# This question doesn't make any sense to me

• Jan 23rd 2011, 09:12 PM
This question doesn't make any sense to me
Whats up people. I'm in Trig/PreCalc in school but I just moved to the United States a couple months ago so I never fully studied absolute value, since we didn't cover that much in Sweden.

We got this review for some California state test or something and this question is kind of weird. If I worked this problem correctly, steps B, C, and D all contain errors. However, the question says only one of them has an error. I provided a picture of the question at the bottom. Can someone tell me the answer as well as how they worked it? Maybe I just solved it wrong.

http://img89.imageshack.us/img89/7448/20091859.png
• Jan 23rd 2011, 11:31 PM
SENTINEL4
At step C it should be 6y>=38 not 6y<=38. When you multiply with a negative number the inequality symbol changes. (Hope you understand what i mean)
• Jan 23rd 2011, 11:41 PM
Quote:

Originally Posted by SENTINEL4
At step C it should be 6y>=38 not 6y<=38. When you multiply with a negative number the inequality symbol changes. (Hope you understand what i mean)

I originally thought that too. However, look at B. It says 7-6y≥24 when it clearly should be 7-6y≥31. Wouldn't that make B the correct answer as well?

I'm confused :(
• Jan 23rd 2011, 11:43 PM
earboth
Quote:

Whats up people. I'm in Trig/PreCalc in school but I just moved to the United States a couple months ago so I never fully studied absolute value, since we didn't cover that much in Sweden.

We got this review for some California state test or something and this question is kind of weird. If I worked this problem correctly, steps B, C, and D all contain errors. However, the question says only one of them has an error. I provided a picture of the question at the bottom. Can someone tell me the answer as well as how they worked it? Maybe I just solved it wrong.

1. Use the definition of the absolute value: $\displaystyle |7-6y|=\left\{\begin{array}{lcr}7-6y & if & 7-6y \geq 0 \implies y\leq \frac76 \\ -(7-6y) & if & 7-6y < 0 \implies y>\frac76 \end{array}\right.$

2. That means you'll get:

$\displaystyle 7-6y\geq 31 \ and\ y \leq \frac76 ~~ \bold{or} ~~ -(7-6y) \geq 31 \ and\ y > \frac76$

3. Solve for y. Compare with the given answers. You'll notice that answer b) contains an error because in the 2nd inequality the summand 7 was subtracted at RHS but not at the LHS.
• Jan 23rd 2011, 11:47 PM
Quote:

Originally Posted by earboth
1. Use the definition of the absolute value: $\displaystyle |7-6y|=\left\{\begin{array}{lcr}7-6y & if & 7-6y \geq 0 \implies y\leq \frac76 \\ -(7-6y) & if & 7-6y < 0 \implies y>\frac76 \end{array}\right.$

2. That means you'll get:

$\displaystyle 7-6y\geq 31 \ and\ y \leq \frac76 ~~ \bold{or} ~~ -(7-6y) \geq 31 \ and\ y > \frac76$

3. Solve for y. Compare with the given answers. You'll notice that answer b) contains an error because in the 2nd inequality the summand 7 was subtracted at RHS but not at the LHS.

But don't c and d contain errors as well? I believe the sign in d should be flipped if I am not mistaken.

Thanks for the help, and sorry I ask so many questions xD
• Jan 24th 2011, 04:33 AM
SENTINEL4
Quote:

I originally thought that too. However, look at B. It says 7-6y≥24 when it clearly should be 7-6y≥31. Wouldn't that make B the correct answer as well?

I'm confused :(

I saw that too but i took it as a typographic error because at step c, 7 just go away and doesn't get substracted from both parts of the inequality
• Jan 24th 2011, 04:37 AM
SENTINEL4
Quote:

But don't c and d contain errors as well? I believe the sign in d should be flipped if I am not mistaken.

Thanks for the help, and sorry I ask so many questions xD

At d the right is y>=38/6 but it should be flipped after step b
• Jan 24th 2011, 04:39 AM
DrSteve
Everyone keeps talking about an error "at" a certain step. I think this is confusing. The error occurs in "going" from one step to another. In this question there is an error in going from step a to step b, but I believe that this is just a typo and the 7 should not be written in step b. There is a genuine error in going from step b to step c.
• Jan 24th 2011, 05:25 AM
Quote:

I originally thought that too. However, look at B. It says 7-6y≥24 when it clearly should be 7-6y≥31. Wouldn't that make B the correct answer as well?

I'm confused :(

Think of "modulus" as being the distance of a point on the vertical number line from 0.

Then $\displaystyle |7-6y|\ge\ 31$

means the point on the y-axis is either at $\displaystyle +31$ or above, or at $\displaystyle -31$ or below.

Then step "a" is fine.

There is an error at step "b" as you correctly pointed out.
Subtract 7 from both sides for the part on the left,
but they've only taken 7 from one side for the part on the right.

Further errors are at part "c".

$\displaystyle -38\ge\ -6y\Rightarrow\ 6y-38\ge\ 0\Rightarrow\ 6y\ge\ 38$

However, they have "cancelled" their original error from "b" with the part on the right!

$\displaystyle 7-6y\ge\ 31\Rightarrow\ 7\ge\ 6y+31\Rightarrow\ 7-31\ge\ 6y$