# Simultaneous Equations, Inequalities and Regions

> Linear inequalities, simultaneous equations by substitution and elimination, word problems, and regions on the number plane.

From HSC Maths Advanced on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-advanced/simultaneous-equations/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The two phone plans
Maya is choosing between two phone plans. Plan A costs 20 dollars a month plus 10 cents per minute. Plan B costs 35 dollars a month with free calls. Which is cheaper depends on how much she talks.

Each plan is a straight line on a graph of cost against minutes. Where the lines cross, the plans cost the same. Finding that crossing point means solving two equations **at the same time**, which is what "simultaneous" means. Shading the part of the graph where Plan A is cheaper is a **region**.

**In short:** Simultaneous equations are two equations that are both true at once, and their solution is the point where their graphs cross. Solve them by substitution or elimination. Solve inequalities like equations, but flip the sign when you multiply or divide by a negative number.

## Starting point: Years 9 and 10

### Linear inequalities

An inequality uses $<$, $>$, $\leq$ or $\geq$. Solve it like an equation, with one extra rule:

### Key idea: Flip the sign
When you multiply or divide both sides by a **negative** number, reverse the inequality sign.

$$3 - 2x > 9 \;\Rightarrow\; -2x > 6 \;\Rightarrow\; x < -3$$

On a number line, an **open circle** means the end is not included ($<$ or $>$). A **closed circle** means it is included ($\leq$ or $\geq$).

### Simultaneous equations

Two equations with two unknowns. The solution is the point that makes **both** true, which is where their graphs cross.

**Substitution** works well when one equation already says $y = \dots$ or $x = \dots$.

### Worked example: Substitution
Solve $y = 2x - 1$ and $3x + y = 14$.

Replace $y$ in the second: $3x + (2x - 1) = 14$, so $5x = 15$ and $x = 3$. Then $y = 2(3) - 1 = 5$. The solution is $(3, 5)$.

**Elimination** works well when the equations line up. Add or subtract them to remove one letter. You may need to multiply first.

### Worked example: Elimination
Solve $2x + 3y = 13$ and $5x - 2y = 4$.

Multiply the first by 2 and the second by 3: $4x + 6y = 26$ and $15x - 6y = 12$. Add: $19x = 38$, so $x = 2$. Then $4 + 3y = 13$, so $y = 3$.

### Calculator tip: Check, do not just solve
Some calculators (such as the Casio fx-100AU) have an equation mode that solves two linear equations at once. Use it to check, not instead of working: the marks are for the method.

Any calculator can check an answer. Store $x$ in `A` and $y$ in `B` with `STO`, then type the left side of each equation using `A` and `B`. Both must give the right-hand side.

### Word problems

1. Name the two unknowns.
2. Write two equations from the facts.
3. Solve and answer in words.

### Worked example: Maya's plans
Let $m$ be minutes. Plan A: $C = 20 + 0.1m$. Plan B: $C = 35$. Equal when $20 + 0.1m = 35$, so $m = 150$. Under 150 minutes, Plan A is cheaper.

### Regions on the number plane

The line $y = x + 1$ splits the plane into two **half-planes**. $y > x + 1$ is everything above the line, and $y < x + 1$ is everything below it.

- Draw the boundary **dashed** for $<$ or $>$ (not included).
- Draw it **solid** for $\leq$ or $\geq$ (included).
- Test a point such as $(0, 0)$ to see which side to shade.

*Diagram: The lines y equals x plus 1, dashed, and x plus y equals 5, solid, meeting at 2, 3, with the region above the first and below the second shaded*
*The shaded part has y > x + 1 (dashed edge, not included) and x + y ≤ 5 (solid edge, included).*

### Watch out: Divide by a negative
$-x < 4$ gives $x > -4$, not $x < -4$.

### Exam tip: Check in both equations
Substitute your answer into **both** original equations. If one fails, there is an arithmetic slip.

