MYP Inquiry Framework
Key Concept Logic & Systems
Related Concepts Equivalence, Intersection, Models
Global Context Fairness and development (Market equilibrium & manufacturing efficiency)
Statement of Inquiry Logical systems of simultaneous constraints establish points of equilibrium and reveal optimal operational conditions.
5.1

Classification & Geometry of Linear Systems

  • The Meaning of a Solution: A solution to a system of equations is a set of values $(x, y)$ that satisfies all equations simultaneously, corresponding to points of intersection on the Cartesian coordinate plane.
  • Three Geometric Possibilities:
    • Consistent & Independent (Unique Solution): Lines have different gradients ($m_1 \ne m_2$) and cross at exactly one point $(x^*, y^*)$.
    • Inconsistent (No Solution): Lines are strictly parallel ($m_1 = m_2, c_1 \ne c_2$). There are no intersection points; equations contradict each other (e.g., $0 = 7$).
    • Dependent (Infinitely Many Solutions): Lines are identical/coincident ($m_1 = m_2, c_1 = c_2$). Every point on the line is a solution ($0 = 0$).
Linear Intersection (1 Unique Root) (x*, y*) Non-Linear Line & Parabola (Δ Cases) Δ > 0 (Secant) Δ = 0 (Tangent) Δ < 0 (None)
5.2

Algebraic Methods: Substitution & Elimination

  • Method of Substitution:
    1. Isolate one variable in terms of the other from the simplest equation (e.g., $x = 5 - 2y$).
    2. Substitute this expression into the other equation, eliminating that variable to yield a single-variable equation.
    3. Solve for the remaining variable, then back-substitute to find the partner value.
    4. Check coordinates in both original equations.
  • Method of Elimination:
    1. Multiply one or both equations by non-zero constants so that coefficients of one variable are equal or opposite.
    2. Add equations (if coefficients are opposite, e.g. $+3y$ and $-3y$) or subtract equations (if identical, $+3y$ and $+3y$) to eliminate that variable.
    3. Solve the resulting linear equation, then back-substitute.
5.3

Non-Linear Systems: Intersection of Lines & Quadratics

  • Form of Non-Linear System: $$\begin{cases} y = mx + c & \text{(Linear)} \\ y = ax^2 + bx + d & \text{(Quadratic)} \end{cases}$$
  • Equating Equations: Substitute $y = mx + c$ into the parabola equation: $$ax^2 + bx + d = mx + c \implies ax^2 + (b - m)x + (d - c) = 0$$
  • Classification via Discriminant $\Delta = (b - m)^2 - 4a(d - c)$:
    • $\Delta > 0$: Line is a secant line; cuts the parabola at two distinct points.
    • $\Delta = 0$: Line is a tangent line; grazes the parabola at exactly one contact point.
    • $\Delta < 0$: Line and parabola do not intersect.
5.4

Real-World Simultaneous Modeling & Break-Even Analysis

  • Break-Even Economic Point: Total Cost $C(x) = \text{Fixed Costs} + (\text{Variable Cost per unit}) \cdot x$. Total Revenue $R(x) = (\text{Selling Price per unit}) \cdot x$. The break-even point is found by setting $R(x) = C(x)$. When $R(x) > C(x)$, the business makes a profit.
  • Mixture Problems: Set up one equation for total mass/volume and a second equation for total mass of the active ingredient (e.g., $x + y = 50$ and $0.20x + 0.50y = 0.32(50)$).

Worked Examples (Criterion A & C)

Example 1 Solving a Non-Linear System (Line & Parabola)

Problem: Find all points of intersection between the line $y = 2x + 1$ and the parabola $y = x^2 - x - 9$.

Step 1: Equate the expressions for $y$ $$x^2 - x - 9 = 2x + 1$$
Step 2: Collect all terms to one side to form standard quadratic $$x^2 - 3x - 10 = 0$$
Step 3: Factorise the quadratic equation $$(x - 5)(x + 2) = 0 \implies x_1 = 5, \quad x_2 = -2$$
Step 4: Find corresponding $y$-coordinates using $y = 2x + 1$

For $x = 5$: $y = 2(5) + 1 = 11 \implies \mathbf{(5, 11)}$.

For $x = -2$: $y = 2(-2) + 1 = -3 \implies \mathbf{(-2, -3)}$.

Criterion A: Methodical substitution yielding two exact coordinate pairs
Example 2 Break-Even Point Modeling

Problem: A student enterprise produces hand-crafted journals. Fixed setup costs are $\$450$, and manufacturing materials cost $\$6$ per journal. They sell each journal for $\$15$.
(a) Set up simultaneous cost and revenue equations.
(b) Calculate the break-even number of journals.

Step 1: Define variables and formulate functions

Let $n$ be the number of journals produced and sold.

Cost function: $C(n) = 6n + 450$.

Revenue function: $R(n) = 15n$.

Step 2: Set Cost equal to Revenue $$15n = 6n + 450 \implies 9n = 450 \implies \mathbf{n = 50\text{ journals}}$$
Step 3: State the financial break-even value

At $n = 50$, $R(50) = 15(50) = \mathbf{\$750}$.

