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znT396
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Algebra 6th ed pg 133 #5 Please Help!!

by znT396 Wed Sep 07, 2016 1:34 pm

5. If |10y-4|>7 and y< 1, which of the following could be y?

(A) -0.8
(B) -0.1
(C) 0.1
(D) 0
(E) 1

I am confused about the answer. How can the answer be -0.8 if after solving the problem y<-0.3 and y>1.1 and y< 1. I understand that considering the inequality y<-0.3 and y>1.1, the answer can only be -0.8. However, what about y< 1. Wouldn't that include the values -0.1, 0.1 and 0?
RonPurewal
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Re: Algebra 6th ed pg 133 #5 Please Help!!

by RonPurewal Wed Sep 14, 2016 2:14 am

if you had ONLY the first inequality (|10y-4|>7), then any value less than –0.3 would work, and any value greater than 1.1 would also work.

however, you need a value that ALSO satisfies the inequality y < 1 ... so, that rules out all the values greater than 1.1.

that leaves only the values less than –0.3.

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Re: Algebra 6th ed pg 133 #5 Please Help!!

by RonPurewal Wed Sep 14, 2016 2:14 am

you could also just take what's probably the most straightforward approach to this problem, which is backsolving:
• take the answer choices
• plug them into the inequalities
• just see which choice satisfies BOTH inequalities.

only y = –0.8 will satisfy both of them, so ... there you go.
znT396
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Re: Algebra 6th ed pg 133 #5 Please Help!!

by znT396 Wed Sep 14, 2016 12:17 pm

Thank you for the clarification Ron!

When you say backsolving, do you mean plugging in -0.8 for y in |10y-4|>7and y< 1?
Because the answer I got after plugging in -0.8 in the first inequality is -8>11 and 8>11 doesn't make sense...
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Re: Algebra 6th ed pg 133 #5 Please Help!!

by RonPurewal Tue Sep 20, 2016 4:33 pm

znT396 wrote:Thank you for the clarification Ron!

When you say backsolving, do you mean plugging in -0.8 for y in |10y-4|>7and y< 1?
Because the answer I got after plugging in -0.8 in the first inequality is -8>11 and 8>11 doesn't make sense...


hm?
math symbols don't mean more than one thing, so it's impossible for "plugging in" to give two results.

| 10y – 4 | > 7

substitute y = –0.8:
| 10(–0.8) – 4 | > 7 ...this is something that will turn out to be either true or false, we just don't know until we simplify it
| –8 – 4 | > 7
| –12 | > 7
12 > 7
this is TRUE (12 is, indeed, greater than 7), so the inequality WORKS.

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i don't know what your concept of "plugging in" entails, but, if your idea of "plugging in" is something that gives you two different results (and neither of them has the numbers 12 and 7 on the left and right, respectively), then you definitely need to ditch your whole notion of "plugging in" and re-learn it from scratch.

"plugging in" is literally just throwing a number into something, doing the arithmetic, and seeing whether the result is true or false.