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# Analyzing structure with linear inequalities: balls

Given a real-world context about the number of balls in a bag, we find a linear inequality that correctly depicts the situation.

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• In the previous video, sal solved the problem by substituting numbers, and I wondered why he did not take a more abstract or mathematically rigorous path. In this video, he did that. How does he make that choice? Can you also solve these problems graphically? Would this make the problem more intuitive ?
• I think Sal was just demonstrating the two methods. And personally I think it's best to first try substituting if your fairly new to algebraic manipulation, but if you're more capable, using the abstract/general method of manipulating the given information can be more useful and intuitive.
• We can also just algebraically manipulate the choices, right?

2b > b+g = 2b-b > b+g-b = b>g, which contradicts given b<g

2b = b+g is 2b-b = b+g-b is b = g, which contradicts b<g

2b < b+g = 2b-b < b+g-b = b<g, which corresponds to the given.
• One of the practice questions I got was a variation on this.

a > b
b > c
a > b + c

The problem was to compare b+c & 2a-b. I used a different method to come to the correct answer but I'm not sure if it was just luck. Since a > b + c I replaced the 'a' in the second statement with b + c:

b + c ? 2a - b
b + c ? 2(b + c) - b
b + c ? b + 2c
b + c < b + 2c

Because I know a is actually greater than the value I substituted it with, 2a - b will actually be greater still, but regardless definitely greater than b + c.

This seems sound to me, but I'm not 100% sure. Any input is appreciated.
• You shouldn't substitute inequalities as values, because you don't know which way the inequality falls (and also a just doesn't equal b + c).

Here's how I would go about doing it:
a > (b + c), so 2a > 2(b + c), so (2a - b) > b + 2c.
b + 2c > b + c (assuming c > 0), so (2a - b) > (b + c).
• Why did we add b to both sides and not g, and why did we not cancel out anything?
(1 vote)
• If you look at three of the 4 answers, it is trying to compare 2b to b+g. So you start with the fact that g > b. Adding g would give 2g > b + g, but this does not help to choose. Adding b to both sides does give a comparison which matches the answers. Noting cancels because you do not have any additive inverses to cancel.
• I think it should be like not enough info.to tell
(1 vote)
• In mathematics, a statement is called "True" if and only if it is true in all the cases, and clearly, this isn't the case here. So shouldn't we choose the last option?
(1 vote)
• You are trying to treat the problem as a logic statement and proving that is is a tautology. That is not what it is asking you to do.

Sal walks you thru the process of figuring out which statement is correct. It is based upon the given info: g>b and using the property of inequality that tells us that adding the same value to both sides of an inequality does not impact the validity or truth of the inequality. For example: 4>1 is true.
4+5>1+5 is still true. Adding a common value to both sides gives us a new and equivalent inequaltiy. Sal is doing the same thing, except the values are in variable form.
Hope this helps.
(1 vote)
• I still don't get it, can anyone please explain it to me a bit more in detail?
(1 vote)
• A bag has more green balls than blue balls:

`g > b`

There is at least one blue ball:

`b ≥ 1`

Therefore, in regards to "2b":

`2b ≥ 2`

and in regards to g+b:

b ≥ 1 and g>b, so g ≥ 2 ("g" can never be 1 and there is presumably no such thing as a fraction of a ball)

so:

`b+g ≥ 3`

So we know 2b is at least 2 (that is, 2b ≥ 2) and we know that b + g is at least 3 (that is, b + g ≥ 3), it follows that:

`2b < b + g` (because 2 is less than 3 for the base scenario of b =1; there can be no b = 0. Also g will always be more than any multiple of b because g will exceed that multiple by at least one; there will always be more green balls than blue balls).