The equation 3x=15 is solved correctly by isolating x, yielding x=5. The preferred method involves multiplying both sides by 1/3 rather than dividing by 3, as this emphasizes the use of multiplicative inverses in algebra. Factoring the right side, 15 into 3*5, can aid understanding but is not strictly necessary for solving the equation. The discussion highlights the importance of understanding algebraic principles over rote calculation methods.
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Why do you need to factor 15? What if it si 3x=16?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##
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.nasu said:Why do you need to factor 15? What if it si 3x=16?
This can be done with far fewer steps.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##
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
In other words, you can depend on the basic algebra rules.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.
Why times by 1/3 rather than divide by 3?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.
/3 is certainly the way I would look at it. I also don't understand why people are saying *(1/3)pinball1970 said:Why times by 1/3 rather than divide by 3?
I mean in terms of isolating x for the op
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.phinds said:/3 is certainly the way I would look at it. I also don't understand why people are saying *(1/3)
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.pinball1970 said:Why times by 1/3 rather than divide by 3?
I mean in terms of isolating x for the op
Is there something a little more far reaching the maths guys automatically think about in terms of next steps with algebra?phinds said:/3 is certainly the way I would look at it. I also don't understand why people are saying *(1/3)
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).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.
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.pinball1970 said:Is there something a little more far reaching the maths guys automatically think about in terms of next steps with algebra?
Same thing.pinball1970 said:Why times by 1/3 rather than divide by 3?
I mean in terms of isolating x for the op
That is, except for my not understanding that stuff about the linear algebra you tried to describe.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.
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.
Not really. If the coefficient is zero, then its reciprocal is undefined.sysprog said:For one thing, in general, multiplying by the reciprocal of the coefficient safeguards against accidentally allowing a divisor to be zero
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:Mark44 said:Not really. If the coefficient is zero, then its reciprocal is undefined.
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.
Given that a and b are integers such that a = b + 1, Prove: 1 = 0 1. a = b + 1 1. Given 2. (a-b)a = (a-b)(b+1) 2. Multiplication Prop. of = 3. a2 - ab = ab + a - b2 - b 3. Distributive Property 4. a2 - ab -a = ab + a -a - b2 - b 4. Subtraction Prop. of = 5. a(a - b - 1) = b(a - b - 1) 5. Distributive Property 6. a = b 6. Division Property of = 7. b + 1 = b 7. Transitive Property of = (Steps 1, 7) 8. Therefore, 1 = 0 8. Subtraction Prop. of =
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.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.
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.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.'.
No need to divide by anything, as 1 is the multiplicative identity: ##1 \cdot a = a## for any a.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.