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I'm sorry if this question (or possibly multiple depending on how long this I intend for this to be) is a little too elementary, but I've been seriously struggling with this for the past week. I never really questioned this until recently and it has me stumped and I can't move on with math without knowing the answer.

Say I have the equation $3x+5=20$

So, I understand that subtracting 5 from both sides and dividing both sides by 3 would keep the equality true. What I don't understand is how doing all this somehow doesn't change the value of x. The equality is true but what if x is not equal to what it was in the original equation anymore? The solution is of course x=5, I have no problem mechanically solving the equation but this intuitively is hard for me to grasp. How can we really know that x=5 is true and have complete confidence in the answer without substitution? It's even harder to trust the solution after a series of these operations, especially when they're more complex like taking the log of both sides, exponentiation, etc.

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    $\begingroup$ Well, sometimes you do end up with other values of $x$ that are not equal to the desired value but do not satisfy the original equation. Take the simple equation $x=5$, for example. If you square it, you get $x^2=25$. $x=5$ still works, but now we have an extra solution, $x=-5$, which obviously does not satisfy $x=5$. This tends to happen when you introduce non-bijective operations (like squaring), for which the only remedy is to keep in mind exactly what you're doing at each step, taking note of any extraneous solutions introduced so that you can exclude them in your conclusions. $\endgroup$ Commented May 25 at 5:23
  • $\begingroup$ @H.sapiensrex that's a wonderful example of an equation where one needs to be careful while simplifying. I'd also like to mention here that $√(x^2)$ is $mod(x)$ and not $x$ and $-x$. $\endgroup$ Commented May 25 at 5:35
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    $\begingroup$ @H.sapiensrex Actually, surjectivity is not a concern in solving equations, as injective functions are left-invertible. $\endgroup$
    – Alex Jones
    Commented May 25 at 18:44
  • $\begingroup$ @AlexJones I didn't say surjectivity, but bijectivity $\endgroup$ Commented May 27 at 5:57
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    $\begingroup$ @H.sapiensrex ... and surjectivity is exactly half of bijectivity. I'm not saying your statement was incorrect, only that you could replace "non-bijective" with "non-injective" and not lose anything. $\endgroup$
    – Alex Jones
    Commented May 27 at 14:42

5 Answers 5

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I'm sorry if this question (or possibly multiple depending on how long this I intend for this to be) is a little too elementary, but I've been seriously struggling with this for the past week. I never really questioned this until recently and it has me stumped and I can't move on with math without knowing the answer.

This question might be "elementary", but is an important one nevertheless. This is crucial for a student to understand algebra well.

Mathematics is based on certain statements which are assumed to be true: the so- called "axioms".

Two of the many such axioms, is:

When equals are added to equals, the wholes are equal.

When equals are subtracted from equals, the remainders are equal.

This means that, if the initial condition, that $3x+5=20$ is true, then $$(3x+5)-5=20-5$$ must be true.

It is also trivial to see that when you multiply the same quantity on both sides, the products must be equal.

It's even harder to trust the solution after a series of these operations, especially when they're more complex like taking the log of both sides, exponentiation, etc.

You need to be more careful while performing such operations on equations. Consider the equation: $$y^x=1$$

Recklessly taking logarithm with the base y, you would get $x=0$. But that's not the only solution. Instead, if one takes a logarithm (with base e) on both sides:

$$x \ln y = 0$$ Which clearly has 2 solutions, $x=0$ or $y=1$. So the lesson one must learn from this example is, always be careful while manipulating algebraic equations and inequalities; else, you may end up with extraneous roots or miss a few of them.

I would encourage you to try out more such algebraic operations, and see which ones work well and which ones you need to be careful with- especially if you're dealing with equations involving inverse trigonometric functions, or with functions whose domain/range is restricted.

