The last line is the equation of a circle centered at (1/2, 0), of radius 1/2.

Last edited: Nov 2, 2016
10. Nov 3, 2016

### Stephen Tashi

Yes, and in general, you have to keep in mind that you may be losing or gaining solutions, no matter what you do to both sides of an equation. It isn't even completely safe to add and subtract things to both sides of an equation. For example:

$x - \frac{1}{(x-3)} = 3 - \frac{1}{(x-3)}$ can be transformed to $x = 3$ by adding $\frac{1}{(x-3)}$ to both sides.

11. Nov 3, 2016

### Staff: Mentor

That almost goes without saying. If you add or subtract, or multiply by or divide by an expression that might be undefined, you have to be careful.

12. Nov 3, 2016

### Kajan thana

I don't understand this question :

secθ(cosecθ-2)=(cosecθ-2)
For the question above we know that cosecθ can't be zero so we can divide RHS by (cosecθ-2) and we will not solution but truly if we take the (cosecθ-2) from RHS to LHS and factorise it we will get two sets of solutions. I don't understand where we will loose a solution, is there any way to determine if I am going to loose some solutions when I divide. At the start I thought if zero is not a solution then we can do whatever we want and still won't loose any solution.

Thank you.

13. Nov 3, 2016

### Stephen Tashi

If you divide by (cosec$\theta$-2) , what you need to consider is whether there is value of $\theta$ such that (cosec$\theta$ - 2) = 0.

There is no general way to solve equations that can be done without thinking. However, in textbooks and in many real life situations, we encounter equations that aren't particularly tricky and we can solve them by "doing the same thing to both sides" several times. People, including myself, often use the procedure of solving the equation with manipulations without thinking much about them. Then they check their solutions by substitution in the original equal to see if all their answers actually work. They use graphs to see if they left out any solutions.

Solving an equation by pure logical deduction would involve considering a number of "cases" instead of writing out a singe sequence of steps. Very few people want to teach solving equations deductively because mastering logical deduction that splits into several cases is hard. Furthermore, it would be a pain in the neck to grade homework that went into that much detail. (Homework that's easy is to grade has is a single sequence of steps to check. )

From the point of view of logical deduction, when you want to divide both sides of an equation by a factor F, you actually need to consider the cases 1) F is a non-zero number 2) F is zero. You need to consider how each of those cases provides with a solution to the equation or eliminates certain numbers from being solutions.

For example:

$(x + 2) \sqrt{x - 2} = \sqrt{x-2}$

case 1) Assume $\sqrt{x-2}$ is a non-zero number.
Then we may divide both sides by $\sqrt{x-2}$ obtaining the equation $(x+2) = 1$, which has the solution x = -1. However, x = -1 makes $\sqrt{x-2}$ undefined, which contradicts the assumption of case 1) that $\sqrt{x-2}$ is a non-zero number. So x =-1 is not a solution to case 1)

case 2) Assume $\sqrt{x-2} = 0$.
This implies x = 2. By substituting x = 2 in the original equation we see that x = 2 is a solution.

14. Nov 3, 2016

### Kajan thana

Thank you made much more sense now, but I feel like case 1 fall under the principle of case 2 for some reason. Or am I thinking in a wrong way?

15. Nov 3, 2016

### Staff: Mentor

It's much simpler to rearrange the equation by bringing all terms to one side.
secθ(cosecθ-2)=(cosecθ-2)
secθ(cosecθ-2) - (cosecθ-2) = 0
(cosecθ-2)(secθ - 1) = 0

Now, either cosecθ = 2 or secθ = 1
The rest is just solving for θ from the two equations above.

16. Nov 4, 2016

### Stephen Tashi

You're thinking the wrong way if you think that assumption made in case 1 applies to case 2. "Non-zero" means $\neq 0$.

17. Nov 4, 2016

### Kajan thana

The way way I think of this question is, if I want to divide the equation by factor F then I will equal the F to zero to find the value of X; then I will pluck it into the original equation. If the LHS = RHS then i will know that we can not divide by that factor so i will make the equation equal to zero then factories it out.

18. Nov 4, 2016

### Stephen Tashi

The usual slang is to "plug" it in. ( "To pluck" means "to pull out".)

You are saying that you do the cases in order. Case 1) first. Case 2) second. That doesn't mean that Case 2) "is part of" Case 1).

You consider the case when the factor is zero. You solve for those value(s) of X that make the factor zero. If any of those values make the LHS=RHS in the original equation, you append those values to the list of solutions.

Next you do case 2. Assume the factor is not zero and divide both sides of the equation by the factor. You proceed to find other solutions of the reduced equation. However if a solution of the reduced equation makes the factor zero, it is not appended to the list of solutions of the original equation because it contradicts the assumption of case 2.

There are also cases to consider in the simple operation of subtracting an expression from both sides of an equation.

Case 1): The expression is a number.
Case 2) The expression does not represent a number.

Consider the following steps, which omit any mention of cases:

original equation: $(x+2)\sqrt{x-2} = \sqrt{x-2}$

step 1) $(x+2)\sqrt{x-2} - \sqrt{x-2} = 0$

step 2) $( (x+2) - 1) (\sqrt{x-2}) = 0$

step 3) $(x+1)(\sqrt{x-2}) = 0$

It would seem the solutions to the original equation are x = -1 and x = 2. However x = -1 is not a solution. The flaw in the work is step 2) where $\sqrt{x-2}$ is subtracted from both sides. To subtract an expression from both sides, we must assume the expression is a number. In order for $\sqrt{x-2}$ to be a number, we must assume $x \geq 2$.

As I mentioned in a previous post, it is tedious work to solve equations by a deductive process. To do so correctly, you must consider cases. Most teachers don't expect students to solve equations by precise logical reasoning. They give guidelines that correct the results of flawed steps such as "Always substitute your answers in the original equation to see if they actually work". That is good practical advice.

19. Nov 4, 2016

### Kajan thana

Thank you for the help, it makes much more sense now.