Simultaneous Equations and Graphs (Edexcel GCSE Maths): Revision Notes

Worked Example: Two linear equations

When solving linear simultaneous equations like $y = 2x + 3$ and $y = 6 - 4x$, you'll be working with two straight lines that intersect at exactly one point.

Step 1: Plot both lines on the same coordinate grid

• The first equation, $y = 2x + 3$, represents a line with gradient 2 and y-intercept 3
• The second equation, $y = 6 - 4x$, can be rewritten as $y = -4x + 6$, showing a line with gradient -4 and y-intercept 6

Step 2: Find the intersection point When you draw both lines carefully, you'll see they cross at the point where $x = \frac{1}{2}$ and $y = 4$.

Solution: $x = \frac{1}{2}$, $y = 4$

Worked Example: Linear-quadratic systems

Sometimes you'll encounter a system where one equation is linear and the other is quadratic. For instance, when solving $x² + y² = 16$ and $y = 2x + 1$, you're working with a circle and a straight line.

Step 1: Identify the shapes

• The equation $x² + y² = 16$ represents a circle with its centre at the origin $(0, 0)$ and radius 4
• The equation $y = 2x + 1$ represents a straight line with gradient 2 and y-intercept 1

Step 2: Plot and find intersections When you plot both graphs, the line intersects the circle at two points.

Solutions: $x = 1.4$, $y = 3.8$ and $x = -2.2$, $y = -3.4$ (rounded to 1 decimal place)

Worked Example: Working with quadratic equations

Consider the situation where you have the graph of $y = x² - 4x + 3$ and need to solve $x² - 5x + 3 = 0$.

Step 1: Rearrange the equation Starting with $x² - 5x + 3 = 0$, you can add $x$ to both sides to get $x² - 4x + 3 = x$.

Step 2: Identify what to draw This means you're looking for points where the parabola $y = x² - 4x + 3$ equals the line $y = x$.

Step 3: Find the solution By drawing the line $y = x$ on the same graph as the given parabola, the intersection points will give you the solutions to the original equation $x² - 5x + 3 = 0$.