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IB MYP 3 Mathematics 4.2 Equations with Variables on Both Sides and Rational Coefficients Study Notes - New Syllabus

IB MYP 3 Mathematics 4.2 Equations with Variables on Both Sides and Rational Coefficients Study Notes

IB MYP 3 Mathematics 4.2 Equations with Variables on Both Sides and Rational Coefficients Study Notes at IITian Academy focus on specific topics and types of questions asked in the actual exam. Study Notes focus on the IB MYP 3 Mathematics syllabus with guiding questions of

Rational coefficient: A coefficient that is a rational number, including integers, fractions and terminating decimals.
Rational number: A number that can be written in the form \(\frac{a}{b}\), where \(a\) and \(b\) are integers and \(b\neq0\).
Variables on both sides: An equation in which the unknown variable appears in terms on both sides of the equation.
Least common denominator (LCD): The smallest common denominator used to eliminate fractions from an equation.
One solution: An equation that simplifies to a statement such as \(x=a\), giving exactly one value of the variable.
No solution: An equation that simplifies to a false statement such as \(7=12\).
Infinitely many solutions: An equation that simplifies to a true statement such as \(9=9\), meaning every value of \(x\) works.
Standard procedure: Expand brackets, collect like terms, move variable terms to one side, move constants to the other side, divide by the coefficient, and check the solution.
Key rule: Perform the same operation on both sides of the equation to maintain balance, and when fractions are present, the LCD can be used to clear the denominators.

IB MYP 3 Mathematics – Study Notes – All Topics

4.2 – Equations with Variables on Both Sides and Rational Coefficients

In the previous section, the variable appeared on only one side of the equation. In this section, the variable may appear on both sides, and the coefficients may be rational numbers, including fractions, integers, and terminating decimals.

The main goal remains the same: isolate the variable on one side while maintaining the balance of the equation.

The textbook emphasizes the same fundamental principle: when an unknown appears more than once, we perform a sequence of operations on both sides until the unknown is isolated. :contentReference[oaicite:0]{index=0}

What Are Rational Coefficients?

A coefficient is the number multiplying a variable.

A rational number can be written in the form \(\frac{a}{b}\), where \(a\) and \(b\) are integers and \(b\neq0\).

CoefficientTypeExample
\(5\)Integer\(5x\)
\(-3\)Integer\(-3x\)
\(\frac{2}{3}\)Fraction\(\frac{2}{3}x\)
\(-\frac{5}{4}\)Fraction\(-\frac{5}{4}x\)
\(0.6\)Terminating decimal\(0.6x\)

The Main Strategy

For an equation of the form

\(ax+b=cx+d\)

we want to collect all the \(x\)-terms on one side and all constant terms on the other side.

⭐ Standard Procedure

  1. Expand brackets if necessary.
  2. Collect like terms on each side.
  3. Move all variable terms to one side.
  4. Move all constant terms to the other side.
  5. Divide by the coefficient of the variable.
  6. Check the solution in the original equation.

This follows the textbook’s general sequence: expand and collect like terms, remove the unknown from one side if necessary, and then use inverse operations to isolate it. :contentReference[oaicite:1]{index=1}

Variables on Both Sides

Example: Solve \(5x+7=2x+25\).

\(5x+7=2x+25\)
\(5x-2x+7=2x-2x+25\)
\(3x+7=25\)
\(3x=18\)
\(x=6\)

We chose to subtract \(2x\) from both sides so that the variable remained only on the left.

💡 Important: You could instead move the \(5x\) to the right. Either method is valid as long as the same operation is performed on both sides.

 Choosing Which Variable Term to Move

Consider:

\(8x+3=5x+21\)

There are two valid approaches.

Method 1 — Move \(5x\) to the left:

\(8x-5x+3=21\)
\(3x+3=21\)
\(3x=18\)
\(x=6\)

Method 2 — Move \(8x\) to the right:

\(3=21-3x\)
\(-18=-3x\)
\(x=6\)

Both methods produce the same solution.

 Equations with Fractional Coefficients

Fractions can make equations look more complicated than they actually are. There are two main approaches:

  • Method 1: Work directly with the fractions.
  • Method 2: Multiply every term by the least common denominator (LCD) first.

Example: Solve

\(\dfrac{x}{3}+2=\frac{x}{6}+7\)

Method 1 — Work directly with fractions

\(\dfrac{x}{3}+2=\frac{x}{6}+7\)
\(\dfrac{x}{3}-\frac{x}{6}=7-2\)
\(\dfrac{2x}{6}-\frac{x}{6}=5\)
\(\dfrac{x}{6}=5\)
\(x=30\)

Method 2 — Clear the fractions first

The denominators are \(3\) and \(6\), so the LCD is \(6\).

