MYP Inquiry Framework
Key Concept Relationships & Logic
Related Concepts Change, Equivalence, Models
Global Context Fairness and development (Economic break-even analysis & decision boundaries)
Statement of Inquiry Constant rates of change define linear relationships, enabling logical constraint boundaries to guide fair resource allocation.
2.1

Solving Multi-Step & Fractional Linear Equations

  • Balance Principle: Whatever operation is applied to one side of the equation must be identically applied to the other side to preserve equality.
  • Clearing Denominators: Multiply both sides of the equation by the Lowest Common Multiple (LCM) of all denominators. For example, to solve $\frac{2x-1}{3} - \frac{x+4}{4} = 1$, multiply every term by $12$: $$12 \left(\frac{2x-1}{3}\right) - 12\left(\frac{x+4}{4}\right) = 12(1) \implies 4(2x-1) - 3(x+4) = 12$$
  • Classification of Solutions:
    • Unique Solution: Standard linear equation simplifying to $x = k$ (e.g., $3x + 5 = 14 \implies x = 3$).
    • Identity (Infinitely Many Solutions): Simplifies to a true statement independent of $x$ (e.g., $2(x+3) = 2x + 6 \implies 6 = 6$). Solution: $x \in \mathbb{R}$.
    • Contradiction (No Solution): Simplifies to a false statement (e.g., $4x - 1 = 4x + 7 \implies -1 = 7$). Solution: $\varnothing$.
2.2

Literal Equations & Changing the Subject of a Formula

  • Literal Equations: Equations where coefficients and variables are expressed as letters (e.g., $v = u + at$, $A = 2\pi r^2 + 2\pi rh$, $S = \frac{a}{1 - r}$).
  • Algorithm for Isolating the Target Variable $x$:
    1. Clear fractions by multiplying by the common denominator.
    2. Expand any brackets containing the target variable.
    3. Collect all terms containing $x$ onto one side, and all non-$x$ terms onto the opposite side.
    4. Factor out $x$ as a common factor: $x(\dots) = \dots$
    5. Divide by the cofactor expression to isolate $x$.
  • Example: Make $x$ the subject of $y = \frac{ax + b}{cx + d}$: $$y(cx + d) = ax + b \implies cyx + dy = ax + b \implies cyx - ax = b - dy$$ $$x(cy - a) = b - dy \implies x = \frac{b - dy}{cy - a}$$
2.3

Linear Inequalities & Number Line Representations

  • Inequality rule: Multiplying or dividing both sides of an inequality by a negative number reverses the inequality symbol: $$-3x \le 12 \iff x \ge -4$$
  • Compound Inequalities: Inequalities of the form $a < x \le b$ represent values lying strictly between $a$ and $b$, inclusive of $b$. Operations must be performed across all three parts simultaneously: $$-5 < 2x + 1 \le 9 \iff -6 < 2x \le 8 \iff -3 < x \le 4$$
  • Interval and Set Notations:
    • $-3 < x \le 4$ in interval notation: $(-3, 4]$. Parenthesis $($ is strict; square bracket $]$ is inclusive.
    • In set builder notation: $\{x \in \mathbb{R} \mid -3 < x \le 4\}$.
x y (0, c) Δx Δy y = mx + c -3 4 -3 < x ≤ 4 Interval: (-3, 4]
2.4

Cartesian Linear Equations & 2D Inequality Regions

  • Forms of a Linear Equation:
    • Gradient-Intercept Form: $y = mx + c$, where $m$ is gradient and $c$ is the $y$-intercept $(0, c)$.
    • Point-Slope Form: $y - y_1 = m(x - x_1)$, ideal when given gradient $m$ and a point $(x_1, y_1)$.
    • General Form: $Ax + By + C = 0$, where $A, B, C \in \mathbb{Z}$ and $A \ge 0$.
  • Gradient Properties: $$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\Delta y}{\Delta x}$$
    • Parallel Lines: $m_1 = m_2$.
    • Perpendicular Lines: $m_1 \cdot m_2 = -1 \iff m_2 = -\frac{1}{m_1}$ (negative reciprocal).
  • Graphing 2D Linear Inequalities:
    1. Graph the boundary line $y = mx + c$: use a solid line for $\le$ or $\ge$, and a dashed line for strict inequalities $<$ or $>$.
    2. Choose a test point not on the line (e.g., $(0, 0)$ if line does not pass through origin).
    3. Substitute test coordinates into inequality: if true, shade that side; if false, shade the opposite side.

