This is an estimate based on your percent correct on our practice tests. The official GED score conversion is not public, so your real score may differ. Scores of 145+ pass; 165+ is College Ready.
What values of x make x - 9 \ge -14 true? Choose the solution set.
Hint
What is being done to x? Undo it on both sides. Does the negative number on the right affect the direction of the sign?
Step-by-step solution
9 is subtracted from x, so do the opposite: add 9 to both sides.
x - 9 + 9 \ge -14 + 9
Simplify. Adding a positive number to -14 moves 9 units to the right on the number line:
x \ge -5
You added a number, so the sign does not change. The negative constant on the right side does not matter: only multiplying or dividing by a negative number flips the sign.
Answer: x \ge -5
Check: Boundary x = -5: -5 - 9 = -14, and -14 \ge -14 is true, so -5 is included. Inside the set, x = 0: 0 - 9 = -9 \ge -14 ✓. Outside, x = -6: -15 \ge -14 is false ✓.
Tip: A negative number sitting on the right side is never a reason to flip. Flip only when you multiply or divide both sides by a negative.
Question 2 of 10 · Solve linear inequalities
No calculator
When x is isolated, what does \dfrac{x}{4} > -2.5 become?
Hint
x is divided by a positive 4. The number on the right is negative — does that matter?
Step-by-step solution
x is divided by 4, so multiply both sides by 4.
You are multiplying by positive 4, so the sign stays >. (The -2.5 is only a number on one side; it does not cause a flip.)
\frac{x}{4} \cdot 4 > -2.5 \cdot 4
Simplify:
x > -10
Answer: x > -10
Check: Try x = 0: \frac{0}{4} = 0 > -2.5 ✓. Try x = -12: \frac{-12}{4} = -3 and -3 > -2.5 is false ✓ (-3 is further left). The boundary x = -10 gives -2.5 > -2.5, false, so the sign stays strict.
Tip: Ask "what am I multiplying or dividing by?" — only a negative multiplier or divisor flips the sign. The signs of the other numbers do not matter.
Question 3 of 10 · Solve linear inequalities
No calculator
Which inequality gives every solution of \dfrac{x + 8}{3} \le 5?
Hint
The whole expression x + 8 is divided by 3. Which operation should you undo first?
Step-by-step solution
The fraction bar groups x + 8, so x + 8 is divided by 3. Undo the division first: multiply both sides by 3 (positive, no flip):
x + 8 \le 15
Subtract 8 from both sides:
x \le 7
Answer: x \le 7
Check: Boundary x = 7: \frac{7 + 8}{3} = \frac{15}{3} = 5 \le 5 ✓. Inside, x = 1: \frac{9}{3} = 3 \le 5 ✓. Outside, x = 10: \frac{18}{3} = 6 \le 5 is false ✓.
Tip: A fraction bar acts like parentheses around the top. When the +8 is inside the fraction, clear the denominator first; when it is outside (as in \frac{x}{3} + 8), undo the +8 first.
Question 4 of 10 · Graph solutions on a number line
No calculator
Which situation could be represented by the graph below?
Hint
Write the inequality the graph shows. Then translate each situation into an inequality and compare.
Step-by-step solution
Read the graph: closed circle at 30, shading to the left → x \le 30. The value 30 is included, and anything smaller is allowed.
Translate each situation:
Speed no faster than 30 mph → s \le 30 ✓ (30 is allowed).
At least 30 → \ge 30 (shades right).
Less than 30 → < 30 (open circle).
More than 30 → > 30 (open circle, shades right).
Only the speed-limit situation matches x \le 30.
Answer: A driver may go no faster than the 30 mph speed limit.
Check: Driving exactly 30 mph is legal (closed circle ✓); 25 mph is legal (shaded ✓); 35 mph is not (unshaded ✓).
Tip: "No faster than", "no more than", "at most" and "maximum" all include the boundary and shade left (\le).
Question 5 of 10 · Graph solutions on a number line
No calculatorFill in
The solution set of -7 < 2x + 1 \le 9 is graphed on a number line. How many integers are in the solution set?
Type your answer in the box. Fractions like 3/4 are fine.
Hint
Do the same operation to all three parts of the compound inequality. Then decide which endpoints are included before you count.
Step-by-step solution
Subtract 1 from all three parts:
-8 < 2x \le 8
Divide all three parts by 2 (positive, no flip):
-4 < x \le 4
The graph would have an open circle at -4 and a closed circle at 4, shaded between.
Integers in the set: -3, -2, -1, 0, 1, 2, 3, 4.
Count: 8.
Answer: 8
Check: Endpoint test: x = -4 gives 2(-4) + 1 = -7, and -7 < -7 is false, so -4 is excluded ✓. x = 4 gives 9 \le 9 ✓, so 4 is included. Formula: integers from -4 to 4 inclusive = 4 - (-4) + 1 = 9; minus 1 open endpoint = 8 ✓.
Tip: In a compound inequality, whatever you do to the middle you must do to both outside parts too.
Question 6 of 10 · Real-world inequalities
A pharmacy cooler must keep a vaccine between 2°C and 8°C, inclusive. When the cooler is switched on, the vaccine is at 14°C, and its temperature drops 1.5°C per minute. For which numbers of minutes t after switching on is the vaccine in the safe range?
