ACT Math 4.2 Elementary Algebra
ACT Math Elementary Algebra Drill 4, Problem 2. Solve for x.
| ACT Math | Elementary Algebra |
| ACT Mathematics | Elementary Algebra |
| Basic Operations | Exponents |
| Elementary Algebra | Solving linear inequalities with one variable |
| Foreign Language | Arabic Subtitled Chinese Subtitled Korean Subtitled Spanish Subtitled |
| Intermediate Algebra | Absolute value equations and inequalities |
| Language | English Language |
| Product Type | ACT Math |
Transcript
We multiply both sides by 2 to get x plus 1 is less than 6.
Subtract 1 from both sides and we have x is less than 5, so our potential answers are
only B and C.
Now let's figure out what this problem would be if x were the most negative it could be.
What would make things under the absolute value lines to equal NEGATIVE 3?
We'd have x plus 1 over 2 is less than negative 3, but with no absolute value lines now.
We multiply both sides by 2 and we have x plus 1 is less than negative 6.
Then subtract 1 from both sides and we have x is less than negative 7.
So we have a positive 5 and a negative 7 as the range here.
Which is... this range... and our answer is C.