Equality question

[BoomSchtick]

The problem is this:

(a+1)/5 = (b+4)/2

I have five options as to what MIGHT be true:

a>b
a<b
a=b
b>2a
a<2b

I'm not sure how to approach the question. Do I try to get a LCD of 10 and multiply one side by 5 and the other by 2?

My main source of confusion is that a and b can be ANYTHING to make it tip one way or the other.

Or am I reading too much into it and the fact that there is an equals sign in the original problem that would make the answer a=b.

Thanks for any help you can provide!

 

[Denis]

(a+1)/5 = (b+4)/2

RULE: if a/x = b/y, then ay = bx (tattoo that on your wrist!)

So, your equation:
2(a + 1) = 5(b + 4)
2a + 2 = 5b + 20
2a - 5b = 18

Can you wrap up now?

 

[BoomSchtick]

(a+1)/5 = (b+4)/2

RULE: if a/x = b/y, then ay = bx (tattoo that on your wrist!)

So, your equation:
2(a + 1) = 5(b + 4)
2a + 2 = 5b + 20
2a - 5b = 18

Can you wrap up now?

Hmmmm....

I'm still not sure...

If I solve for a or b, they are just going to end up back over there with the 18 and then I'll have a fraction when I divide by either 2 or 5. If I leave them where they are and just solve pretending that the other does not exist, I get a= 8 and b = 3.6. In that case it would be a>b.

Is that a valid way of working it out?

 

[mmm4444bot]

… pretending that the other does not exist … Is that a valid way of working it out?

Hi BoomSchtick:

In general, dismissing part of an expression is not a good idea. But, testing specific cases is valid.

Since this exercise places no constraints on the values of numbers a and b, we can assume that both of them represent ANY Real numbers, and this in turn means that if a>b for an arbitrary pair of (a, b) values then we must have a>b for all possible pairs of (a, b) values.

Therefore, we can pick ANY Real number for b, solve for the corresponding value of a, and then compare the results.

(I'm focusing on the first of the five multiple choices in this exercise because, except for the choice a=b, all of the other choices involve b being the larger value. In other words, if our test case verifies a>b, then there's no need to consider the three choices where b is the larger value.)

Again, if a > b in the following test case, then a > b for all possible values of a and b.

Try b = 0 (zero is usually a good test value when testing cases because zero tends to make arithmetic easier). Compare zero with the corresponding value of a.


THERE IS ANOTHER APPROACH, as well. I prefer it, and it's called analysis.

2a - 5b = 18

We are subtracting a multiple of 5 away from a multiple of 2 and getting a positive result. Therefore, we must have the following relationship.

2a > 5b

Do you understand why?

If the number 5b were larger than the number 2a, then their different would have to be negative. If 5b were equal to 2a, then their difference would have to be zero. The fact that their difference is +18 tells us that 5b must be smaller than 2a.

Solving this inequality for the number a shows us which of the five multiple choices is correct.

Cheers,

~ Mark

 

[BoomSchtick]

Ok... that makes sense!

If I set b=0 then A=9 and that make a>b for sure!

Thanks for the help everyone! Especially to Mark... that was a great explanation!