greenwich wrote:Is x>0?
(1) |x + 3| = 4x - 3
(2) |x + 1| = 2x - 1
When solving questions involving ABSOLUTE VALUE, there are 3 steps:
1. Apply the rule that says:
If |x| = k, then x = k and/or x = -k
2. Solve the resulting equations
3. Plug in the solutions to check for extraneous roots
Now here's one approach:
Target question: Is x > 0?
Statement 1: |x + 3| = 4x - 3
When we apply the
rule above, we get two equations
case a: x + 3 = 4x - 3
Rearrange: 6 = 3x
Solve: x =
2
Check for extraneous roots by plugging x =
2 into solution
|
2 + 3| = 4(
2) - 3
5 = 5
Great, we'll KEEP the solution x =
2
case b: x + 3 = -(4x - 3)
Expand: x + 3 = -4x + 3
Rearrange: 5x = 0
Solve: x =
0
Check for extraneous roots by plugging x =
0 into solution
|
0 + 3| = 4(
0) - 3
3 = -3
No good. We'll DISREGARD the solution x =
0
So, x MUST equal 2, which means
x is definitely greater than 0
Since we can answer the
target question with certainty, statement 1 is SUFFICIENT
Statement 2: |x + 1| = 2x - 1
When we apply the
rule above, we get two equations
case a: x + 1 = 2x - 1
Rearrange: 2 = x
Solve: x =
2
Check for extraneous roots by plugging x =
2 into solution
|
2 + 1| = 2(
2) - 1
3 = 3
Great, we'll KEEP the solution x =
2
case b: x + 1 = -(2x - 1)
Expand: x + 1 = -2x + 1
Rearrange: 3x = 0
Solve: x =
0
Check for extraneous roots by plugging x =
0 into solution
|
0 + 1| = 2(
0) - 1
1 = -1
No good. We'll DISREGARD the solution x =
0
So, x MUST equal 2, which means
x is definitely greater than 0
Since we can answer the
target question with certainty, statement 2 is SUFFICIENT
Answer =
D
Cheers,
Brent
