B Basic Algebra Question: 3x = 15

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
 
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Math_QED

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Yes, it is correct. More precisely, you multiply both sides with ##1/3##.
 
Thank a million
 
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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##
 
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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?
 

symbolipoint

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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.
 
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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.
 
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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.
 
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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.
 

FactChecker

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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.
 

symbolipoint

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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
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.
 

pinball1970

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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
 

jbriggs444

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/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.
 

FactChecker

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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.
 
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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.
 

pinball1970

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/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.
 
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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.
 
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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).
 
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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.
 

symbolipoint

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symbolipoint

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THIS IS THE GOOD STUFF! WE FEEL GREAT WHEN WE KNOW TO DO THESE THINGS:
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.
 

symbolipoint

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Yes again! This is the stuff that we need:
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.
 
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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.
 

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