Getting Started with Ratio Algebra

Think of ratio algebra as solving puzzles where the missing piece is always 'x'. These problems look intimidating at first, but they follow a simple pattern every time.

The secret is recognising that ratios are just another way of writing fractions. Once you see that connection, you're halfway to the solution!

Key Insight: Every ratio can be written as a fraction, and fractions are much easier to work with when solving for x.

Basic Ratio Problems (Type 1)

When you see something like $2x+1$ : $3x-5$ = 2 : 5, you're looking at a Type 1 problem. These always give you one ratio equal to another ratio.

Your first move is converting this to fractions: $2x+1$/$3x-5$ = 2/5. Now you can cross-multiply to get rid of those annoying fractions.

Cross-multiplying gives you: 5$2x+1$ = 2$3x-5$. From here, it's just expanding brackets, collecting like terms, and solving for x like any other equation.

Pro Tip: Always check your answer by substituting it back into the original ratio - it's the quickest way to spot mistakes!

Worked Examples and Practice

Let's walk through that example: $2x+1$ : $3x-5$ = 2 : 5. After cross-multiplying, you get 10x + 5 = 6x - 10.

Collecting like terms: 4x = -15, so x = -3.75. See how straightforward it becomes once you've got rid of the ratio format?

The practice problems follow exactly the same method. Whether it's $5x-2$ : $2x+3$ = 3 : 5 or any other combination, the approach never changes.

Remember: Cross-multiply, expand, collect like terms, solve - that's your four-step process every time.

Challenge Questions and Exam Prep

Now we're getting into the good stuff - questions that might actually appear on your exams. These problems test whether you really understand the method or you're just following steps blindly.

Look at question 5: $y + x$ : $y - x$ = k : 1. This one's asking you to rearrange for y, not solve for a number. It's the same technique, but you need to be comfortable with algebra.

Question 6 is particularly sneaky: $x + 2$ : $x + 4$ = $x + 6$ : $x + 9$. Both sides have expressions in x, so you'll need to expand brackets carefully and watch your algebra closely.

Exam Strategy: These harder questions often give you expressions on both sides - don't panic, just cross-multiply as usual.

Solutions and Working

Here's where you can check your working and see the complete solutions. Notice how question 5 shows the full algebraic manipulation - this is what examiners want to see.

The key step in question 5 is recognising that y$1-k$ = -x$1+k$ can be rearranged to give the required form. Don't worry if this feels tricky - it's testing A-level style algebra.

Question 6 demonstrates why checking is so important. After cross-multiplying and expanding, you get x² + 11x + 18 = x² + 10x + 24, which simplifies beautifully to x = 6.

Study Tip: Cover up the answers and try working through these yourself - you'll learn much more than just reading the solutions.

Introduction to Type 2 Problems

Right, now we're stepping up a gear. Type 2 ratio problems are where things get properly interesting because you'll end up with quadratic equations.

The setup looks similar to Type 1, but instead of getting a simple linear equation, your algebra leads to something like x² - x - 6 = 0.

Don't let this put you off - the initial steps are identical. It's only at the end that you need to factorise or use the quadratic formula.

Heads Up: Type 2 problems usually give you two possible values for x, not just one.

Setting Up Type 2 Problems

Look at this example: $3x + 5$ : $x + 4$ = $2x + 4$ : $x + 2$. Notice how both sides have different expressions? That's your clue this is Type 2.

You still start the same way: convert to fractions and cross-multiply. But now you're dealing with $x + 2$$3x + 5$ = $2x + 4$$x + 4$.

When you expand these brackets, you'll get x² terms that don't cancel out. That's what creates the quadratic equation you'll need to solve.

Watch Out: Be extra careful when expanding brackets - small mistakes here will mess up your final answer.

Solving Type 2 Problems

Let's follow through that example completely. After expanding $x + 2$$3x + 5$ = $2x + 4$$x + 4$, you get 3x² + 11x + 10 = 2x² + 12x + 16.

Rearranging gives you x² - x - 6 = 0. Now you can factorise: $x - 3$$x + 2$ = 0, so x = 3 or x = -2.

The 'Your Turn' example works exactly the same way, giving you x = 2 or x = 4. Two solutions is completely normal for these problems.

Key Point: Always write both solutions clearly - marks get dropped for missing one of the answers.

Practice Problems

These six questions will test everything you've learned about ratio algebra. Start with question 1 - it's the most straightforward of the bunch.

Question 6 is particularly interesting because it involves x² in the ratio itself. Don't let this throw you - the method stays exactly the same.

Work through these systematically. If you get stuck, go back to the examples and check you're following the same steps.

Study Method: Try to do these without looking at the answers first - you'll learn much more from making mistakes and fixing them.