x² + y² = 29
y - x = 3
When you encounter simultaneous equations in your GCSE maths exam, you'll typically see two types: two linear equations, or one linear and one quadratic equation. Today, we're going to focus on the second type, using a common exam question as our example.
The Question
Let's solve the simultaneous equations:
$$x² + y² = 29$$
$$y - x = 3$$
Understanding the Equations
Before we start solving, let's understand what we're working with:
$x² + y² = 29$ is a quadratic equation that represents a circle with centre at the origin (0,0) and radius $√29$
- $y - x = 3$ is a linear equation that represents a straight line
When we solve these simultaneously, we're finding the points where the line intersects the circle.
Step-by-Step Solution
Step 1: Rearrange the Linear Equation
The linear equation is easier to work with, so let's rearrange it to make $y$ the subject:
$$y - x = 3$$
$$y = x + 3$$
Step 2: Substitute into the Quadratic Equation
Now we substitute $y = x + 3$ into the quadratic equation $x² + y² = 29$:
$$x² + (x + 3)² = 29$$
Step 3: Expand and Simplify
Let's expand $(x + 3)²$:
$$(x + 3)² = (x + 3)(x + 3) = x² + 3x + 3x + 9 = x² + 6x + 9$$
Now substitute this back into our equation:
$$x² + (x² + 6x + 9) = 29$$
$$2x² + 6x + 9 = 29$$
Step 4: Rearrange to Form a Quadratic Equation
Subtract 29 from both sides to set the equation equal to zero:
$$2x² + 6x + 9 - 29 = 0$$
$$2x² + 6x - 20 = 0$$
We can simplify this by dividing all terms by 2:
$$x² + 3x - 10 = 0$$
Step 5: Solve the Quadratic Equation
We now have a quadratic equation in the form $ax² + bx + c = 0$. We can solve this by factorising:
$$x² + 3x - 10 = 0$$
We need two numbers that multiply to give -10 and add to give 3. These numbers are 5 and -2:
$$(x + 5)(x - 2) = 0$$
This gives us two possible solutions for $x$:
$$x + 5 = 0 \quad or \quad x - 2 = 0$$
$$x = -5 \quad or \quad x = 2$$
Step 6: Find the Corresponding y-values
For each $x$-value, we use the linear equation $y = x + 3$ to find the corresponding $y$-value:
When $x = -5$:
$$y = -5 + 3 = -2$$
When $x = 2$:
$$y = 2 + 3 = 5$$
Step 7: Write the Complete Solutions
We have two solution pairs:
$$x = -5, \quad y = -2$$
$$x = 2, \quad y = 5$$
We can write these as coordinate pairs: $(-5, -2)$ and $(2, 5)$.
Checking Our Solutions
It's always good practice to check your solutions in the original equations:
Check $(-5, -2)$:
$x² + y² = (-5)² + (-2)² = 25 + 4 = 29$ ✓
- $y - x = -2 - (-5) = -2 + 5 = 3$ ✓
Check $(2, 5)$:
$x² + y² = 2² + 5² = 4 + 25 = 29$ ✓
- $y - x = 5 - 2 = 3$ ✓
Both solutions satisfy both original equations.
Visualising the Solution
flowchart TD
A["Start: x² + y² = 29 and y - x = 3"] --> B["Rearrange linear equation: y = x + 3"]
B --> C["Substitute into quadratic: x² + (x+3)² = 29"]
C --> D["Expand and simplify: 2x² + 6x - 20 = 0"]
D --> E["Divide by 2: x² + 3x - 10 = 0"]
E --> F["Factorise: (x+5)(x-2) = 0"]
F --> G["Solve for x: x = -5 or x = 2"]
G --> H["Find y for each x: y = x + 3"]
H --> I["Solutions: (-5,-2) and (2,5)"]
I --> J["Check solutions in original equations"]
Common Mistakes to Avoid
Forgetting to find both $x$-values: Quadratic equations often give two solutions.
Not finding the corresponding $y$-values: Each $x$-value needs its matching $y$-value.
Making errors when expanding brackets: Remember $(x + 3)² = x² + 6x + 9$, not $x² + 9$.
- Not checking your solutions: Always substitute back into the original equations to verify.
Alternative Methods
While we used substitution and factorisation in this example, you could also:
Use the quadratic formula if the quadratic doesn't factorise easily
- Rearrange the linear equation differently (e.g., $x = y - 3$ and substitute)
Exam Tips
Show all your working - even if you make a calculation error, you can earn method marks.
Write your final answer clearly - either as coordinate pairs or with $x$ and $y$ values clearly labelled.
Remember the context - these solutions represent intersection points between a circle and a line.
- Time management - this type of question typically takes 5-7 minutes in an exam.
Practice Questions
Try these similar questions to test your understanding:
Solve: $x² + y² = 25$ and $y = x + 1$
Solve: $x² + y² = 13$ and $y = x - 1$
- Solve: $x² + y² = 10$ and $y = 2x$
Summary
Solving quadratic simultaneous equations involves:
Rearranging the linear equation
Substituting into the quadratic equation
Solving the resulting quadratic equation
Finding the corresponding $y$-values
- Checking your solutions
This method works for any pair of simultaneous equations where one is linear and one is quadratic. With practice, you'll be able to solve these quickly and accurately in your GCSE maths exam.
- Good luck with your revision!