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**TryingMyBest50****Member**- Registered: 2013-07-12
- Posts: 13

Hello Forum, after i find the solution to this equation 5m+10=8m-2, how do i check to make sure the answer is correct. Greatly appreciate any help!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,309

Hi TryingMyBest50;

Welcome to the forum.

Subtract 5m from both sides.

Add 2 to both sides.

Divide both sides by 3.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**TryingMyBest50****Member**- Registered: 2013-07-12
- Posts: 13

Hi bobbym, i first need to say that math is not my subject. With that said, is that how i check that the solution is a true statement?

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,380

hi TryingMyBest50

Welcome to the forum.

And then check by working out if 5x4 + 10 gives the same result as 8x4 -2

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,309

You check by substituting the 4 into the m.

5 x 4 = 20 and 8 x 4 = 32

20 + 10 = 30 and 32 - 2 = 30

Checked and you are done.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,380

hi bobbym

how are you today?

That latex error was strange. \times (4) worked for me in a preview.

Test

curious

Maybe it's all in the spaces?

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,309

Hi Bob;

I am fine, how is everything?

Until it gets into the cache, some latex will not work. It is functioning okay now.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**TryingMyBest50****Member**- Registered: 2013-07-12
- Posts: 13

Hi bob, i want to thank both you and bobbym for your feed back. So i just replace the answer with what ever letter is presented to check my that the sollution is correct, right?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,309

Replace the letter ( variable ) that you solved for and do the arithmetic correctly. Both sides should be the same.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**TryingMyBest50****Member**- Registered: 2013-07-12
- Posts: 13

Variable, i knew the letter had a name i just coudln't think of it. Thank you so much for your help! You guy's are great! Have a Blessed day!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,309

You are welcome.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,380

hi bobbym,

While your latex was producing an error I copied it so I could check the ascii codes of your symbols. Then when that revealed nothing I did my test post 6. And your original latex is still there producing an error. So I don't think it's a cache problem.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,309

Hi Bob;

On mine it is working fine. Does it look like this for you?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,380

hi bobbym,

Yes, it does.

But look at post 6

Bob

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,309

Hi;

Sorry, I did not see that.

Removing the spaces causes it to work.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,380

Oh yes. But this is still weird for me:

Wont work version: 5 \times (4)+10 = 8 \times (4)-2

De-spaced version works: 5 \times (4)+10=8 \times (4)-2

But my working version had loads of spaces:

5 \times (4) + 10 = 8 \times (4) -2

Just got to put my head in some ice. Back later.

Bob

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,309

That is true. I can remove one space and everything is fine.

Just got to put my head in some ice.

Sounds cooling!

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**TryingMyBest50****Member**- Registered: 2013-07-12
- Posts: 13

Hi guy's it's me again. How do i turn this linear equation 4x+11=6x-3 into an linear inequality? Would i write it like 4x<+11=6x<-2, and the lesser sign is underlined?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,309

Hi;

I have never heard of that...

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**TryingMyBest50****Member**- Registered: 2013-07-12
- Posts: 13

Hi Bobbym, I need to change the linear equation 4x+11=6x-3 into a linear inequality (using < or >). Can it be done?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,309

Why couldn't you just replace the = with a < or a >?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**TryingMyBest50****Member**- Registered: 2013-07-12
- Posts: 13

ok, so how would you solve the equation using < or >.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,309

Which one?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**TryingMyBest50****Member**- Registered: 2013-07-12
- Posts: 13

I need to turn this linear equation 4x+11=6x-3 into a linear inequality.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,309

4x+11<6x-3 or 4x+11>6x-3 or 4x+11≤ 6x-3 or 4x+11 ≥ 6x-3

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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