Given that |x - 5| + |x - 7| < 4 and we need to find the range of values of x

Let's solve the problem using three methods

Method 1: Substitution

Let's take values in each answer choice and substitutive in the equation and check which one satisfies the equation.

If you glance through the answer choices then you will see that if u take x=0 and are able to prove it wrong then we can eliminate B and E

(B), (E) : x=0
=> |x - 5| + |x - 7| < 4 => |0 - 5| + |0 - 7| < 4 => 5 + 7 = 12 NOT LESS THAN 4
=> FALSE

(A) 4 < x < 8 x = 7
=> |7 - 5| + |7 - 7| < 4 => |2| + 0 < 4
=> TRUE => Possible

(C) 3 < x < 5 x = 4
=> |4 - 5| + |4 - 7| < 4 => |-1| + |-3| = 4 NOT LESS THAN 4
=> FALSE

(D) -8 < x < -4 x = -5
=> |-5 - 5| + |-5 - 7| < 4 => |-10| + |-12| < 4 => 10 + 12 NOT LESS THAN 4
=> FALSE

Method 2: Algebra

|x - 5| + |x - 7| < 4
Let's find the points where the values inside the Absolute Value will change signs
=> x-5 = 0 => x = 5
=> x-7 = 0 => x = 7

So, we can divide the number line into three parts

x < 5, 5 <= x < 7 , x > 7



Case 1: x < 5

|x - 5| = -(x-5) (as x-5 will be negative)
|x - 7| = -(x-7) (as x-7 will be negative)
=> -(x - 5) + (-(x - 7)) < 4 => -x + 5 -x +7 < 4 => -2x + 12 < 4
=> -2x < 4-12 => -2x < -8 => x > \(\frac{-8}{-2}\) => x > 4
But the condition was x < 5
So, intersection is 4 < x < 5

Case 2: 5 <= x < 7

|x - 5| = x-5 (as x-5 will be non-negative)
|x - 7| = -(x-7) (as x-7 will be negative)
=> x - 5 + (-(x - 7)) < 4 => x - 5 -x +7 < 4 => 2 < 4
Which is always true
=> 5 <= x < 7

Case 3: x >= 7

|x - 5| = x-5 (as x-5 will be non-negative)
|x - 7| = x-7 (as x-7 will be non-negative)
=> x - 5 + x - 7 < 4 => -x + 5 + x -7 < 4 => 2x - 12 < 4
=> 2x < 4+12 => 2x < 16 => x < \(\frac{16}{2}\) => x < 8
But the condition was x >= 7
So, intersection is 7 <= x < 8

So, solution is 4 < x < 5, 5 <= x < 7, 7 <= x < 8
Which is nothing but 4 < x < 8

Method 3: Graphical Method

| x - 5 | is the distance between x and 5
| x - 7 | is the distance between x and 7

Case 1: If we take a value of x between 5 and 7



Then, the distance between x and 5 + Distance between x and 7 = Distance between 5 and 7 = 2 < 4
=> All values between 5 and 7 inclusive are possible

Case 2: If we take a value of x < 5



Then Distance between x and 7 = Distance between x and 5 + Distance between 5 and 7
=> Distance between x and 7 = Distance between x and 5 + 2
So, for Distance between x and 5 + Distance between x and 7 < 4
=> Distance between x and 5 + Distance between x and 5 + 2 < 4
=> 2 * Distance between x and 5 < 4-2
=> Distance between x and 5 < 1
So, If x is at a distance of less than 1 from 5 on the left hand side then it is possible
=> 4 < x <= 5

Case 3: If we take a value of x > 7



Then Distance between x and 5 = Distance between x and 7 + Distance between 5 and 7
=> Distance between x and 5 = Distance between x and 7 + 2
So, for Distance between x and 5 + Distance between x and 7 < 4
=> Distance between x and 7 + 2 + Distance between x and 7 < 4
=> 2 * Distance between x and 7 < 4-2
=> Distance between x and 7 < 1
So, If x is at a distance of less than 1 from 7 on the right hand side then it is possible
=> 7 =< x < 8

So, solution is 4 < x < 5, 5 <= x < 7, 7 <= x < 8
Which is nothing but 4 < x < 8

So, Answer will be A
Hope it helps!

Watch the following video to learn How to Solve Absolute Value Problems