Criterion C & D: Clear units, defined variables, and economic interpretation

MYP 4–5 Practice Exercises: Systems of Equations

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

Question 1 Tier 1: Standard 3 Marks

Solve the linear system by substitution: $$\begin{cases} y = 3x - 4 \\ 2x + 3y = 21 \end{cases}$$

View Full Method Markscheme
[M1] Substitute $y = 3x - 4$ into the second equation: $2x + 3(3x - 4) = 21$.
[M1] Expand and solve for $x$: $2x + 9x - 12 = 21 \implies 11x = 33 \implies x = 3$.
[A1] Find $y$: $y = 3(3) - 4 = 5$. Solution: $\mathbf{(3, 5)}$.
Question 2 Tier 1: Standard 3 Marks

Solve the linear system by elimination: $$\begin{cases} 4x + 5y = 22 \\ 3x - 5y = -1 \end{cases}$$

View Full Method Markscheme
[M1] Add the two equations directly to eliminate $y$: $(4x + 3x) + (5y - 5y) = 22 + (-1)$.
[M1] $7x = 21 \implies x = 3$.
[A1] Back-substitute into first equation: $4(3) + 5y = 22 \implies 12 + 5y = 22 \implies 5y = 10 \implies y = 2$. Solution: $\mathbf{(3, 2)}$.
Question 3 Tier 1: Standard 3 Marks

Determine without graphing whether the system has a unique solution, no solution, or infinitely many solutions: $$\begin{cases} 2x - 3y = 7 \\ 6x - 9y = 15 \end{cases}$$

View Full Method Markscheme
[M1] Multiply equation 1 by $3$: $6x - 9y = 21$.
[M1] Compare with equation 2: $6x - 9y = 15$. Left sides are identical ($m_1 = m_2 = \frac{2}{3}$), but right constants differ ($21 \ne 15$).
[A1] Subtracting gives $0 = 6$, an impossible contradiction. The lines are parallel, so there is $\mathbf{\text{No solution}}$.
Question 4 Tier 2: Intermediate 4 Marks

Solve by elimination: $$\begin{cases} 3x + 4y = 7 \\ 5x - 2y = 16 \end{cases}$$

View Full Method Markscheme
[M1] Multiply the second equation by $2$ to align coefficients of $y$: $10x - 4y = 32$.
[M1] Add to the first equation: $(3x + 10x) + (4y - 4y) = 7 + 32 \implies 13x = 39$.
[A1] Solve for $x$: $x = 3$.
[A1] Back-substitute into $3x + 4y = 7$: $3(3) + 4y = 7 \implies 9 + 4y = 7 \implies 4y = -2 \implies y = -0.5$. Solution: $\mathbf{(3, -0.5)}$.
Question 5 Tier 2: Intermediate 4 Marks

Solve the non-linear system: $$\begin{cases} y = x^2 - 4x + 3 \\ y = 2x - 6 \end{cases}$$

View Full Method Markscheme
[M1] Equate expressions: $x^2 - 4x + 3 = 2x - 6$.
[M1] Rearrange to quadratic: $x^2 - 6x + 9 = 0$.
[M1] Factorise: $(x - 3)^2 = 0 \implies x = 3$ (repeated root).
[A1] Find $y$: $y = 2(3) - 6 = 0$. The line is tangent to the parabola at the single contact point $\mathbf{(3, 0)}$.
Question 6 Tier 2: Intermediate 4 Marks

A chemist needs to create $60\text{ litres}$ of a $30\%$ alcohol solution by mixing an $18\%$ solution with a $45\%$ solution. How many litres of each solution must be mixed?

View Full Method Markscheme
[M1] Let $x$ be litres of $18\%$ solution and $y$ be litres of $45\%$ solution. Total volume: $x + y = 60$.
[M1] Alcohol content equation: $0.18x + 0.45y = 0.30(60) = 18$.
[M1] Substitute $x = 60 - y$: $0.18(60 - y) + 0.45y = 18 \implies 10.8 - 0.18y + 0.45y = 18 \implies 0.27y = 7.2$.
[A1] Solve: $y = \frac{7.2}{0.27} = \mathbf{26.67\text{ L}}$ (or $26\frac{2}{3}\text{ L}$) of $45\%$, and $x = 60 - 26.67 = \mathbf{33.33\text{ L}}$ (or $33\frac{1}{3}\text{ L}$) of $18\%$.
Question 7 Tier 2: Intermediate 4 Marks

Find the value(s) of $k$ such that the line $y = 4x + k$ intersects the parabola $y = 2x^2$ at exactly one point.

View Full Method Markscheme
[M1] Equate equations: $2x^2 = 4x + k \implies 2x^2 - 4x - k = 0$.
[M1] For exactly one intersection, discriminant must equal zero: $\Delta = b^2 - 4ac = 0$.
[M1] Substitute $a = 2, b = -4, c = -k$: $(-4)^2 - 4(2)(-k) = 0 \implies 16 + 8k = 0$.
[A1] Solve: $8k = -16 \implies \mathbf{k = -2}$.
Question 8 Tier 3: Extended 5 Marks

Find all intersection points between the circle $x^2 + y^2 = 25$ and the line $x - y = 1$.

