4|3x - 1| + 1 = 8
4|3x - 1| = 8 - 1
4|3x - 1| = 7
|3x - 1| = 7/4
Now lets think, we have 3x - 1 in module, so either its 7/4 or -7/4, we will have 7/4 at the end because of it, so we may have 2 solutions in this case:
3x - 1 = 7/4
and
3x - 1 = -7/4
So let's see:
3x - 1 = 7/4
3x = 7/4 + 1
3x = 11/4
x = 11/12
3x - 1 = -7/4
3x = -7/4 + 1
3x = -3/4
x = -1/4
So we have two possible answers, x = 11/2 and x = -1/4
The answer is B. hope i helped
Given, IJ = 3x + 3, HI = 3x - 1, and HJ = 3x + 8.
Since I is a point on line segment HJ, we can write


Put x=2 in HJ=3x+8.

Therefore, the numerical length of HJ is 14.
Solve the equation with the following steps:
8n-12=6n+6
Subtract 6n from both sides
2n-12=6
Add 12 to both sides
2n=18
Divide both sides by 2
n=9
Because all 3 sides are given for both triangles you can use the SSS theorem