Basic Algebra Question: 3x = 15

In summary, a problem may ask for one or more steps to be done in a certain way, but the steps themselves are not always required to be justified.
  • #1
koiuuuuuuuuuuu
7
0
TL;DR Summary
Please show me the method if I'm solving the equation wrong.
Can you help me with this question.
3x=15
Do you have to divide both sides by 3 which, is x=5
Am I right. If not explain please,
Thank-you
 
Last edited by a moderator:
Mathematics news on Phys.org
  • #2
Yes, it is correct. More precisely, you multiply both sides with ##1/3##.
 
  • Like
Likes symbolipoint
  • #3
Thank a million
 
  • #4
Step by step:

Given the equation: ## 3 x = 15##
Factoring 15: ##3 x = 3 * 5##
Given the identity: ##1/3 = 1/3##

Multiply both sides: ##( 1/3 ) * ( 3 x ) = ( 1/3 ) * ( 3 * 5 )##
Apply associativity: ##( 1/3 * 3) * ( x ) = ( 1/3 * 3 ) * ( 5 )##
Reduce inverses: ##( 1 ) * ( x ) = ( 1 ) * ( 5 )##

Answer: ##x = 5##
 
  • Like
Likes QuantumQuest and sysprog
  • #5
jedishrfu said:
Step by step:

Given the equation: ## 3 x = 15##
Factoring 15: ##3 x = 3 * 5##
Given the identity: ##1/3 = 1/3##

Multiply both sides: ##( 1/3 ) * ( 3 x ) = ( 1/3 ) * ( 3 * 5 )##
Apply associativity: ##( 1/3 * 3) * ( x ) = ( 1/3 * 3 ) * ( 5 )##
Reduce inverses: ##( 1 ) * ( x ) = ( 1 ) * ( 5 )##

Answer: ##x = 5##
Why do you need to factor 15? What if it si 3x=16?
 
  • #6
I did for the benefit of the OP for this problem.
 
  • Like
Likes symbolipoint
  • #7
nasu said:
Why do you need to factor 15? What if it si 3x=16?
Factoring the 15 is an option, and a good one. One knows the factorization for 15 anyway. When both sides are multiplied by (1/3) one either knows the basic multiplication fact of 3*5=15 and does the right-side multiplication easily, or one "does not " know the basic fact and may choose to factorize as 15=3*5, which then makes the right-side multiplication by (1/3) very obvious.
 
  • Like
Likes jedishrfu
  • #8
It's a good option when you are familiar with the algebra. May not be so good when someone struggles to understand the general procedure.
 
  • #9
But just knowing you can write numbers as factors is a fantastic leap in solving these problems.

Often teachers skip this because they want you to think long division or have you use a calculator. We know that school problems often have these ideas baked in whereas in the real world you may well have to resort to long division.

But then again you could apply an estimate (real world skill) and get an approximate value for x by saying 16 is close to 15 and solve it mentally.

I don’t think we should be arguing these points here. @nasu If you still feel strongly about it then open a new thread to discuss teaching methods. However, I do understand your point of view.

There are many times in math that we must balance the need to calculate vs the need to have insight into a problem and do the calculation in the final step. Admittedly here the 15/3 is the final step but the factoring provides the insight to say 5 and not whatever value the calculator gives ( more true when the answer is x=4/3 vs 1.333333...) but you get my point.
 
  • Like
Likes sysprog and symbolipoint
  • #10
jedishrfu said:
Step by step:

Given the equation: ## 3 x = 15##
Factoring 15: ##3 x = 3 * 5##
Given the identity: ##1/3 = 1/3##

Multiply both sides: ##( 1/3 ) * ( 3 x ) = ( 1/3 ) * ( 3 * 5 )##
Apply associativity: ##( 1/3 * 3) * ( x ) = ( 1/3 * 3 ) * ( 5 )##
Reduce inverses: ##( 1 ) * ( x ) = ( 1 ) * ( 5 )##

Answer: ##x = 5##
This can be done with far fewer steps.
##3x = 15##
Multiply both sides by 1/3: ##\frac 1 3 (3x) = \frac 1 3 (15)##
Simplify: ##x = 5##

Unless the problem requirements specifically stated that each step must be justified by citing some property of equations or addition or multiplication of real numbers (such as associativity, multiplicative identity, etc.), the sequence of operations I showed would suffice.
 
  • Like
Likes nasu, Klystron, sysprog and 2 others
  • #11
One should not depend on a solution method that requires the desired factor to be there. The answer should be found even if it is an improper factor or a mixed number.
 
