Solution of equations

Q: What is a linear equation and what is the primary goal when solving it?

A linear equation contains variables raised only to the power of 1 (such as plain \(x\)), and the goal is to isolate the unknown on one side of the equals sign.

Q: How should you handle brackets and fractions when solving linear equations?

Expand brackets by multiplying the term outside by everything inside, and clear fractions by multiplying every single term in the equation by the denominator.

Q: What technique is used to make a variable the subject of a formula if it appears in two different terms?

Collect all terms containing the desired variable on one side of the equation and then factorise it out.

Q: What are the three main algebraic methods for solving quadratic equations?

The three methods are factorising into brackets, applying the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), and completing the square.

Q: How do you solve an equation that is a quadratic in a function of an unknown, such as \(x^4 - 5x^2 + 4 = 0\)?

Introduce a substitution (such as \(u = x^2\)) to transform the equation into a standard quadratic in \(u\), solve for \(u\), and then solve back for the original variable \(x\).

Q: What does the discriminant \(b^2 - 4ac\) indicate about the roots of a quadratic equation?

It determines root nature: if \(b^2 - 4ac > 0\), there are two distinct real roots; if \(b^2 - 4ac = 0\), one repeated root; and if \(b^2 - 4ac

Q: When is the substitution method preferred over elimination for simultaneous equations?

Substitution is preferred when at least one of the equations is non-linear (such as a quadratic).

Q: What do the geometric points of intersection of two graphs represent algebraically?

The coordinates of the points of intersection represent the simultaneous solutions to the equations of the two graphs.

Mathematics B (MEI) - H630 · Cambridge OCR AS Level · 6 min read · Updated: 12 Oct 2026

Welcome to the World of Equations!

In this chapter, we are going to learn how to find "hidden" numbers. Think of an equation like a balanced scale. Whatever we do to one side, we must do to the other to keep it level. Solving equations is just like being a detective—we are looking for the value of the unknown (usually \(x\)) that makes the statement true. Whether you are comfortable with algebra or find it a bit intimidating, these notes will guide you through the process step-by-step.

1. Solving Linear Equations

A linear equation is one where the unknown (like \(x\)) isn't squared or cubed—it’s just a plain old \(x\). Our goal is to "isolate" \(x\), which means getting it all by itself on one side of the equals sign.

Handling Brackets and Fractions

Sometimes equations look messy because of brackets or fractions. Don't worry! We just need to tidy them up first.

Brackets: Multiply the term outside the bracket by everything inside.
Fractions: The easiest way to deal with a fraction is to multiply every term in the equation by the number on the bottom (the denominator). This "clears" the fraction.

Example: Unknowns on Both Sides

Solve: \(3(x + 2) = 10 - x\)

Step 1 (Expand): \(3x + 6 = 10 - x\)
Step 2 (Collect \(x\) terms): Add \(x\) to both sides: \(4x + 6 = 10\)
Step 3 (Isolate): Subtract 6 from both sides: \(4x = 4\)
Step 4 (Solve): Divide by 4: \(x = 1\)

Common Mistake: When multiplying to clear a fraction, students often forget to multiply the terms that aren't fractions. Make sure you multiply every single term!

Key Takeaway: Always expand brackets and clear fractions first. Then, move all your \(x\) terms to one side and your numbers to the other.

2. Rearranging Formulas (Changing the Subject)

Rearranging a formula is exactly like solving an equation, but instead of getting a number for an answer, you get a new formula. We call the letter we want to isolate the subject.

The "Factorising" Trick

A common challenge is when the letter you want (the new subject) appears in two different places. To fix this, you need to get both terms onto one side and factorise.

Example: Make \(x\) the subject of \(y = \frac{x+a}{x-b}\)

Multiply by \((x-b)\): \(y(x-b) = x + a\)
Expand: \(yx - yb = x + a\)
Move all \(x\) terms to one side: \(yx - x = a + yb\)
Factorise \(x\) out: \(x(y - 1) = a + yb\)
Divide: \(x = \frac{a + yb}{y - 1}\)

Quick Review: To "undo" a square (\(x^2\)), use a square root (\(\sqrt{x}\)). To "undo" a reciprocal (\(\frac{1}{x}\)), multiply by \(x\).

Key Takeaway: If the new subject appears twice, move those terms to one side and factorise them out.

3. Solving Quadratic Equations

A quadratic equation contains an \(x^2\) term. It usually looks like \(ax^2 + bx + c = 0\). Unlike linear equations, quadratics can have two solutions!

Methods to Solve:

Factorising: Putting the equation into two brackets, e.g., \((x + 2)(x - 3) = 0\). This works best when the numbers are simple.
The Quadratic Formula: The "safety net" that works for every quadratic.
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Completing the Square: Rewriting the equation in the form \((x + p)^2 + q = 0\). This is very useful for finding the turning point of a graph!

