\(y = |x – 1|\) and \(y = 3x + 3,\)

Let's substitute the y = 3x + 3 in \(y = |x – 1|\), we get
3x + 3 = |x – 1|
=> | x-1 | 3x + 3 ..(1)

Let's solve it using two methods

Method 1: Algebra

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As we have |x-1| in the equation so we will have two cases
-Case 1: x - 1 ≥ 0 => x ≥ 1 => | x-1 | = x-1 => x-1 = 3x + 3 (From (1)) => 2x = -4 => x = -2 But our condition was x ≥ 1 => NO SOLUTION	-Case 2: x ≤ 1 => | x-1 | = -(x-1) => -(x-1) = 3x + 3 (From (1)) => -x + 1 = 3x + 3 => 4x = -2 => x = -2/4 = -0.5 But our condition was x ≤ 1, and -0.5 ≤ 1 => x = -0.5 is a SOLUTION

Method 2: Substitution

| x-1 | = 3x + 3
Let's pick values in the range of each option choice and see if it satisfies the above equation

(A) 2 and 3
x = 2.5
=> | 2.5-1 | = 3*2.5 + 3 => 1.5 = 10.5 => NOT POSSIBLE

(B) 1 and 2
x = 1.5
=> | 1.5-1 | = 3*1.5 + 3 => 0.5 = 7.5 => NOT POSSIBLE

(C) 0 and 1
x = 0.5
=> | 0.5-1 | = 3*0.5 + 3 => 0.5 = 4.5 => NOT POSSIBLE

(D) -1 and 0
x = -0.5
=> | -0.5-1 | = 3*-0.5 + 3 => 1.5 = 1.5 => POSSIBLE
We don't need to solve further, but solving to complete the solution

(E) -2 and -1
x = -1.5
=> | -1.5-1 | = 3*(-2.5) + 3 => 2.5 = -4.5 => NOT POSSIBLE

So, Answer will be D
Hope it helps!

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