  • Like
Likes nasu and sysprog
  • #12
koiuuuuuuuuuuu said:
Summary: Please show me the method if I'm solving the equation wrong.

Can you help me with this question.
3x=15
Do you have to divide both sides by 3 which, is x=5
Am I right. If not explain please,
Thank-you
FactChecker said:
One should not depend on a solution method that requires the desired factor to be there. The answer should be found even if it is an improper factor or a mixed number.
In other words, you can depend on the basic algebra rules.
3x=15

(1/3)3x=(1/3)*15

x=15/3
and reduce 'if possible', here which is 5.
 
  • Like
Likes FactChecker
  • #13
symbolipoint said:
In other words, you can depend on the basic algebra rules.
3x=15

(1/3)3x=(1/3)*15

x=15/3
and reduce 'if possible', here which is 5.
Why times by 1/3 rather than divide by 3?
I mean in terms of isolating x for the op
 
  • #14
pinball1970 said:
Why times by 1/3 rather than divide by 3?
I mean in terms of isolating x for the op
/3 is certainly the way I would look at it. I also don't understand why people are saying *(1/3)
 
  • Like
Likes pinball1970
  • #15
phinds said:
/3 is certainly the way I would look at it. I also don't understand why people are saying *(1/3)
I think the idea is to stay more into the abstract algebra side of things. "Dividing by 3" is an abbreviation meaning "multiply by the right-hand inverse of 3 on the right" which is not necessarily the same thing as multiplying by the left hand inverse of 3 on the left.

However, responding to a question about 3x=15 with such quibbles strikes me as unnecessary.
 
  • Like
Likes nasu and symbolipoint
  • #16
pinball1970 said:
Why times by 1/3 rather than divide by 3?
I mean in terms of isolating x for the op
That was not the point of that post (#12). The point was that expecting to be able to factor the 15, as suggested by a prior post, is not a good general assumption.
 
  • #17
For simpler equations, and for many more complicated single-variable ones, I was taught to first ask myself 'what is the last thing that was done to ##x##?', then undo that operation on both sides, then repeat as necessary until ##x## appears alone on only one side of the equation, and then ensure that the expression on the other side is fully simplified.

For the equation, ##3x=15##, that would mean seeing that on the left-hand side, ##x## has been multiplied by 3, so divide both sides by 3, which would immediately yield ##x=5##.

I think the procedure specified by @jedishrfu has more general applicability, but for getting the answer quickly for a simple equation like ##3x=15##, the simpler approach is preferable.
 
  • Like
Likes pinball1970 and jedishrfu
  • #18
phinds said:
/3 is certainly the way I would look at it. I also don't understand why people are saying *(1/3)
Is there something a little more far reaching the maths guys automatically think about in terms of next steps with algebra?
At 13 or 14 I was told to isolate an unknown by reversing the operation.
 
  • #19
I think the basic subtlety is that we like to define algebraic fields in terms of addition and multiplication with additive and multiplicative inverses.

That means we don't subtract as its a form of addition of an additive inverse and we don't divide as its a form of multiplication of a multiplicative inverse.

The other part is to discourage students from applying arithmetic operations too quickly meaning we should simplify and then use arithmetic operations to reduce to the answer.

I remember this coming up in Linear Algebra where the prof wanted us to add and subtract rows in matrix row reduction to get a 1 in a cell and to then use multiples of that row to simplify other rows and eventually come to a solution. Other methods of row reduction would lead to hideous fractions that caused a real mess in your solution.

In other words avoid arithmetic ops until the last step.
 
  • Like
Likes Klystron, pinball1970, symbolipoint and 1 other person
  • #20
pinball1970 said:
Is there something a little more far reaching the maths guys automatically think about in terms of next steps with algebra?
At 13 or 14 I was told to isolate an unknown by reversing the operation.
For one thing, in general, multiplying by the reciprocal of the coefficient safeguards against accidentally allowing a divisor to be zero (ref: https://www.pleacher.com/mp/mhumor/onezero2.html).
 
  • Like
Likes jedishrfu
  • #21
pinball1970 said:
Is there something a little more far reaching the maths guys automatically think about in terms of next steps with algebra?
If one understands the concepts of additive inverse and multiplicative inverse, then the four arithmetic operations really boil down to just two: addition and multiplication. Instead of subtracting 3, you can add -3, the additive inverse of 3; instead of dividing by 4, you can multiply by 1/4, the multiplicative inverse of 4. Two numbers that are additive inverses of each other add up to 0. Two numbers that are multiplicative inverses of each other multiply to 1.

