I have a question here, the general definition is -a when a<0. But here we are using -a when a<=0. The answer comes out right, but is this correct? And if it is, is this some exception to the rule?
We need to find How many integer values of x satisfy the equation |x-3|+|x-20|<17?Let's solve the problem using two methods:
Method 1: AlgebraAs we have two absolute values so we will have three cases and the number line will be divided into three points by equating following equations to 0
x - 3 =0 and x - 20 = 0
| Number line is divided into three parts x ≤ 3, 3 ≤ x ≤ 20, x ≥ 20 |
-Case 1: x ≤ 3 Now when x ≤ 3 then both x - 3 and x - 20 will be non-positive => | x-3| = -(x-3) and |x-20| = -(x-20) => -(x-3) - (x-20) < 17 => -2x + 23 < 17 => 2x > 23 - 17 => x > 6/2 => x > 3
But the range was x ≤ 3 => NO SOLUTION in this case | -Case 2: 3 ≤ x ≤ 20 Now when 3 ≤ x ≤ 20 then x - 3 will be non-negative and x - 20 will be non-positive => | x-3| = (x-3) and |x-20| = -(x-20) => (x-3) - (x-20) < 17 => 17 < 17
=> NO SOLUTION in this case | Case 3: x ≥ 20 Now when x ≥ 20 then both x - 3 and x - 20 will be non-negative => | x-3| = (x-3) and |x-20| = (x-20) => (x-3) + (x-20) < 17 => 2x - 23 < 17 => 2x < 40 => x < 40/2 => x < 20
But the range was x ≥ 20 => NO SOLUTION in this case |
Method 2: Graph|x-3| = Distance between x and 3
|x-20| = Distance between x and 20
=> |x-3| + |x-20| = Distance between x and 3 + Distance between x and 20Now, there can be 5 cases
Case 1: x is smaller than 3Distance between x and 20 will be greater than the distance between 3 and 20, so will be more than 17
=>
Distance between x and 3 + Distance between x and 20 > 17 => it is NOT < 17
Attachment:
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Case 2: x = 3Distance between x and 20 = 17 and distance between x and 3 = 0
=> Distance between x and 3 + Distance between x and 20 = 17 => it is NOT < 17Attachment:
The attachment case 2.jpg is no longer available
Case 3: x is between 3 and 20Distance between x and 3 + Distance between x and 20 = 17 => it is NOT < 17 Attachment:
The attachment case-3.jpg is no longer available
Case 4: x = 20Distance between x and 20 = 0 and distance between x and 3 = 17
=> Distance between x and 3 + Distance between x and 20 = 17 => it is NOT < 17Attachment:
The attachment case 4.jpg is no longer available
Case 5: x is greater than 20Distance between x and 3 will be greater than the distance between 3 and 20, so will be more than 17
=>
Distance between x and 3 + Distance between x and 20 > 17 => it is NOT < 17
Attachment:
The attachment case-5.jpg is no longer available
So,
Answer will be CHope it helps!
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