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  • $\begingroup$ Hm, so, I understand the addition, subtraction, multiplication, etc. properties. My question is, how does applying them not change the solution to the equation? I guess a better and clearer way to state my question is: How are (3x+5)-5=20-5 and (3x+5)=20 equivalent to each other? And how would they yield the same solution? It's hard for me personally to conceive the idea that you can do whatever you want to the equation without changing it's value. Ever since last week I for some reason just can't comprehend it. $\endgroup$
    – Lucas
    Commented May 25 at 5:48
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    $\begingroup$ @Lucas Think of the variable "x" as a placeholder for a number. So $(3x+5)$ IS A NUMBER for any value of $x$. Now, if it were to satisfy this equation, then the given equation must be true. In essence, both the LHS and the RHS are numbers, nothing more, nothing less. So, if the numbers are equal, (i.e., they are the SAME NUMBER, then you can add the same number to both the numbers, and the resultant numbers on LHS and RHS would still be the same! P.S. Take your time to figure it out. Keep thinking about it, and let me know if you could figure it out. $\endgroup$ Commented May 25 at 6:07
  • $\begingroup$ But why would the value of x remain the same even if they're still equal to each other? I don't understand why the solution would satisfy both the original and the modified considering the values are not the same. $\endgroup$
    – Lucas
    Commented May 25 at 6:22
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    $\begingroup$ @Lucas The original number which satisfies the LHS is obtained when X takes a certain value. Now, when you subtracted 5 on both sides, the values on the LHS and the RHS must remain equal. So the same value of X must satisfy the new equation as well. Consider the equation $3(2)+5=11$. Subtract some number, say 6, on both sides, you get $3(2)-1 = 5$. This equation is still true for the same value of $x$ (that is, 2). Just try to imagine x as a NUMBER, not as an unknown entity. Think about it. $\endgroup$ Commented May 25 at 6:24
  • $\begingroup$ Thank you! I get it now. $\endgroup$
    – Lucas
    Commented May 25 at 6:44
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The process of solving an equation typically involves some kind of "do something to both sides" step(s). The concept of "do something" in math is called a function. A function is a thing which takes some input and gives you some output, with the caveat that the same input always gives the same output. This is important, because it allows us to trust that every application has the same effect.

When we apply a function to both sides of an equation, what happens?

Well, if the equation were true, then both sides would be the same. And by the definition of a function, when you put the same thing in, you get the same thing out, and so both sides will still be the same. Thus, the new equation will also be true.

But if it were false, then we don't know! It's possible that two different values go into a function, but two same values pop out. Consider the function that squares its inputs; $(1)^2 = 1$ and $(-1)^2 = 1$, so the obviously false $1 = -1$ becomes the true statement $1=1$. As a result, $x = -1$ can become $x^2 = 1$, which has $x = 1$ as an additional solution. So, doing something to both sides may actually change the solution, in particular by creating new solutions.

That's not all, though. See, a function doesn't have to accept everything as input. For example, the square root function doesn't accept negative values, while the reciprocal function never accepts zero. If you apply such a function carelessly, you may actually destroy valid solutions. For example, $x^2 = x$ is true when $x = 0$ or when $x = 1$, but divide both sides by $x$ and the resulting equation $x = 1$ is no longer true for $x = 0$. This is particularly a danger when the function involves any of the equation variables.

All this is to say, you are absolutely right to be wary of the solutions changing as you "do something to both sides", since it can both create and destroy solutions, so you can end up with an entirely different solution set.

So when is it safe to apply a function to both sides? The first issue was that two different inputs gave two same outputs, so let's say our function doesn't do this; this is called a one-to-one function. The next issue was that our function didn't take all inputs, so let's make sure our function can take all real inputs. Note that, if any of our equation variables are part of the function, we're actually defining a whole family of functions, and so we would need to verify that this is true for all of them; as such, let's avoid this as well.

Thus, a sufficient (but not necessary) condition would be that any function which is defined everywhere and is one-to-one, which doesn't include any of our equation variables, can be safely applied to both sides while preserving the solution set.

