Learn with Amar
Teaching video
A teaching recording for this chapter has not been published yet. Start with the worked example below and explore the related animations where available.
Grade 9 chapters and video availability01 · Read and understand
What you will learn
- Multiply by a positive common denominator before solving a linear inequality.
- Justify the method and check its domain, units, or logical conditions.
Before you start
Signed arithmetic, fraction and decimal operations, one-step equations, and introductory coordinate graphs.
Keep paper nearby. Read the question once for the context, then again to identify what is known and what you need to find.
Start with a question
Solve 2(x−3)/3 ≤ (x+4)/2 over the real numbers.
Why this math matters
Clear fixed fractional coefficients while preserving the direction and domain of an inequality. The example connects a stated condition to a conclusion and then checks whether the result satisfies that condition.

Set up the model
A useful answer starts with clear assumptions:
- Use the real-number system and the restrictions or data model stated in the question unless another domain is specified.
- Treat provided measurements as exact classroom-model values unless an approximation or uncertainty is stated.
02 · Work through the example
Follow the reasoning, one step at a time.
Try to predict the next step before reading it. After each calculation, explain why the operation makes sense and how it helps answer the original question.

Work through it with Amar
See this example unfold.
The complete worked example, one idea at a time.
Clear fractions in a linear inequality
PausedQuestion: Start with the question. Paused.
Question
Start with the question
Solve 2(x−3)/3 ≤ (x+4)/2 over the real numbers.
Before you calculate
Read what is known and what you need to find. Make a prediction before moving to the first calculation.
Starts paused. Play advances through the full text at a reading pace; pause whenever you need more time. Previous, Next, and the phase buttons let you set your own pace. Playback pauses when this walkthrough leaves the screen or you switch tabs.
Your device’s reduced-motion setting keeps each phase still. Manual controls remain available. The full written solution stays below.
Represent the conditions
Multiply both sides by 6: 4(x−3) ≤ 3(x+4)
The positive common denominator clears both fractions without reversing the inequality.
Develop the calculation
4x−12 ≤ 3x+12
Distribute both multipliers before collecting terms.
Check and interpret
x ≤ 24
Subtract 3x and add 12; at x=24 both original sides equal fourteen, so the boundary is included.
The result
x ≤ 24
Subtract 3x and add 12; at x=24 both original sides equal fourteen, so the boundary is included.
Common mistakes to catch
- Multiply every term on both sides by the common denominator.
- Do not use this shortcut with a variable denominator before checking its sign and excluded inputs.
03 · Practice independently
Try it before revealing the answer.
Use paper or a calculator as needed. Write your units and reasoning, then open the hint or explanation to check your approach.
Practice 1
Solve (x−2)/3 > (x+1)/4.
Show a hint
Multiply both sides by positive twelve.
Reveal answer and explanation
x > 11
4x−8 > 3x+3 gives x>11; equality at eleven is excluded.
Practice 2
Why can a positive common denominator be used without reversing the inequality here?
Show a hint
The denominators two and three are fixed positive numbers.
Reveal answer and explanation
The multiplier six is positive for every allowed input
A variable denominator could change sign or be zero, so its sign and domain would need separate analysis before multiplication.
Take the idea with you
Clear fixed fractional coefficients while preserving the direction and domain of an inequality.
04 · Reflect and continue
Can you explain it in your own words?
Before moving on, explain the main idea without looking at the worked example. Try both practice questions, check your reasoning, and name one mistake you now know how to avoid. Return to a step if you still need support.
Up next: Solve a two-sided bound
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