Worked Examples (Criterion A & C)

Example 1 Fractional Linear Equation with Brackets

Problem: Solve for $x$: $\dfrac{3(x - 2)}{4} - \dfrac{2x + 1}{6} = \dfrac{1}{3}$.

Step 1: Identify the Lowest Common Multiple of denominators

The denominators are $4$, $6$, and $3$. Their LCM is $12$.

Step 2: Multiply every term by $12$ to clear fractions $$12 \cdot \frac{3(x - 2)}{4} - 12 \cdot \frac{2x + 1}{6} = 12 \cdot \frac{1}{3}$$ $$3 \cdot 3(x - 2) - 2(2x + 1) = 4$$ $$9(x - 2) - 2(2x + 1) = 4$$
Step 3: Expand brackets and collect like terms $$9x - 18 - 4x - 2 = 4 \implies 5x - 20 = 4$$ $$5x = 24 \implies \mathbf{x = \frac{24}{5} = 4.8}$$ Criterion A: Complete algebraic precision without rounding error
Example 2 Perpendicular Bisector Line Equation

Problem: Points $A(1, 4)$ and $B(5, -2)$ are given. Find the equation of the perpendicular bisector of segment $AB$ in general form $Ax + By + C = 0$.

Step 1: Find the midpoint of $AB$ $$M = \left(\frac{1 + 5}{2}, \frac{4 + (-2)}{2}\right) = (3, 1)$$
Step 2: Find the gradient of line segment $AB$ $$m_{AB} = \frac{-2 - 4}{5 - 1} = \frac{-6}{4} = -\frac{3}{2}$$
Step 3: Find the perpendicular gradient $$m_\perp = -\frac{1}{m_{AB}} = -\frac{1}{-3/2} = \frac{2}{3}$$
Step 4: Use point-slope form with $M(3, 1)$ $$y - 1 = \frac{2}{3}(x - 3) \implies 3(y - 1) = 2(x - 3) \implies 3y - 3 = 2x - 6$$ $$\mathbf{2x - 3y - 3 = 0}$$ Criterion C: Logical deduction with correct general form coefficients

MYP 4–5 Practice Exercises: Linear Equations & Inequalities

Tier 1 (Core skills), Tier 2 (Reasoning), and Tier 3 (Extension and Criterion D applications).

Question 1 Tier 1: Standard 3 Marks

Solve the linear equation: $5(2x - 1) = 3(x + 4) + 4$.

View Full Method Markscheme
[M1] Expand brackets: $10x - 5 = 3x + 12 + 4 \implies 10x - 5 = 3x + 16$.
[M1] Rearrange to isolate $x$: $10x - 3x = 16 + 5 \implies 7x = 21$.
[A1] Divide by $7$: $\mathbf{x = 3}$.
Question 2 Tier 1: Standard 3 Marks

Solve the inequality and express the solution in interval notation: $7 - 3x \le 19$.

View Full Method Markscheme
[M1] Subtract $7$ from both sides: $-3x \le 12$.
[M1] Divide by $-3$ and reverse the inequality sign: $x \ge -4$.
[A1] Express in interval notation: $\mathbf{[-4, \infty)}$.
Question 3 Tier 1: Standard 3 Marks

Find the gradient and $y$-intercept of the straight line with equation $3x - 4y = 12$.

View Full Method Markscheme
[M1] Rearrange into slope-intercept form $y = mx + c$: $-4y = -3x + 12$.
[M1] Divide each term by $-4$: $y = \frac{3}{4}x - 3$.
[A1] Gradient $\mathbf{m = \frac{3}{4}}$, $y$-intercept is $\mathbf{(0, -3)}$ or $c = -3$.
Question 4 Tier 2: Intermediate 4 Marks

Rearrange the physics formula to make the acceleration $a$ the subject: $s = ut + \frac{1}{2}at^2$.