\(6\left(\frac{x}{3}+2\right)=6\left(\frac{x}{6}+7\right)\)
\(2x+12=x+42\)
\(x+12=42\)
\(x=30\)
⭐ Why multiply by the LCD?
Multiplying every term on both sides by the LCD eliminates the fractions while preserving the equality.

Equations with Several Fractions

When an equation contains several fractional coefficients, clearing the denominators first is often the most efficient method.

Example: Solve

\(\dfrac{2x}{3}-\frac{1}{2}=\frac{x}{6}+5\)

The denominators are \(3\), \(2\), and \(6\), so the LCD is \(6\).

\(6\left(\frac{2x}{3}-\frac{1}{2}\right)=6\left(\frac{x}{6}+5\right)\)
\(4x-3=x+30\)
\(3x=33\)
\(x=11\)

The important point is that the \(6\) must multiply every term, not just the fractions.

Decimal Coefficients

Terminating decimals can also be rational coefficients.

Example: Solve

\(0.8x+3=0.3x+13\)

\(0.8x-0.3x=13-3\)
\(0.5x=10\)
\(x=20\)

Alternatively, multiply the entire equation by \(10\):

\(8x+30=3x+130\)
\(5x=100\)
\(x=20\)
💡 Exam Tip: If decimal coefficients are awkward, multiply the whole equation by \(10\), \(100\), \(1000\), etc., depending on the number of decimal places.

 Equations with Brackets and Variables on Both Sides

If brackets are present, expand them before collecting the variable terms.

Example: Solve

\(3(x+2)=2x+14\)

\(3x+6=2x+14\)
\(3x-2x=14-6\)
\(x=8\)

For more complicated equations, the same sequence applies: expand → collect → move variable terms → isolate → check. The textbook uses this exact progression when solving equations involving repeated unknowns. :contentReference[oaicite:2]{index=2}

 Three Possible Types of Solutions

An equation with variables on both sides does not necessarily have exactly one solution. It may have:

TypeResult after simplifyingMeaning
One solution\(x=a\)Exactly one value works.
No solution\(a=b\), where \(a\neq b\)The equation is impossible.
Infinitely many solutions\(a=a\)Every value of \(x\) works.

Example — No solution:

\(4x+7=4x+12\)
\(7=12\)

Since \(7=12\) is false, there is no solution.

Example — Infinitely many solutions:

\(3(x+2)=3x+6\)
\(3x+6=3x+6\)

Since the resulting statement is always true, there are infinitely many solutions.

🚨 Do not automatically stop when the variable disappears.
If you get a true statement such as \(8=8\), there are infinitely many solutions.
If you get a false statement such as \(8=11\), there is no solution.

 Checking Solutions

For equations with variables on both sides, substitution into the original equation is an excellent way to verify the result.

For example, for \(5x+7=2x+25\), we found \(x=6\).

LHS \(=5(6)+7=37\)
RHS \(=2(6)+25=37\)

Therefore, LHS \(=\) RHS and \(x=6\) is verified.

Common Mistakes

  • Moving a term to the other side without changing its sign.
  • Combining unlike terms such as \(3x+5\) into \(8x\).
  • Multiplying only some terms by the LCD.
  • Forgetting that a negative coefficient must be included when dividing.
  • Expanding brackets incorrectly.
  • Failing to distinguish between no solution and infinitely many solutions.
  • Not checking the final answer.

Example 1: 

\(7x-5=3x+19\)

Give your answer in its simplest form.

▶️ Answer/Explanation

Solution

\(7x-5=3x+19\)
\(7x-3x=19+5\)
\(4x=24\)
\(x=6\)

Check:

LHS \(=7(6)-5=37\)
RHS \(=3(6)+19=37\)

Therefore, \(\boxed{x=6}\).

Example 2: 

\(\frac{3x}{4}+2=\frac{x}{2}+8\)

▶️ Answer/Explanation

Solution

The denominators are \(4\) and \(2\), so the LCD is \(4\).

\(4\left(\frac{3x}{4}+2\right)=4\left(\frac{x}{2}+8\right)\)
\(3x+8=2x+32\)
\(3x-2x=32-8\)
\(x=24\)

Check:

LHS \(=\frac{3(24)}{4}+2=18+2=20\)
RHS \(=\frac{24}{2}+8=12+8=20\)

Therefore, \(\boxed{x=24}\).

Example 3:

Determine whether each equation has one solution, no solution, or infinitely many solutions.

a) \(5x+9=5x+9\)

b) \(4x-7=4x+2\)

▶️ Answer/Explanation

a)

\(5x+9=5x+9\)
\(9=9\)

This is always true, so there are infinitely many solutions.

b)

\(4x-7=4x+2\)
\(-7=2\)

This is false, so there is no solution.

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