This way of thinking has practical advantages, as well. For example, it can be less costly, in terms of CPU cycles, to multiply a number by 1/4 (or .25) than to divide by 4. Furthermore, some RISC (reduced instruction set computing) processors are able to eliminate a certain subtraction instruction, because the same result can be had by adding a negative value. IOW, instead of subtracting 3, we can add -3.
 
  • Like
Likes Klystron, pinball1970 and symbolipoint
  • #22
pinball1970 said:
Why times by 1/3 rather than divide by 3?
I mean in terms of isolating x for the op
Same thing.
Study of Basic Algebra
 
  • #23
THIS IS THE GOOD STUFF! WE FEEL GREAT WHEN WE KNOW TO DO THESE THINGS:
jedishrfu said:
I think the basic subtlety is that we like to define algebraic fields in terms of addition and multiplication with additive and multiplicative inverses.

That means we don't subtract as its a form of addition of an additive inverse and we don't divide as its a form of multiplication of a multiplicative inverse.

The other part is to discourage students from applying arithmetic operations too quickly meaning we should simplify and then use arithmetic operations to reduce to the answer.

I remember this coming up in Linear Algebra where the prof wanted us to add and subtract rows in matrix row reduction to get a 1 in a cell and to then use multiples of that row to simplify other rows and eventually come to a solution. Other methods of row reduction would lead to hideous fractions that caused a real mess in your solution.

In other words avoid arithmetic ops until the last step.
That is, except for my not understanding that stuff about the linear algebra you tried to describe.
 
  • #24
Yes again! This is the stuff that we need:
Mark44 said:
If one understands the concepts of additive inverse and multiplicative inverse, then the four arithmetic operations really boil down to just two: addition and multiplication. Instead of subtracting 3, you can add -3, the additive inverse of 3; instead of dividing by 4, you can multiply by 1/4, the multiplicative inverse of 4. Two numbers that are additive inverses of each other add up to 0. Two numbers that are multiplicative inverses of each other multiply to 1.

This way of thinking has practical advantages, as well. For example, it can be less costly, in terms of CPU cycles, to multiply a number by 1/4 (or .25) than to divide by 4. Furthermore, some RISC (reduced instruction set computing) processors are able to eliminate a certain subtraction instruction, because the same result can be had by adding a negative value. IOW, instead of subtracting 3, we can add -3.
 
  • #25
sysprog said:
For one thing, in general, multiplying by the reciprocal of the coefficient safeguards against accidentally allowing a divisor to be zero
Not really. If the coefficient is zero, then its reciprocal is undefined.
 
  • #26
Mark44 said:
Not really. If the coefficient is zero, then its reciprocal is undefined.
It may sometimes be easier to notice that the coefficient is zero if we aim to multiply by its reciprocal instead of directly dividing by the coefficient. In the first false proof example from the referenced math humor page:
Given that a and b are integers such that a = b + 1,Prove: 1 = 0
1. a = b + 11. Given
2. (a-b)a = (a-b)(b+1)2. Multiplication Prop. of =
3. a2 - ab = ab + a - b2 - b3. Distributive Property
4. a2 - ab -a = ab + a -a - b2 - b4. Subtraction Prop. of =
5. a(a - b - 1) = b(a - b - 1)5. Distributive Property
6. a = b6. Division Property of =
7. b + 1 = b7. Transitive Property of = (Steps 1, 7)
8. Therefore, 1 = 08. Subtraction Prop. of =
Statement '5.' is vacuously true, but given statement '1.' as true, statement '6.' cannot be true. The false proof works partly by obscuring the zero divisor in the bilateral division operation employed to justify the transition from statement '5.' to statement '6.'. I think it is harder to fail to notice that the divisor is zero if the division operation is seen as multiplying by ##1/(a-b-1)## instead of as simply canceling the ##(a-b-1)##, whatever that quantity might be, from both sides.
 
Last edited:
  • #27
wow. i wonder how we would explain how to solve 1.X = 5?, i.e. divide by 1, or multiply by 1^-1.
 
  • Like
Likes suremarc and jedishrfu
  • #28
mathwonk said:
wow. i wonder how we would explain how to solve 1.X = 5?, i.e. divide by 1, or multiply by 1^-1.
Is the low-positioned dot intended to be the multiplication operation symbol? The person would be expected to recognize the Identity Element 1, and understand that X is 5.

EDIT: Another way to say this is, the person would logically recognize the occurrence of the identity element 1, and understand that X is 5.
 
Last edited:
  • Like
Likes jedishrfu
  • #29
sysprog said:
The false proof works partly by obscuring the zero dividend in the bilateral division operation employed to justify the transition from statement '5.' to statement '6.'.
Except that from statement 1, a = b + 1, which is equivalent to a - b = 1, we can see that a - b - 1 = 0. This means that we are dividing by 0 going from step 5 to step 6.
 