**Mind map**

```text
Simultaneous and regions
  Inequalities
    Flip sign for negatives
    Open and closed circles
  Simultaneous
    Where graphs cross
    Substitution
    Elimination
  Word problems
    Two unknowns, two facts
  Regions
    Dashed for < and >
    Solid for ≤ and ≥
    Test a point
  Line and parabola
    0, 1 or 2 meeting points
```

### Year 11: Lines meeting parabolas
To find where $y = x^2$ meets $y = x + 6$, substitute: $x^2 = x + 6$, so $(x - 3)(x + 2) = 0$. They meet at $(3, 9)$ and $(-2, 4)$.

After substituting, a quadratic $ax^2 + bx + c = 0$ appears. Its discriminant counts the meeting points:
- $\Delta > 0$: the line cuts the parabola twice.
- $\Delta = 0$: the line is a **tangent** (touches once).
- $\Delta < 0$: they never meet.

**Example.** For which $k$ is $y = 2x + k$ a tangent to $y = x^2$? $x^2 - 2x - k = 0$ needs $4 + 4k = 0$, so $k = -1$.

**Regions with curves.** $y \geq x^2$ is everything inside (above) the parabola. $x^2 + y^2 < 9$ is the inside of a circle of radius 3, drawn dashed.

**A line and a product.** Solve $x + y = 6$ and $xy = 8$. From the first, $y = 6 - x$. Then $x(6 - x) = 8$, so $x^2 - 6x + 8 = 0$, giving $x = 2$ or $x = 4$. The solutions are $(2, 4)$ and $(4, 2)$.

**Tangents to a circle.** Substitute the line into the circle and set $\Delta = 0$. For $y = mx + 5$ and $x^2 + y^2 = 5$: $x^2 + (mx + 5)^2 = 5$ gives $(1 + m^2)x^2 + 10mx + 20 = 0$. Then $\Delta = 100m^2 - 80(1 + m^2) = 20m^2 - 80 = 0$, so $m = \pm 2$.

**Lines through a fixed point.** Every line through $(0, -4)$ is $y = mx - 4$. It meets $y = x^2$ where $x^2 - mx + 4 = 0$. Two meeting points need $\Delta = m^2 - 16 > 0$, so $m < -4$ or $m > 4$. Set $\Delta = 0$ instead to find the gradients of the two tangents from that point: $m = \pm 4$.

**Two parabolas.** $y = kx^2$ and $y = x^2 + 2$ meet where $(k - 1)x^2 = 2$. That needs $k - 1 > 0$, so they never meet when $k \leq 1$.

### Year 12: Regions and integration
The area of a region bounded by two curves is found by integration: top minus bottom, between their meeting points. The simultaneous equations step is always first, because it gives the limits of the integral.

The region where $x + 2 \geq y \geq x^2$ lies between $x = -1$ and $x = 2$, and its area is $\displaystyle\int_{-1}^{2}(x + 2 - x^2)\,dx = 4.5$.

## Quick summary

- Solve inequalities like equations, but flip the sign when multiplying or dividing by a negative.
- Simultaneous equations: the point that makes both true, where the graphs cross.
- Substitution when one letter is already alone. Elimination when the equations line up.
- Regions: dashed boundary for $<$ and $>$, solid for $\leq$ and $\geq$. Test a point to choose the side.
- A line and a parabola meet 0, 1 or 2 times. Use the discriminant.

## Quiz

Try these questions from the chapter quiz.

1. Solve $4x - 3 \leq 9$.
   - A. $x \geq 3$
   - B. $x \leq 3$
   - C. $x \leq 1.5$
   - D. $x < 3$
   - Answer: B. $x \leq 3$. $4x \leq 12$.

2. Solve $-3x > 12$.
   - A. $x > -4$
   - B. $x < -4$
   - C. $x > 4$
   - D. $x < 4$
   - Answer: B. $x < -4$. Dividing by $-3$ flips the sign.

3. On a number line, $x \geq 2$ is shown with:
   - A. a closed circle at 2 and an arrow to the right
   - B. an open circle at 2 and an arrow to the right
   - C. a closed circle at 2 and an arrow to the left
   - D. an open circle at 2 and an arrow to the left
   - Answer: A. a closed circle at 2 and an arrow to the right. Closed because 2 is included.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-advanced/simultaneous-equations/#quiz