View Full Method Markscheme
[M1] Subtract $ut$ from both sides: $s - ut = \frac{1}{2}at^2$.
[M1] Multiply both sides by $2$: $2(s - ut) = at^2$.
[A1] Divide by $t^2$ (assuming $t \ne 0$): $\mathbf{a = \frac{2(s - ut)}{t^2}}$ or $\mathbf{a = \frac{2s - 2ut}{t^2}}$.
Question 5 Tier 2: Intermediate 4 Marks

Solve the compound inequality and state all integer solutions for $x$: $-7 < 3x + 2 \le 14$.

View Full Method Markscheme
[M1] Subtract $2$ from all three sections: $-7 - 2 < 3x \le 14 - 2 \implies -9 < 3x \le 12$.
[M1] Divide all three sections by $3$: $-3 < x \le 4$.
[A1] Solution interval: $(-3, 4]$.
[A1] Integer values of $x$: $\mathbf{\{-2, -1, 0, 1, 2, 3, 4\}}$.
Question 6 Tier 2: Intermediate 4 Marks

Find the equation of the line passing through point $P(-2, 5)$ that is parallel to the line $4x + 2y = 9$. Write your answer in slope-intercept form.

View Full Method Markscheme
[M1] Find gradient of $4x + 2y = 9$: $2y = -4x + 9 \implies y = -2x + 4.5$. Thus $m = -2$.
[M1] Since lines are parallel, target line gradient is $m_{||} = -2$.
[M1] Apply point-slope form with $(-2, 5)$: $y - 5 = -2(x - (-2)) \implies y - 5 = -2(x + 2)$.
[A1] Expand and simplify: $y - 5 = -2x - 4 \implies \mathbf{y = -2x + 1}$.
Question 7 Tier 2: Intermediate 4 Marks

Solve for $x$: $\dfrac{x + 1}{2} - \dfrac{2x - 3}{5} = \dfrac{x}{10} + 1$.

View Full Method Markscheme
[M1] Multiply every term by the LCM $10$: $$10\left(\frac{x+1}{2}\right) - 10\left(\frac{2x-3}{5}\right) = 10\left(\frac{x}{10}\right) + 10(1)$$
[M1] Expand: $5(x + 1) - 2(2x - 3) = x + 10 \implies 5x + 5 - 4x + 6 = x + 10$.
[M1] Simplify left side: $x + 11 = x + 10$.
[A1] Subtracting $x$ gives $11 = 10$, a contradiction! Hence, there is $\mathbf{\text{No solution } (\varnothing)}$.
Question 8 Tier 3: Extended 5 Marks

Make $x$ the subject of the formula: $y = \dfrac{3x - 5}{2x + 7}$.

View Full Method Markscheme
[M1] Clear fraction: $y(2x + 7) = 3x - 5$.
[M1] Expand brackets: $2xy + 7y = 3x - 5$.
[M1] Collect $x$-terms on one side: $7y + 5 = 3x - 2xy$.
[M1] Factor out $x$: $7y + 5 = x(3 - 2y)$.
[A1] Divide to isolate $x$: $\mathbf{x = \dfrac{7y + 5}{3 - 2y}}$ (for $y \ne \frac{3}{2}$).
Question 9 Tier 3: Extended 5 Marks

Criterion D Real-World Modeling: A solar panel installation company offers two pricing structures:
• Plan A: Flat setup fee of $\$1200$ plus $\$150$ per kilowatt-peak (kWp) capacity installed.
• Plan B: No setup fee, but $\$225$ per kWp installed.
(a) Write linear cost functions $C_A(x)$ and $C_B(x)$ in terms of capacity $x$ in kWp.
(b) Determine algebraically the capacity range for which Plan A is strictly more economical than Plan B.