View Full Method Markscheme
[M1] Express $y$ in terms of $x$: $y = x - 1$.
[M1] Substitute into circle equation: $x^2 + (x - 1)^2 = 25$.
[M1] Expand and collect: $x^2 + x^2 - 2x + 1 = 25 \implies 2x^2 - 2x - 24 = 0 \implies x^2 - x - 12 = 0$.
[M1] Factorise: $(x - 4)(x + 3) = 0 \implies x = 4$ or $x = -3$.
[A1] Find corresponding $y$-values:
  • If $x = 4$, $y = 4 - 1 = 3 \implies \mathbf{(4, 3)}$
  • If $x = -3$, $y = -3 - 1 = -4 \implies \mathbf{(-3, -4)}$
Question 9 Tier 3: Extended 5 Marks

A rowing team travels $24\text{ km}$ downstream with the river current in $2\text{ hours}$. The return journey upstream against the current takes $6\text{ hours}$.
(a) Let $v$ be rowing speed in still water and $c$ be river current speed. Set up a system of linear equations.
(b) Solve for $v$ and $c$.

View Full Method Markscheme
[M1] Downstream speed is $(v + c)$: $\text{Distance} = \text{Speed} \times \text{Time} \implies 2(v + c) = 24 \implies v + c = 12$.
[M1] Upstream speed is $(v - c)$: $6(v - c) = 24 \implies v - c = 4$.
[M1] Add the two equations: $(v + c) + (v - c) = 12 + 4 \implies 2v = 16$.
[A1] Rowing speed in still water: $\mathbf{v = 8\text{ km/h}}$.
[A1] Current speed: $c = 12 - 8 = \mathbf{4\text{ km/h}}$.
Question 10 Tier 3: Extended 5 Marks

Solve the $3 \times 3$ linear system: $$\begin{cases} x + y + z = 6 \\ 2x - y + z = 3 \\ x + 2y - z = 2 \end{cases}$$

View Full Method Markscheme
[M1] Add equation (1) and equation (3) to eliminate $z$: $(x+y+z) + (x+2y-z) = 6 + 2 \implies 2x + 3y = 8$. (Eq 4)
[M1] Add equation (2) and equation (3) to eliminate $z$: $(2x-y+z) + (x+2y-z) = 3 + 2 \implies 3x + y = 5 \implies y = 5 - 3x$. (Eq 5)
[M1] Substitute Eq 5 into Eq 4: $2x + 3(5 - 3x) = 8 \implies 2x + 15 - 9x = 8 \implies -7x = -7 \implies \mathbf{x = 1}$.
[A1] Find $y$: $y = 5 - 3(1) = \mathbf{2}$.
[A1] Find $z$ using Eq 1: $1 + 2 + z = 6 \implies \mathbf{z = 3}$. Unique solution is $\mathbf{(1, 2, 3)}$.

Systems of Equations Formulas

  • 2x2 System Form: $a_1 x + b_1 y = c_1, \quad a_2 x + b_2 y = c_2$
  • Condition for Unique Solution: $\dfrac{a_1}{a_2} \ne \dfrac{b_1}{b_2}$
  • Condition for No Solution (Parallel): $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2}$
  • Condition for Infinitely Many Solutions: $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$
  • Line-Parabola Tangency: $\Delta = (b - m)^2 - 4a(d - c) = 0$
  • Break-Even Condition: $R(x) = C(x) \implies \text{Profit } P(x) = 0$

Top Common MYP Pitfalls

1. Solving for Only One Variable

Stopping after finding $x = 3$ and forgetting that a system solution requires the ordered pair $(x, y)$.

2. Sign Errors During Elimination Subtraction

When subtracting equations like $(2x - 3y) - (2x + 5y)$, writing $-3y - 5y = -8y$. Students often forget to distribute the negative to all terms.

3. Forgetting the Second Root in Non-Linear Systems

Line-parabola intersections typically produce two coordinate points. Never stop after finding just one branch of a quadratic.

4. Mixing Up Elimination Multiplication

Multiplying only the left side of the equation and forgetting to multiply the constant term on the right.

Interactive Mini Test: Systems of Equations

Answer the questions to see your score and feedback.

Question 1 of 5 Linear System Solution

What is the solution to $\begin{cases} 2x + y = 7 \\ x - y = 2 \end{cases}$?

Question 2 of 5 Inconsistent Systems

Which system of equations has NO solutions?

Question 3 of 5 Line-Parabola Intersection

How many times does the line $y = 3$ intersect the parabola $y = x^2 - 1$?

Question 4 of 5 Tangency Condition

If the line $y = mx$ is tangent to $y = x^2 + 9$, what are the possible values of $m$?

Question 5 of 5 Economic Break-Even Units

A boutique bakery has fixed costs of $\$600/\text{month}$ and produces cakes for $\$10$ each. If each cake sells for $\$25$, how many cakes must be sold to break even?