Clearing decimals by multiplication in an equation

[DS2C]

Homework Statement

In my text, it had a quick side note on solving basic equations. It related to clearing decimals by multiplying by an appropriate factor of ten.

For example,
0.25x + 0.10(x-3) = 1.10

can be cleared of decimals by multiplying the decimals by 100, which would result in

25x + 10(x-3) = 110

Homework Equations

NA

The Attempt at a Solution

[/B]
The solution is x=4, but my question is not how to solve it.
My question is in regards to clearing the decimals. How can this equation remain balanced when the left side was multiplied by 100 twice, and the right side was multiplied by 100 once?

 

[berkeman]

My question is in regards to clearing the decimals. How can this equation remain balanced when the left side was multiplied by 100 twice, and the right side was multiplied by 100 once?

Because of the Distributive property of multiplication. Start by multiplying ALL of both sides by 100:

100[0.25x + 0.10(x-3)] = 100[1.10]

Then distribute the 100 on the LHS to both terms, and keep going. Makes sense? Are you familiar with the Distributive property of Multiplication?

 

[DS2C]

Thanks for the response berkeman.
I'm familiar with the distributive property, but I'm still a little confused with this particular situation.

Say we were to use the distributive property, which would look like 100(0.25x + 0.10(x-3)) = (1.10)100. This would turn into 25x + 10(x-3) = 110.

Now say we were to multiple the two individual terms on the left side by 100, which would look like 100(0.25x) + 100(0.10(x-3)) = (1.10)100. This would turn into
25x + 10(x-3) = 110.

So isn't using the distributive property still multiplying the LHS by 100 twice? Maybe I'm assuming something that I shouldn't be, or I'm making a dumb mistake.

 

[berkeman]

Now say we were to multiple the two individual terms on the left side by 100, which would look like 100(0.25x) + 100(0.10(x-3)) = (1.10)100. This would turn into
25x + 10(x-3) = 110.

Which is the same thing, you have just skipped the explicit Distribution step.

So isn't using the distributive property still multiplying the LHS by 100 twice?

No, multiplying the LHS by 100 twice would look like this:

100{100[0.25x + 0.10(x-3)]} = 100[1.10]

Which would be wrong...

 

[DS2C]

Ok, so even though two terms are being multiplied by 100, that only counts as multiplying by 100 once?

 

[Mark44]

Ok, so even though two terms are being multiplied by 100, that only counts as multiplying by 100 once?

I think you might be thinking that the distributive property somehow applies to multiplication. The distributive property is about how multiplication distributes over addition. I.e., that a(b + c) = a * b + a * c. There is no similar property for pure multiplication.

There is, however, the associative property of multiplication (and a property of the same name applies to pure addition). This associative property of multiplication says that a(b * c) = (a * b) * c.

Relative to the problem you posted, 100 * 0.25 * x can be grouped (associated) in two ways: (100 * 0.25) * x, or 100 * (0.25*x). Either way, these simplify to 25x. But, 100 (.25*x) is NOT the same as 100(.25) * 100(x), which is what I think you are trying to do.

 

[DS2C]

Apologies for the last response. Been heavily busy with school work.

I think you might be thinking that the distributive property somehow applies to multiplication. The distributive property is about how multiplication distributes over addition. I.e., that a(b + c) = a * b + a * c. There is no similar property for pure multiplication.

There is, however, the associative property of multiplication (and a property of the same name applies to pure addition). This associative property of multiplication says that a(b * c) = (a * b) * c.

Relative to the problem you posted, 100 * 0.25 * x can be grouped (associated) in two ways: (100 * 0.25) * x, or 100 * (0.25*x). Either way, these simplify to 25x. But, 100 (.25*x) is NOT the same as 100(.25) * 100(x), which is what I think you are trying to do.

Actually this is exactly what I thought. I still can't get around thinking of it that way as it looks like 100 is being multiplied by A and then by B, but in looking at it the reverse way I can see how it's really just once. Not sure why this has confused me. I suppose I just took it at face value when I learned it and didn't think through it. What if there were more than two terms? For example 100(a + b + c + d)? This would just equal 100a + 100b + 100c + 100d?

 

[Mark44]

What if there were more than two terms? For example 100(a + b + c + d)? This would just equal 100a + 100b + 100c + 100d?

Yes.

 

[DS2C]

Thanks for the help guys, its greatly appreciated.