  • #30
mathwonk said:
wow. i wonder how we would explain how to solve 1.X = 5?, i.e. divide by 1, or multiply by 1^-1.
No need to divide by anything, as 1 is the multiplicative identity: ##1 \cdot a = a## for any a.
 
  • #31
Mark44 said:
Except that from statement 1, a = b + 1, which is equivalent to a - b = 1, we can see that a - b - 1 = 0. This means that we are dividing by 0 going from step 5 to step 6.
Well, we can also start by recognizing that 1 ≠ 0, and not bother finding which step in the false proof is in error. In the false proof example, in which "we are dividing by 0 going from step 5 to step 6" is quod erat obscurandum (what was to be obscured), multiplying by the reciprocal instead of just canceling out the unevaluated parenthetical expressions, can give us pause to examine their content, and so be more apt to not fail to notice that it sums to zero.
 
  • #32
sysprog said:
Well, we can also start by recognizing that 1 ≠ 0, and not bother finding which step in the false proof is in error.
Well, clearly 1 ≠ 0, but sorting out this contradiction requires finding the exact step that is invalid. I don't think we're in disagreement, since both dividing by zero or multiplying by the reciprocal of something that is zero are invalid operations.

For the record, when I taught algebra, I avoided the use of the term "cancel," as too often students equated cancalling things with just crossing them off, such as "cancelling" the 5s in this expression, ##\frac {2 + 1} 2##, obtaining 1.
 
  • #33
I would have never thought that the question "How do you solve ##3x=15##?" would bring about so much discussion.

In addition to the discussion of multiplying by 1/3 versus dividing by 3, I would like to know if anyone else visualizes the operations as I do. I understand (I think :smile:) the formal mathematics of the transformation from ##3x = 15## to ##x = 15/3##, and I am careful, for example, when discussing it with my son to always talk about it in terms of applying the same operation to both sides of the equation, but this is not how I picture it when I do algebra. For me, the symbols move from one side of the equation to the other, as in
algebra001.jpg

Likewise, if we had ##x + 3 = 15##, I would "see" the 3 going from the lhs to the rhs and acquiring a minus sing in the process.
 
  • Like
Likes ChemAir, lomidrevo, Klystron and 2 others
  • #34
symbolipoint said:
Same thing.
Study of Basic Algebra
Thanks.
I was asking why chose to multiply rather than divide when the unknown is already being multiplied by something.
symbolipoint said:
Same thing.
Study of Basic Algebra
Yes I get that I was asking why choose to multiply over division.
It looks like things have moved on a bit since then.
 
  • #35
Really @DrClaude? You would call the fraction bar in ##\frac {15} 3## a minus sign? 😉
 
  • Haha
Likes DrClaude
<h2>1. What is the value of x?</h2><p>The value of x is 5.</p><h2>2. How do you solve for x?</h2><p>To solve for x, we need to isolate it on one side of the equation. In this case, we can divide both sides by 3 to get x = 15/3 = 5.</p><h2>3. Can you check if the solution is correct?</h2><p>Yes, we can substitute the value of x into the original equation to check. 3(5) = 15, so the solution is correct.</p><h2>4. What is the meaning of the "3x" in the equation?</h2><p>The "3x" represents three times the value of x. In other words, it is a way to express multiplication.</p><h2>5. Can this equation be solved using any other method?</h2><p>Yes, we can also solve this equation by subtracting 15 from both sides and then dividing by 3. This will give us the same solution of x = 5.</p>

1. What is the value of x?

The value of x is 5.

2. How do you solve for x?

To solve for x, we need to isolate it on one side of the equation. In this case, we can divide both sides by 3 to get x = 15/3 = 5.

3. Can you check if the solution is correct?

Yes, we can substitute the value of x into the original equation to check. 3(5) = 15, so the solution is correct.

4. What is the meaning of the "3x" in the equation?

The "3x" represents three times the value of x. In other words, it is a way to express multiplication.

5. Can this equation be solved using any other method?

Yes, we can also solve this equation by subtracting 15 from both sides and then dividing by 3. This will give us the same solution of x = 5.

Similar threads

  • General Math
Replies
4
Views
3K
  • General Math
Replies
22
Views
399
Replies
1
Views
854
Replies
1
Views
1K
  • General Math
2
Replies
47
Views
2K
Replies
2
Views
684
  • General Math
Replies
7
Views
759
Replies
2
Views
1K
Replies
4
Views
1K
  • General Math
Replies
23
Views
1K
Back
Top