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September 24[edit]

how to solve |5-x|+7<|x+4|?[edit]

Hey everyone,

I tried to solve : |5-x|+7<|x+4| but i'm not sure if the equality are right:

x	positive\negative	equality
x>5	x=6 (-)-(+)<-7	(5-x)--(x+4)<7
(-4<x<5)	x=-3 (+)-(-)<7	(5-x)--(x+4)<7
x<-4	x=-5 (x)-(-)<-7	-(5-x)-(x+4)<7

I think one should be positive. one negative and one negative and positive but not sure how should I write it. — Preceding unsigned comment added by 132.64.177.219 (talk) 09:01, 24 September 2017 (UTC)[reply]

My approach is to graph both sides of the inequality, giving immediately that it is true for x>4. →109.155.51.78 (talk) 10:19, 24 September 2017 (UTC)[reply]

Combining inequality with absolute values is ugly, all right. Another way is to plug in different numbers and try them out. I rewrote it as |5-x|+7<|4+x| first:

 x     Plugged in          Step 1           Step 2        Step 3    T? 
---  ---------------    -------------      --------       ------   ---
 -7  |5+7|+7 < |4-7|    |12|+7 < |-3|      12+7 < 3       19 < 3    F
 -6  |5+6|+7 < |4-6|    |11|+7 < |-2|      11+7 < 2       18 < 2    F
 -5  |5+5|+7 < |4-5|    |10|+7 < |-1|      10+7 < 1       17 < 1    F
 -4  |5+4|+7 < |4-4|    | 9|+7 < | 0|       9+7 < 0       16 < 0    F
 -3  |5+3|+7 < |4-3|    | 8|+7 < | 1|       8+7 < 1       15 < 1    F
 -2  |5+2|+7 < |4-2|    | 7|+7 < | 2|       7+7 < 2       14 < 2    F
 -1  |5+1|+7 < |4-1|    | 6|+7 < | 3|       6+7 < 3       13 < 3    F
  0  |5  |+7 < |4  |    | 5|+7 < | 4|       5+7 < 4       12 < 4    F
  1  |5-1|+7 < |4+1|    | 4|+7 < | 5|       4+7 < 5       11 < 5    F
  2  |5-2|+7 < |4+2|    | 3|+7 < | 6|       3+7 < 6       10 < 6    F
  3  |5-3|+7 < |4+3|    | 2|+7 < | 7|       2+7 < 7        9 < 7    F
  4  |5-4|+7 < |4+4|    | 1|+7 < | 8|       1+7 < 8        8 < 8    F
  5  |5-5|+7 < |4+5|    | 0|+7 < | 9|       0+7 < 9        7 < 9    T
  6  |5-6|+7 < |4+6|    |-1|+7 < |10|       1+7 <10        8 <10    T
  7  |5-7|+7 < |4+7|    |-2|+7 < |11|       2+7 <11        9 <11    T

So, based on this, I also get x>4. You can also plug in 4.1 an 3.9, if you feel the need for more precision. StuRat (talk) 16:55, 24 September 2017 (UTC)[reply]

This "method" combines excessive amounts of work with a complete failure to guarantee correctness against any moderate standard of skepticism. --JBL (talk) 11:46, 25 September 2017 (UTC)[reply]

It's only a lot of work to write it all down. I'd do it in my head, using values of x around the transition point (4), obtained by other methods, such as graphing. It's a good, quick check. StuRat (talk) 15:05, 25 September 2017 (UTC)[reply]

You should note that if x>5 then the inequality is always satisfied. But for x<-4 it is never satisfied. For intermediate values of x you get x>4. Therefore, the final solution is x>4. Ruslik_Zero 20:27, 24 September 2017 (UTC)[reply]

Hello Jerusalem! It is good to know many ways to solve a given problem, but you should also know how to do it using the method you were taught, which is presumably how you were presenting it above. However, there are at least three errors in your table.

You appear to have rearranged your inequality as |5-x|-|x+4|<-7, which is fine, but how did your -7 become positive in the middle row "(+)-(-)<7"? Presumably that was a typo. Also, your -7 became a 7 when going from the second to the third column in the both the first and last rows, presumably by you multiplying both sides of the inequalities by -1, but doing so should also change the direction of the inequality sign. Negating both sides isn't really necessary, but if you do so you need to do it right. (See Inequality (mathematics)#Multiplication and division.) Finally, where your middle row says "(+)-(-)<-7" (or would once you fixed the typo), it should read "(+)-(+)<-7" since x+4 is positive when -4<x<5.

So, if you fix those problems, what do you do next? Presumably, you add a fourth column where you simplify the inequality, and a fifth where you combine that with the domain of your first column, then combine all three rows into a single answer. -- ToE 21:41, 24 September 2017 (UTC)[reply]

The graph of y=|5-x|-|x+4|+7 is polygonal with corners at (-4, 16) and (5, -2). It has slope 0 for x<-4 and x>5, so it's only x-intercept is between -4 and 5. A quick calculation shows the x-intercept is at x=4 giving y<0 for x>4. This idea works for any linear combination of absolute values, so |8x-16|+75<|11x+44|+|9x-45| could be solved the same way. More generally, you only need to plot local extrema points to narrow down where the intercepts are, and this can be done with a bit of calculus keeping in mind that critical points include those where a function is not differentiable. In general, absolute values can be eliminated from an inequality using 1) |a|<b iff a<b and -a<b, 2) |a|>b iff a>b or -a>b. This rapidly gets unwieldy if there are more than a few absolute values, but it might be the way to go if there are several variables involved. --RDBury (talk) 05:31, 25 September 2017 (UTC)[reply]