View Full Method Markscheme
[M1] Formulate functions: $C_A(x) = 150x + 1200$ and $C_B(x) = 225x$.
[M1] Set up the strict inequality for Plan A to be cheaper: $C_A(x) < C_B(x)$.
[M1] Substitute: $150x + 1200 < 225x$.
[M1] Rearrange: $1200 < 225x - 150x \implies 1200 < 75x$.
[A1] Divide by $75$: $x > \frac{1200}{75} = 16$. Plan A is more economical for installations strictly exceeding $\mathbf{16\text{ kWp}}$ ($x > 16$).
Question 10 Tier 3: Extended 5 Marks

A triangular region on the Cartesian plane is bounded by the inequalities: $$x \ge 0, \quad y \ge 0, \quad 2x + y \le 8, \quad x + 2y \le 10$$ (a) Find the coordinates of all four vertices of this feasible polygonal region.
(b) Find the maximum value of the objective function $P = 3x + 4y$ across this region.

View Full Method Markscheme
[M1] Identify boundary intersections with coordinate axes:
  • Origin: $(0, 0)$
  • $y$-axis intercept ($x = 0$): $y \le 8$ and $2y \le 10 \implies y \le 5$, giving $(0, 5)$
  • $x$-axis intercept ($y = 0$): $2x \le 8 \implies x \le 4$, giving $(4, 0)$
[M1] Find intersection of $2x + y = 8$ and $x + 2y = 10$: Multiply first by $2$: $4x + 2y = 16$. Subtract second: $3x = 6 \implies x = 2$. Then $y = 8 - 2(2) = 4$. Vertex is $(2, 4)$.
[A1] Four vertices: $\mathbf{(0,0), (4,0), (2,4), (0,5)}$.
[M1] Evaluate objective function $P = 3x + 4y$ at all 4 vertices:
  • At $(0, 0)$: $P = 0$
  • At $(4, 0)$: $P = 3(4) + 0 = 12$
  • At $(0, 5)$: $P = 0 + 4(5) = 20$
  • At $(2, 4)$: $P = 3(2) + 4(4) = 6 + 16 = 22$
[A1] Maximum value is $\mathbf{P = 22}$ occurring at vertex $(2, 4)$.

Linear Algebra Formula Reference

  • Gradient formula: $m = \frac{y_2 - y_1}{x_2 - x_1}$
  • Slope-Intercept form: $y = mx + c$
  • Point-Slope form: $y - y_1 = m(x - x_1)$
  • General form: $Ax + By + C = 0 \quad (A \ge 0, \, A,B,C \in \mathbb{Z})$
  • Parallel lines condition: $m_1 = m_2$
  • Perpendicular lines condition: $m_1 \cdot m_2 = -1 \iff m_2 = -\frac{1}{m_1}$
  • Midpoint coordinates: $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$
  • Distance between points: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Top Common MYP Pitfalls

1. Forgetting to Flip Inequality Signs

Dividing or multiplying by a negative number without reversing the inequality symbol (e.g., $-2x < 6 \implies x < -3$ is wrong; it must be $x > -3$).

2. Confusing Boundary Line Styles

Using a solid line for strict inequalities ($<$ or $>$). Strict inequalities require a dashed line to indicate that points on the line are excluded!

3. Sign Errors When Clearing Fractions

In $- \frac{2x - 3}{5}$, the negative sign belongs to the entire binomial: $-(2x - 3) = -2x + 3$. Forgetting brackets turns $+3$ into $-3$.

4. Perpendicular Slope Calculation

Taking the reciprocal without negating (e.g., perpendicular to $3$ is $-\frac{1}{3}$, not $+\frac{1}{3}$).

Interactive Mini Test: Linear Equations & Inequalities

Answer the questions to see your score and feedback.

Question 1 of 5 Solving Linear Inequalities

What is the solution set to the inequality $4 - 5x \ge 24$?

Question 2 of 5 Perpendicular Gradient

What is the gradient of a straight line perpendicular to $2x + 5y = 10$?

Question 3 of 5 Literal Equation Subject

If $T = 2\pi \sqrt{\dfrac{L}{g}}$, what is $L$ expressed in terms of $T$, $\pi$, and $g$?

Question 4 of 5 Compound Inequality Solutions

How many integers satisfy the double inequality $-4 \le 2x - 1 < 7$?

Question 5 of 5 Types of Linear Solutions

Which statement correctly describes the equation $3(2x - 4) = 6x - 12$?