Chapter 9: Simultaneous Linear Equations - The Great Detective Challenge!

Hello future Maths detectives! Welcome to one of the most useful topics in algebra: Simultaneous Linear Equations. Doesn't that sound complicated? Don't worry, it's just a fancy way of saying "solving a mystery with two clues".

In this chapter, you'll learn how to take two related problems (our 'clues') with two unknown values (our 'suspects', which we'll call x and y) and find the one and only solution that makes both clues true! This is a super-power you can use to solve real-life puzzles, like figuring out the price of a movie ticket and a popcorn, or planning a budget.

Ready to put on your detective hat? Let's get started!


What Are We Dealing With? A Quick Recap

Before we jump in, let's remember what a linear equation in two unknowns is. It's an equation with two variables (like x and y) where the variables aren't squared or cubed or anything tricky. When you draw its graph, you get a perfect straight line.

Example: $$y = 2x + 1$$ is a linear equation. Every point (x, y) that sits on the line of this graph is a solution to this one equation.

A "simultaneous" equation is when we have a pair of these linear equations, and we want to find the one special pair of (x, y) values that is a solution for BOTH lines at the same time. This special point is where the two lines cross!


1. Solving with Pictures: The Graphical Method

The most visual way to solve simultaneous equations is to simply draw their graphs and see where they meet. The point of intersection is your answer!

How to do it, step-by-step:

Let's solve this system of equations:
Equation 1: $$y = x + 2$$
Equation 2: $$y = -x + 4$$

Step 1: Draw the line for the first equation.
Find a few points for $$y = x + 2$$. A good way is to create a small table.
If x = 0, y = 0 + 2 = 2. So, one point is (0, 2).
If x = 1, y = 1 + 2 = 3. So, another point is (1, 3).
Plot these points on a graph and draw a straight line through them.

Step 2: Draw the line for the second equation ON THE SAME GRAPH.
Now find a few points for $$y = -x + 4$$.
If x = 0, y = -0 + 4 = 4. So, one point is (0, 4).
If x = 2, y = -2 + 4 = 2. So, another point is (2, 2).
Plot these points and draw the second straight line.

Step 3: Find the point of intersection.
Look at your graph. The two lines should cross at a single point. In this case, they cross at the coordinates (1, 3).

Step 4: State your solution.
The solution is $$x = 1, y = 3$$. This is the only pair of values that makes both equations true!

Important Note:

The graphical method is great for understanding the concept, but it can be tricky to get an exact answer if you don't draw perfectly, or if the answer isn't a nice whole number. That's why we need other, more precise methods!

Key Takeaway:

The graphical solution to a pair of simultaneous linear equations is the coordinate point (x, y) where their two lines intersect.


2. Solving with Algebra: The Substitution Method

This is our first algebraic method, and it's like being a clever substitute teacher. You take what one variable is "equal to" from one equation and substitute it into the other equation.

Analogy: Swap It Out!

Imagine you know that 1 apple = 2 bananas. If you have a basket with 1 apple and 3 bananas, you can "substitute" the apple for 2 bananas. Now your basket just has 2 bananas + 3 bananas = 5 bananas. You've gotten rid of the apples! We do the same thing with x and y.

How to do it, step-by-step:

Let's solve this system:
Equation 1: $$y = x + 1$$
Equation 2: $$2x + y = 7$$

Step 1: Find an easy variable to isolate.
Look at Equation 1. It's already telling us exactly what 'y' is: $$y = x + 1$$. Perfect!

Step 2: Substitute this expression into the *other* equation.
In Equation 2, wherever you see 'y', replace it with '(x + 1)'.
$$2x + (x + 1) = 7$$

Step 3: Solve the new equation.
You now have an equation with only 'x'!
$$2x + x + 1 = 7$$
$$3x + 1 = 7$$
$$3x = 6$$
$$x = 2$$
Great! You've found the first part of the solution.

Step 4: Find the value of the other variable.
Take your answer, $$x=2$$, and plug it back into the easiest of the original equations to find 'y'. Equation 1 looks easy!
$$y = x + 1$$
$$y = 2 + 1$$
$$y = 3$$

Step 5: Check your answer!
Your solution is $$x = 2, y = 3$$. Does this work in BOTH original equations?
Check in Eq 1: Is 3 = 2 + 1? Yes!
Check in Eq 2: Is 2(2) + 3 = 7? Is 4 + 3 = 7? Yes!
It works! We found the correct solution.

Common Mistake to Avoid:

Always substitute into the equation you didn't use in Step 1. If you substitute back into the same equation, you'll just end up with something like 1 = 1, which doesn't help you find the answer!

Key Takeaway:

The Substitution Method involves isolating one variable in one equation and plugging its value into the second equation to solve.


3. Solving with Algebra: The Elimination Method

This method is fantastic when the equations are neatly lined up. The goal is to add or subtract the two equations from each other in a way that makes one of the variables disappear or get "eliminated".

