"if 5y+3 = 53 = 5(10) + 3 then y + 3 = 10 + 3"

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Agent Smith

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ALiteralFlute said:
When solving for Y in the expression below, I got the wrong answer by dividing 5 from both sides after the first simplification. I then got the right answer by subtracting 3 from both sides after the first simplification. Although I did the same thing in both attempts, I only got it right in the second attempt. How was I supposed to know that I had to subtract 3 before I could divide 5?

2+(5y+1)=53
5y+3=53
y+3=10.6
y=7.6

2+(5y+1)=53
5y+3=53
5y=50
y=10


(I cannot ask my teacher this question because I do not have a teacher. I never received an education in math. If this question is too simple then what are some resources I can use to understand this instead?)
Click to expand...
[imath]5y + 3 = 53[/imath]
[imath]5y + 3 = 5(10) + 3[/imath]
[imath]y + 3 = 10 + 3[/imath] (dividing by 5)
[imath]y = 10[/imath]
 
Agent Smith said:
[imath]5y + 3 = 53[/imath]
[imath]5y + 3 = 5(10) + 3[/imath]
[imath]y + 3 = 10 + 3[/imath] (dividing by 5)
[imath]y = 10[/imath]
Click to expand...
NO! To get from the second line to the third line, you have NOT divided each side by 5.

Your third line should be

\(\displaystyle y+\frac{3}{5} = 10 + \frac{3}{5}\).

The only reason your method "appears" to work is that there is a +3 on each side. What if the original RHS was 52 not 53? Your "method" wouldn't work. Check it for yourself.
 
Harry_the_cat said:
NO! To get from the second line to the third line, you have NOT divided each side by 5.

Your third line should be

\(\displaystyle y+\frac{3}{5} = 10 + \frac{3}{5}\).

The only reason your method "appears" to work is that there is a +3 on each side. What if the original RHS was 52 not 53? Your "method" wouldn't work. Check it for yourself.
Click to expand...
Ok, ok, calm down.

5x + 3 = 13
13 = 5(2) + 3
5x + 3 = 5(2) + 3, no?
Divide (partially) by 5 and you get ...
x + 3 = 2 + 3
x = 2
Have I violated any rule?
 
Agent Smith said:
Have I violated any rule
Click to expand...
Those examples are not real algebra, so no rule worries. Easy making incorrect maths give correct result when using partial steps or other trickery.

Here is piece of real algebra,
Distributive Property: a(b+c)=(a)(b)+(a)(c)
You used a part of it: a(b+c)=(a)(b)+c

Correct way to multiply or divide polynomials is:
multiply or divide every term :)
 
Bruce said:
Those examples are not real algebra, so no rule worries. Easy making incorrect maths give correct result when using partial steps or other trickery.

Here is piece of real algebra,
Distributive Property: a(b+c)=(a)(b)+(a)(c)
You used a part of it: a(b+c)=(a)(b)+c

Correct way to multiply or divide polynomials is:
multiply or divide every term :)
Click to expand...
[imath]3(4 + 1) = 3(4) + x[/imath]
 
Agent Smith said:
[imath]3(4 + 1) = 3(4) + x[/imath]
Click to expand...
@Agent Smith .........Are you proposing a new example or continuing from a given response in this thread?

If it is "continuing from a given response in this thread" please provide a reference to the 'response # '.

If it is a new example, please state the complete problem .
 
khansaheb said:
@Agent Smith .........Are you proposing a new example or continuing from a given response in this thread?

If it is "continuing from a given response in this thread" please provide a reference to the 'response # '.

If it is a new example, please state the complete problem .
Click to expand...
Ok,

a(b + c) = ab + d. (partial distribution).
 
Note: This thread has been split from a student's question (answered a couple of months ago), so as to keep the mathematically incorrect assertions from causing confusion to students.
 
You can easily write down two equations that have the same solution but you can say that you can go from any equation to the next while making mistakes!
5x + 3 = 10 + 3. The solution is x=2.
x+ 3 = 2 + 3. The solution is x = 2.
The reason that you can divide the first equation by 5 the way you did it is because the 3's are not really there--they cancel out to 0.

Here is what is really happening.

