Respuesta :
y = |x - 1|
and
y = 3x + 2
Equating the values of y, we can write:
|x - 1| = 3x + 2
This means:
A) x - 1 = 3x + 2
x - 3x = 2 + 1
-2x = 3
⇒ x = -3/2 = -1.5
B) - ( x - 1) = 3x + 2
3x + x = 1 - 2
4x = -1
⇒ x= -1/4 = - 0.25
For x = -0.25
y = 3(-0.25) + 2 = 1.25
x = -1.5 is an extraneous root and is not applicable.
So the solution set is (-0.25, 1.25)
Option A gives the approximate solution set.
and
y = 3x + 2
Equating the values of y, we can write:
|x - 1| = 3x + 2
This means:
A) x - 1 = 3x + 2
x - 3x = 2 + 1
-2x = 3
⇒ x = -3/2 = -1.5
B) - ( x - 1) = 3x + 2
3x + x = 1 - 2
4x = -1
⇒ x= -1/4 = - 0.25
For x = -0.25
y = 3(-0.25) + 2 = 1.25
x = -1.5 is an extraneous root and is not applicable.
So the solution set is (-0.25, 1.25)
Option A gives the approximate solution set.