4x + 3 < 16 or –2x < 3
one-Variable Compound Inequalities
4x + 3 < 16
4x < 13
x < 13/4
–2x < 3
-3 < 2x
-3/2 < x
-3/2 < x < 13/4
This is equivalent to
4x + 3 < 16 AND –2x < 3
Compound inequalities always imply AND.
As you wrote it, x can be anything, because it will always be less than 13/4 OR greater than -3/2
How would you explain step by step to get the answer?
How do you graph this?
I showed the solutions. As for graphing, just plot the two solutions (points) and shade to the left or right, as appropriate.
If that doesn't clear it up, review the use of the number line.
To solve the given compound inequalities, let's break it down into two separate inequalities:
1. 4x + 3 < 16
2. –2x < 3
1. Solving 4x + 3 < 16:
Subtract 3 from both sides to isolate the variable:
4x + 3 - 3 < 16 - 3
4x < 13
Divide both sides by 4 to solve for x:
(4x) / 4 < 13 / 4
x < 13/4
The solution to the first inequality is x < 13/4.
2. Solving –2x < 3:
Divide both sides by -2 while remembering to flip the inequality sign:
(-2x) / -2 > 3 / -2
x > -3/2
The solution to the second inequality is x > -3/2.
Therefore, the solution to the compound inequality 4x + 3 < 16 or –2x < 3 is the combined solution of the two inequalities:
x < 13/4 or x > -3/2.