SOLUTION: Please help with the following inequality. I have the problem and the answer but nothing in between. (y-3)/2 > 1/2 - (y-3)/4 interval notation: (11/3, %) Thanks for your

Algebra ->  Inequalities -> SOLUTION: Please help with the following inequality. I have the problem and the answer but nothing in between. (y-3)/2 > 1/2 - (y-3)/4 interval notation: (11/3, %) Thanks for your      Log On


   

Question 118198: Please help with the following inequality. I have the problem and the answer but nothing in between.
(y-3)/2 > 1/2 - (y-3)/4 interval notation: (11/3, %)
Thanks for your help just can't seem to get this question. It's the only problem in my unit lesson I couldn't complete.
Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Given:
.
%28y+-+3%29%2F2+%3E+1%2F2+-+%28y-3%29%2F4
.
To make things easier, get rid of the denominators by multiplying both sides (all terms) by +4.
This multiplication becomes:
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%284%2A%28y+-+3%29%29%2F2+%3E+%284%2A1%29%2F2+-+%284%2A%28y-3%29%29%2F4
.
In each term divide the denominator into the +4 multiplier. This eliminates each of the denominators
and reduces the problem to:
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%282%2A%28y+-+3%29%29+%3E+%282%2A1%29+-+%281%2A%28y-3%29%29
.
Do the indicated multiplications in each of the three terms:
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2y+-+6+%3E+2+-+y+%2B+3
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Get rid of the -6 on the left side by adding +6 to both sides:
.
+2y+%3E+2+-+y+%2B+3+%2B+6
.
Get rid of the -y on the right side by adding +y to both sides:
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+3y+%3E+2+%2B+3+%2B+6
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Add all the terms on the right side:
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3y+%3E+11
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Solve for y by dividing both sides by 3:
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y+%3E+11%2F3
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This answer tells you that the original inequality will be true as long as y is greater than 11%2F3.
.
Let's check. Suppose that y equals 4. That value is greater than 11/3, so when y equals 4,
the inequality should be true. Take the original inequality and substitute +4 for y ...
The original inequality is:
.
%28y+-+3%29%2F2+%3E+1%2F2+-+%28y-3%29%2F4
.
When you substitute +4 for y, it becomes:
.
%284+-+3%29%2F2+%3E+1%2F2+-+%284-3%29%2F4
.
The left side reduces to 1%2F2 and on the right side the negative term reduces to -1%2F4.
This makes the inequality become:
.
1%2F2+%3E+1%2F2+-+1%2F4
.
Subtract 1%2F2+ from both sides and you have:
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0+%3E+-1%2F4
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This is true. Zero is greater than -1%2F4 because 0 is to the right of -1%2F4 on the
number line.
.
Now try a value of y that is less than 11%2F3. Suppose you let y = 3. That is less than 11%2F3.
.
Start with the original inequality:
.
%28y+-+3%29%2F2+%3E+1%2F2+-+%28y-3%29%2F4
.
Substitute 3 for y and this inequality becomes:
.
%283-+3%29%2F2+%3E+1%2F2+-+%283-3%29%2F4
.
The numerator on the left side and also the numerator of the negative term on the right side
both become zero. So the inequality becomes:
.
0+%3E+1%2F2+-+0
.
The right side reduces to 1%2F2 so the inequality is:
.
0+%3E+1%2F2
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This is NOT true ... zero is not greater than 1/2. So choosing a value for y that is less
than 11%2F3 did not work.
.
From this "quick trial" it appears that y must be greater than 11%2F3 is a good answer.
.
Hope this helps you to understand the problem and how to solve it ...
.