(absolute value of x minus 5) is less than or equal to 2 How many integers satisfy the inequality above? (A) None (B) One (C) Two (D) Three (E) Five
sorry I x-5 I less than or equal to 2
\(|x-5| \le 2 \) for |a| <=b, we have a<=b or a>=-b so what about \(|x-5| \le 2 \) ??
how many integers satisfy the equation?
lets find out! \(|x-5|\le 2 \implies (x-5)\le2 \quad or \quad (x-5)\ge -2 \) can you soilve them individually ?
**solve
i think so... not so sure...
x-5 <= 2 add 5 to both sides, what u get ?
you get x = 7
not x=7 \(x-5 \le 2\) adding 5 to both sides, \(x-5+5 \le 2+5 \implies x \le7 \) you don't loose <= sign
oh ya i forgot. *facepalm*
similarly what u get after adding 5 to both sides of x-5 >= -2 ??
you would get +3
x>=3, right ? so, x>=3 and x<=7 gives you, x=3,4,5,6,7 ----> 5 values
A great way of solving this problem would be to visually interpret it. When we say: \[\bf |x-5| \le 2\]We're really finding values of 'x' which are within a distance of 2 unit from 5. The only values within 2 units from 5 are 7 and 3. So really, we are looking at all integers within 2 units to the left of 5 on the number line and within 2 units on the right of the number line.|dw:1374403855680:dw|Which integers are in that range once you look at the number line? @betterwiththelightsoff99
oh i understand now.... thank you hartnn
@genius12 wouldnt that be a little more time-consuming?
Nope. Once you understand how that works, you can work it out in your head without having to write or type a word. Think about it. We are looking at integers within 2 units of 5. How many are there? 3, 4, 6, 7. 2 units to the left and 2 units to the right. Easy. You can always doing it algebraically like @hartnn but understanding what the |x - 5| really means can be really helpful as you can solve the inequality in your head without much work. @betterwiththelightsoff99
oh! ok thank you @genius12
But remember, because its less than or equal to 2, we are finding values that are 2 units or less away from 5. Because its 2 units or less, we include the 3 and 7. If it was just less than 2 units, then the only integers satisfying the inequality would 4 and 6.
alright got it thanks for the help @genius12
np
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