Quadratic in a Function of the Unknown (Disguised Quadratics)

Some equations do not look like standard quadratics at first, but can be rewritten as one because the variable appears as a function of the unknown, such as \(x^2\), \(\sqrt{x}\), or \(e^x\).

Example: Solve \(x^4 - 5x^2 + 4 = 0\)

Notice that \(x^4 = (x^2)^2\). Let \(u = x^2\).
Substitute \(u\) to form a quadratic: \(u^2 - 5u + 4 = 0\).
Factorise: \((u - 1)(u - 4) = 0\), giving \(u = 1\) or \(u = 4\).
Substitute back to find \(x\): \(x^2 = 1 \implies x = \pm 1\), and \(x^2 = 4 \implies x = \pm 2\).

Did you know? The \(\pm\) symbol in the formula is why you get two different answers—one where you add the square root, and one where you subtract it.

Key Takeaway: If you can't see how to factorise a quadratic within 30 seconds, just use the Quadratic Formula to save time! For disguised quadratics, substitute \(u = f(x)\) to reduce the equation to a standard quadratic.

4. The Discriminant: The "Root Detector"

The discriminant is the part of the quadratic formula inside the square root: \(b^2 - 4ac\). It tells us how many solutions (or "roots") a quadratic has without us having to solve it.

The Rules:

If \(b^2 - 4ac > 0\): There are 2 distinct real roots (the graph crosses the x-axis twice).
If \(b^2 - 4ac = 0\): There is 1 repeated root (the graph just touches the x-axis).
If \(b^2 - 4ac < 0\): There are no real roots (the graph never touches the x-axis).

Analogy: Think of the discriminant as a metal detector. It tells you if there's "treasure" (roots) buried under the axis before you start digging!

Key Takeaway: Use the discriminant when a question asks about the number of roots or mentions "distinct," "repeated," or "no real" roots.

5. Simultaneous Equations

This is when you have two equations and two unknowns. You are looking for the pair of values (\(x\) and \(y\)) that work in both equations at the same time.

Method A: Substitution (The "Replacement")

Rearrange one equation to get "\(x = ...\)" or "\(y = ...\)" and plug it into the other equation. This is the best method when one equation is a quadratic.

Method B: Elimination (The "Removal")

Multiply the equations so that the coefficients of \(x\) (or \(y\)) are the same, then add or subtract the equations to "cancel out" that letter. This is usually fastest for two linear equations.

Don't worry if this seems tricky: When solving a linear and a quadratic equation simultaneously, you will usually end up with a new quadratic equation to solve. Just follow the steps and use your "safety net" formula if needed!

Key Takeaway: Substitution is your "all-rounder" tool—it works for almost any pair of equations.

6. The Power of Graphs

Sometimes we forget that equations are just descriptions of lines on a graph. The solution to an equation (or a set of simultaneous equations) is simply the point of intersection—the exact place where the lines cross.

If you graph \(y = f(x)\) and \(y = g(x)\), the \(x\)-values where they cross are the solutions to \(f(x) = g(x)\).
If a line and a curve don't cross, the discriminant of their combined equation will be negative!

Quick Review Box:
- Linear: \(x\) only. Balance the scales.
- Quadratic: \(x^2\). Use factorising, the formula, or a substitution if it is a quadratic in a function of \(x\).
- Discriminant: \(b^2 - 4ac\). Tells you the number of roots.
- Simultaneous: Two equations. Use substitution or elimination.
- Intersections: Where graphs cross = the solution.

References

AS Level GCE Mathematics B (MEI) Specification (H630) — OCR

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is a linear equation and what is the primary goal when solving it?

A linear equation contains variables raised only to the power of 1 (such as plain \(x\)), and the goal is to isolate the unknown on one side of the equals sign.

How should you handle brackets and fractions when solving linear equations?

Expand brackets by multiplying the term outside by everything inside, and clear fractions by multiplying every single term in the equation by the denominator.

What technique is used to make a variable the subject of a formula if it appears in two different terms?

Collect all terms containing the desired variable on one side of the equation and then factorise it out.

What are the three main algebraic methods for solving quadratic equations?

The three methods are factorising into brackets, applying the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), and completing the square.

How do you solve an equation that is a quadratic in a function of an unknown, such as \(x^4 - 5x^2 + 4 = 0\)?

Introduce a substitution (such as \(u = x^2\)) to transform the equation into a standard quadratic in \(u\), solve for \(u\), and then solve back for the original variable \(x\).

What does the discriminant \(b^2 - 4ac\) indicate about the roots of a quadratic equation?

It determines root nature: if \(b^2 - 4ac > 0\), there are two distinct real roots; if \(b^2 - 4ac = 0\), one repeated root; and if \(b^2 - 4ac < 0\), no real roots.

When is the substitution method preferred over elimination for simultaneous equations?

Substitution is preferred when at least one of the equations is non-linear (such as a quadratic).

What do the geometric points of intersection of two graphs represent algebraically?

The coordinates of the points of intersection represent the simultaneous solutions to the equations of the two graphs.

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