5x + 3 = 10 + 3---divide by 5
x + 3/5 = 2 + 3/5---now add 12/5 to both sides
x+ 3 = 2 + 3. You never noticed that you added 12/5 to both sides.

Clearly 8 + 4 = 2 + 12. Now divide "both sides" by 2
8/2 + 4 =2/2 + 12
4+4 = 1 + 12
8 = 13
Is that correct?
 
khansaheb said:
Incorrect.

You did not respond to request for complete problem statement.
Click to expand...
khansaheb said:
in the above work,

in the second line ....... where did 2 come from?
Click to expand...
5x + 3 = 13
Then
5x + 3 = 10 + 3
Then
5x + 3 = 5(2) + 3
x + 3 = 2 + 3 (partial division)
Ergo,
x = 2
The 2 came from 10, 10 = 5 × 2
The logic is quite simple ...

If 5x + 3 = 13 and
5x + 3 = 10 + 3
then
5x = 10
5x = 5(2)
x = 2
If so x + 3 = 2 + 3
5x + 3 = 5(2) + 3 (partial multiplication)
5x + 3 = 13 (we return to the original problem)

By the way, I'm not recommending that math be done this way, it can leadto errors. I'm only trying to show the OP that his method of partial division is not entirely wrong.
 
Harry_the_cat said:
So how would you solve 5x+3 =12 ?
Click to expand...
Easy!

[imath]5x + 3 = 9 + 3[/imath]

[imath]\displaystyle 5x + 3 = 5\left(\frac{9}{5}\right) + 3[/imath]

[imath]\displaystyle x + 3 = \frac{9}{5} + 3[/imath] (divide partially by [imath]5[/imath])

Then

[imath]\displaystyle x = \frac{9}{5}[/imath]


I admit that Agent Smith is a genius professor?‍?
 
Here's food for thought.

10000x + 0.00000001 = 20000
x = 2 (partial division). 0.00000001 is only going to get smaller, so small in fact that we may ignore its existence ?.
 

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Agent Smith said:
[imath]5y + 3 = 53[/imath]
[imath]5y + 3 = 5(10) + 3[/imath]
[imath]y + 3 = 10 + 3[/imath] (dividing by 5)
[imath]y = 10[/imath]
Click to expand...
5y + 3 =53
minus 3 from both sides
5y = 50
divide both sides 5
y = 10
all algebra is based on the premise what you do to one side of the equals sign you must do to the other to maintain the equality
 
helpermonkey said:
5y + 3 =53
minus 3 from both sides
5y = 50
divide both sides 5
y = 10
all algebra is based on the premise what you do to one side of the equals sign you must do to the other to maintain the equality
Click to expand...
Kindly review my previous statements.

x + 3 = 5
x + 2 > 3 (LHS - 1, RHS - 2)
x > 1 (LHS - 2, RHS - 2)
---
x + 3 = 5
x + 1 < 4 (LHS - 2, RHS - 1)
x < 3 (LHS - 1, RHS - 1)

We have now 1 < x < 3.

There is only one integer that satisfies the inequality, viz. x = 2
 
helpermonkey said:
equality
Click to expand...
Agent Smith said:
5x + 3 = 13
Then
5x + 3 = 10 + 3
Then
5x + 3 = 5(2) + 3
x + 3 = 2 + 3 (partial division)
Ergo,
x = 2
The 2 came from 10, 10 = 5 × 2
The logic is quite simple ...

If 5x + 3 = 13 and
5x + 3 = 10 + 3
then
5x = 10
5x = 5(2)
x = 2
If so x + 3 = 2 + 3
5x + 3 = 5(2) + 3 (partial multiplication)
5x + 3 = 13 (we return to the original problem)

By the way, I'm not recommending that math be done this way, it can leadto errors. I'm only trying to show the OP that his method of partial division is not entirely wrong.
Click to expand...

No, math cannot be done that way. "Partial division" and "partial multiplication"
do not even exist. Not only are these non-existent, and "can lead to errors" as you stated, I am stating to you as a fact that these are errors in themselves.

I am reporting this post of yours that I put in the quote box, and I am requesting a temporary ban on you for promoting false methods in solving and hindering/confusing the students seeking help, despite it being pointed out to you repeatedly and by multiple helpers.
 
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