Analogy: Team Up and Cancel Out!

Imagine you have two teams in a tug-of-war. Team A has 3 big players and 2 small players. Team B has 3 big players and 1 small player. If you pit them against each other, the 3 big players on each side cancel each other out, and you're just left with the small players. We do the same to cancel out x or y.

How to do it, step-by-step:

Let's solve this system:
Equation 1: $$2x + 3y = 8$$
Equation 2: $$4x - 3y = 4$$

Step 1: Line up the equations.
Make sure the x's, y's, and equals signs are all stacked neatly above each other. (They already are in our example).

Step 2: Look for variables that can be eliminated.
Notice the 'y' terms. We have a +3y in the first equation and a -3y in the second. If we add the two equations together, they will cancel out! (+3 - 3 = 0).

Step 3: Add (or subtract) the equations.
Let's add Equation 1 and Equation 2 together, term by term.
$$(2x + 4x) + (3y - 3y) = (8 + 4)$$
$$6x + 0y = 12$$
$$6x = 12$$

Step 4: Solve for the remaining variable.
$$6x = 12$$
$$x = 2$$
Easy!

Step 5: Find the value of the other variable.
Take your answer, $$x=2$$, and plug it back into either original equation. Let's use Equation 1.
$$2x + 3y = 8$$
$$2(2) + 3y = 8$$
$$4 + 3y = 8$$
$$3y = 4$$
$$y = 4/3$$

Step 6: State and check your solution.
The solution is $$x=2, y=4/3$$. Let's check it in Equation 2 to be sure.
Is 4(2) - 3(4/3) = 4? Is 8 - 4 = 4? Yes!
It works!

What if nothing cancels out immediately?

Sometimes you need to multiply one or both equations first to make the numbers in front of a variable (the coefficients) match.
Example: For $$x + 2y = 5$$ and $$3x + y = 5$$, you could multiply the second equation by 2. This gives you $$6x + 2y = 10$$. Now you have a '+2y' in both equations, so you can subtract them to eliminate y!

Memory Aid:

When the coefficients you want to eliminate have the...
Same Sign, you Subtract.
Different Signs, you Add.

Key Takeaway:

The Elimination Method involves adding or subtracting equations to make one variable disappear, allowing you to solve for the other.


4. When Things Get Weird: Special Cases

Sometimes, things don't work out perfectly. Just like in a mystery, sometimes there's no single culprit, or everyone's a culprit! There are two special cases you need to know about.

Case 1: No Solution

Graphically: This happens when the two lines are parallel. Since they never, ever cross, there is no point of intersection and therefore no solution.
Algebraically: When you try to solve using substitution or elimination, both variables will disappear, and you'll be left with a false statement.
Example: Solving might lead you to something like $$0 = 5$$. This is impossible! It means there is no solution.

Case 2: Infinite Solutions

Graphically: This happens when both equations actually describe the exact same line. Since the lines are on top of each other, they "intersect" at every single point along the line. This means there are an infinite number of solutions.
Algebraically: When you try to solve, both variables will disappear, and you'll be left with a true statement.
Example: Solving might lead you to $$7 = 7$$ or $$0 = 0$$. This is always true! It means any point on the line is a solution.

Quick Review Box

- One Solution: Lines intersect at one point. (Normal case)
- No Solution: Lines are parallel. (Algebra gives a false statement, e.g., $$0=5$$)
- Infinite Solutions: The lines are identical. (Algebra gives a true statement, e.g., $$7=7$$)


5. Maths in the Real World: Solving Word Problems

This is where your detective skills really shine! The key is to turn a story into two simple equations.

How to tackle word problems, step-by-step:

Problem: At the cinema, 2 adult tickets and 3 child tickets cost $39. 1 adult ticket and 2 child tickets cost $22. What is the price of one adult ticket and one child ticket?

Step 1: Define your variables.
This is the MOST important step! Tell the world what 'x' and 'y' mean.
Let a = the cost of one adult ticket.
Let c = the cost of one child ticket.

Step 2: Translate the sentences into two equations.
"2 adult tickets and 3 child tickets cost $39" becomes: $$2a + 3c = 39$$
"1 adult ticket and 2 child tickets cost $22" becomes: $$a + 2c = 22$$

Step 3: Solve the system of equations.
Let's use substitution. From the second equation, we can easily say $$a = 22 - 2c$$.
Now substitute this into the first equation:
$$2(22 - 2c) + 3c = 39$$
$$44 - 4c + 3c = 39$$
$$44 - c = 39$$
$$5 = c$$
So, a child ticket costs $5! Now find 'a':
$$a = 22 - 2c$$
$$a = 22 - 2(5)$$
$$a = 22 - 10$$
$$a = 12$$
An adult ticket costs $12!

Step 4: Answer the question clearly in a sentence.
An adult ticket costs $12 and a child ticket costs $5.

Key Takeaway:

To solve word problems, first define your variables, then translate the problem's information into two equations, and solve!