Hint
The temperature after t minutes is 14 - 1.5t. Put it between 2 and 8, and remember what happens when you divide by a negative.
Step-by-step solution
Temperature after t minutes: 14 - 1.5t.
It must be between 2 and 8, inclusive:
2 \le 14 - 1.5t \le 8
Subtract 14 from all three parts:
-12 \le -1.5t \le -6
Divide all three parts by -1.5 and flip both signs:
8 \ge t \ge 4
Rewrite from smallest to largest: 4 \le t \le 8.
Answer: 4 \le t \le 8 (from 4 to 8 minutes)
Check: At t = 4: 14 - 6 = 8°C ✓. At t = 8: 14 - 12 = 2°C ✓. At t = 6: 14 - 9 = 5°C, inside the range ✓.
Tip: A compound inequality is solved like a regular one — just do every step to all three parts. Dividing by a negative flips both symbols.
Question 7 of 10 · Write inequalities from context
No calculatorFill in
A storage unit costs $45 per month plus a one-time $30 move-in fee. Ella can spend at most $300 in total. Complete the inequality for the number of months m she can rent the unit.
m\ +\le 300
Type your answer in the box. Fractions like 3/4 are fine.
Hint
Which number is charged every month? Which is charged only once?
Step-by-step solution
Variable:m = number of months.
Rate: $45 per month → 45m. So the first blank is 45.
Fixed amount: $30 one time → +30. So the second blank is 30.
Limit: "at most $300" → \le 300 (already given).
45m + 30 \le 300
Answer: 45 and 30
Check: For 6 months: 45(6) + 30 = 270 + 30 = 300 \le 300 ✓ — she can rent for 6 months and spend exactly $300. For 7 months: 315 + 30 = 345, over budget ✓.
Tip: "Per month", "each", "every" → coefficient of the variable. "One-time", "fee", "already" → the constant.
Question 8 of 10 · Graph solutions on a number line
No calculatorSelect all
Jorge graphed the solution set of \dfrac{x}{-2} + 1 \le 3 as shown.
Select all the errors in Jorge's graph.
Select all that apply.
Hint
Solve the inequality completely. Then compare the boundary, the circle type, and the direction with Jorge's graph one at a time.
Step-by-step solution
Solve. Subtract 1 from both sides:
\frac{x}{-2} \le 2
Multiply both sides by -2 and flip:
x \ge -4
Correct graph: closed circle at -4, shaded right.
Compare with Jorge's graph (closed circle at 4, shaded left):
Boundary: his is 4, it should be -4 → error ✓
Direction: his goes left, it should go right → error ✓
Circle: closed is correct for \ge → not an error.
Answer: "The boundary should be at -4, not 4" and "The shading should go to the right."
Check: Test x = 5, which Jorge left unshaded: \frac{5}{-2} + 1 = -1.5 \le 3 ✓, so 5 is a solution his graph misses. Test x = -6, which he shaded: \frac{-6}{-2} + 1 = 4 \le 3 is false, so it should not be shaded.
Tip: Grade a graph like a checklist: boundary value, circle type, direction. Each one can be right or wrong independently.
Question 9 of 10 · Real-world inequalities
Fill in
A rideshare driver earns a daily bonus plus a fixed amount per ride. The table shows her total earnings for some days.
Rides in a day
5
10
15
Total earnings
$62
$112
$162
She wants to earn at least $400 in one day. What is the least number of rides she must give?
rides
Type your answer in the box. Fractions like 3/4 are fine.
Hint
How much do 5 more rides add? Use that to find the earnings per ride, then the bonus.
Step-by-step solution
Rate: 5 more rides add 112 - 62 = 50 dollars, so each ride pays 50 \div 5 = 10 dollars.
Bonus: 5 rides earn 5 \times 10 = 50, but she made $62, so the bonus is 62 - 50 = 12 dollars.
Earnings for r rides: 12 + 10r. (Check: 12 + 10(15) = 162 ✓.)
Tip: Rate = \dfrac{\text{change in output}}{\text{change in input}}. Then the fixed part = any table output minus rate \times its input.
Question 10 of 10 · Write inequalities from context
No calculator
Maria works two part-time jobs. She earns $16 per hour at a store and $22 per hour tutoring. She must earn at least $600 per week, and she can work at most 35 hours per week in total. Let s be her store hours and t her tutoring hours. Which system of inequalities represents this situation?
Hint
Write one inequality for money and one for time. Which numbers go with money, and which with hours?
Step-by-step solution
Money: store pay 16s plus tutoring pay 22t, and she must earn at least $600:
16s + 22t \ge 600
Time: total hours s + t, and she can work at most 35:
s + t \le 35
Both must be true together.
Answer: 16s + 22t \ge 600 and s + t \le 35
Check: 15 store hours and 18 tutoring hours: pay = 240 + 396 = 636 \ge 600 ✓ and time = 33 \le 35 ✓ — a schedule that works. 30 store hours and 5 tutoring hours: pay = 480 + 110 = 590 ✗.
Tip: In a system, each inequality describes one limit. Check units: dollars with dollars, hours with hours.