An example would be the function which subtracts $5$ from its input. It's clearly a function, as subtracting $5$ does the same thing every time. It's one-to-one because subtracting $5$ from two different numbers always gives two different numbers. It's defined everywhere; there's no numbers that we can't subtract $5$ from. And it doesn't involve any equation variables, it's just subtracting the fixed value $5$ no matter what. As a result, subtracting $5$ from both sides of an equation does not affect the solution set. Similarly, dividing by $3$ on both sides does not affect the solution set, and so the equations $3x+5 = 20$ and $x = 5$ are equivalent.

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Other than the already posted answer, here is another way to look at it. $$\begin{align*} x &=\frac13(3x)\\ &=\frac13(3x+5-5)\\ &=\frac13(20-5)\\ &=\frac13(15)\\ &=5 \end{align*}$$ This is just another way to write the algebraic manipulations, but all you need to do here, is just substitute $3x+5=20$.

Hope this helps. :)

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Think about it in even more basic terms. In an equation the $x$ is just an unknown value.

So let's think less abstractly, in terms of concrete numbers. Suppose you already know the value of $x$ in advance. In the examples below, imagine that the bracketed number plays the role of $x$. You'll see you're not affecting the value of $x$ by these operations.

The assumption is that the equation is satisfied from the start. All the manipulations maintain the equality. Not the overall value on each side (what you get when you multiply and sum up everything together). These change, but they change in the same way and so the two sides always remain equal - it's this fact that remains unaffected, and it's what we're making use of.

What you're doing is you're eliminating any extra fluff on one side, until just the $x$ remains, and then seeing what the value on the right side turns out to be. That is all.

$$ (4) + 2 = 6 \\ \color{grey}{\text{# Now subtract 2 from both sides:}} \\ (4) + 2 \color{red}{- 2} = 6 \color{red}{- 2} \\ (4) = 4 \\ $$

You're just eliminating the extra term (turning it into a zero).

Or with multiplication: $$ 3\cdot(7) = 21 \\ \color{grey}{\text{# Now divide both sides by 3:}} \\ \frac{3}{\color{red}{3}}(7) = \frac{21}{\color{red}{3}} \\ (7) = 7 $$

You're just eliminating the multiplier (turning it into $1$).
Note also that: $\frac{3}{\color{red}{3}}(7)$ and $\frac{3(7)}{\color{red}{3}}$ are two ways to write down the same thing.

In your example, you have:

$$ \color{grey}{\text{# First, each side equals 20}} \\ 3(5) + 5 = 20 \\ 3(5) + 5 \color{red}{- 5} =20 \color{red}{- 5} \\ \color{grey}{\text{# Now each side equals 15}} \\ 3(5) = 15 \\ \frac{3}{\color{red}{3}}(5) = \frac{15}{\color{red}{3}} \\ \color{grey}{\text{# Now each side equals 5}} \\ \color{grey}{\text{# it's just that we can conclude that x = 5}} \\ \color{grey}{\text{# because nothing but x remains on the left}} \\ (5) = 5$$

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To use a simplistic metaphor: I have many bags, each filled with $x$ apples. On one side of my scale, I have $20$ apples, and on the other, I have three bags such bags, plus an extra five loose apples. Saying $$3x + 5 = 20$$ is all so much words to say that my scale is balanced- both sides have the same number of apples. Now, I go about these algebraic manipulations- subtracting five from both sides means I clear out my loose apples on the one side, and take away a matching five apples on the other. Division by three: I partition my apples on the one side into three equal groups, and on the other I focus my attention on just one bag. At the end of this story, I'm left with just one bag on the one side, and five apples on the other, and I happily say my bag had five apples in it all along. $x = 5$.

But, to answer your question, "why didn't $x$ change", notice in this metaphor that I never opened the bags. I simply didn't have the opportunity to change $x$.

This answer is not at the same level of rigor of many of the others here- it's certainly not anything approaching a proof of the fact that $x$ did not change. It only worked because every number involved was a positive integer (the metaphor gets completely unwieldly if $x$ was, say, $-0.12$.) But I hope that this provides you with a mental model in which $x$ doesn't change, like you ask for.

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