Solving Equations & Inequalities

Solving Linear Equations

A linear equation has a variable to the power $1$ only. The goal is to isolate the variable by doing the same operation to both sides until you reach x = ....

Always undo operations in reverse order (inverse of $+$ is $-$, inverse of $\times$ is $\div$).

Worked example Solve $5x - 7 = 3x + 11$.

Collect the $x$ terms on one side and numbers on the other.

$5x - 3x = 11 + 7$, so $2x = 18$, giving $x = 9$.

With brackets: expand first, then solve.

Worked example Solve $3(2x - 4) = 2(x + 6)$.

Expand: $6x - 12 = 2x + 12$.

With fractions: multiply every term by the lowest common denominator to clear the fractions.

Worked example Solve $\dfrac{x+1}{2} + \dfrac{x-2}{3} = 4$.

Watch out When multiplying through by the LCD, multiply the whole numerator. A common error is writing $3 \cdot x + 1$ instead of $3(x+1)$. Use brackets every time.

Linear Inequalities

An inequality uses $<$, $\le$, $>$ or $\ge$. Solve it almost exactly like an equation, with one crucial rule.

Key terms $x < a$: strictly less than (open circle).

$x \le a$: less than or equal to (closed/filled circle).

Multiplying or dividing both sides by a negative number reverses the inequality sign.

Worked example Solve $4 - 3x \le 19$.

Subtract $4$: $-3x \le 15$.

Divide by $-3$ and flip the sign: $x \ge -5$.

You can represent solutions on a number line. Use an open circle $\circ$ for $<$ or $>$, and a filled circle $\bullet$ for $\le$ or $\ge$. An arrow shows the solution continuing.

We also write solution sets using set notation $\{x : -2 < x \le 3\}$ or interval notation $(-2, 3]$, where a round bracket excludes the endpoint and a square bracket includes it.

Exam tip A double inequality such as $-1 \le 2x + 3 < 9$ is solved by doing the same thing to all three parts. Subtract $3$: $-4 \le 2x < 6$. Divide by $2$: $-2 \le x < 3$.

Simultaneous Equations: Two Linear

When two equations share the same two unknowns, solve them together.

Elimination — make the coefficients of one variable match, then add or subtract.

Worked example Solve $3x + 2y = 16$ and $5x - 2y = 8$.

The $y$ coefficients are $+2$ and $-2$, so add the equations to eliminate $y$:

Watch out If the matching coefficients have the same sign, subtract; if opposite signs, add. "Same Signs Subtract."

Substitution — rearrange one equation for a variable, then put it into the other.

Worked example Solve $y = 2x - 1$ and $3x + y = 14$.

Substitute $y = 2x - 1$ into the second: $3x + (2x - 1) = 14$.

Simultaneous Equations: One Linear, One Non-Linear

When one equation is a curve (often containing $x^2$, $y^2$ or $xy$), substitution is the reliable method. Rearrange the linear equation and substitute into the non-linear one. You usually get two pairs of solutions.

Worked example Solve $y = x + 1$ and $x^2 + y^2 = 25$.

Substitute $y = x + 1$:

Exam tip Keep your two answers paired. Each $x$ has its own $y$ — never mix them. Substitute back into the linear equation (it is simpler and safer).

Solving Quadratic Equations

A quadratic equation has the form $ax^2 + bx + c = 0$. There are three methods.

1. By factorising. Write the equation as a product of two brackets equal to zero, then set each bracket to zero.

Worked example Solve $x^2 - 7x + 10 = 0$.

Find two numbers multiplying to $+10$ and adding to $-7$: that is $-2$ and $-5$.

Watch out The equation must equal $0$ before factorising. If you have $x^2 = 3x$, do not divide by $x$ (you lose the solution $x = 0$). Rearrange to $x^2 - 3x = 0$, factor $x(x - 3) = 0$, giving $x = 0$ or $x = 3$.

2. By the quadratic formula. Use this when factorising is hard or impossible:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Worked example Solve $2x^2 + 3x - 4 = 0$, giving answers to $2$ decimal places.

Here $a = 2$, $b = 3$, $c = -4$.

Exam tip Put brackets around negative values of $b$ and $c$ in your calculator. The biggest formula errors come from mishandling the sign inside $b^2 - 4ac$, especially $-4ac$ when $c$ is negative.

3. By completing the square. Rewrite $x^2 + bx + c$ as $(x + \frac{b}{2})^2 - (\frac{b}{2})^2 + c$. This is useful for exact answers and for finding turning points.

Worked example Solve $x^2 + 6x - 4 = 0$ by completing the square.

Half of $6$ is $3$, so $x^2 + 6x = (x + 3)^2 - 9$.

Setting Up Equations from Words

Many marks come from forming equations yourself. Define a letter for the unknown, translate each sentence into algebra, then solve.

Worked example The length of a rectangle is $3\,\text{cm}$ more than its width. The area is $40\,\text{cm}^2$. Find the width.

Let the width be $w$ cm, so the length is $(w + 3)$ cm.

Area: $w(w + 3) = 40$, so $w^2 + 3w - 40 = 0$.

Real world Quadratics describe projectile paths, profit curves and areas. The negative root is often rejected because a length, time or quantity cannot be negative — always check your answer makes sense in context.

Exam tip State what your variable represents, keep units consistent, and finish by answering the actual question asked (e.g. "the width is $5\,\text{cm}$"), not just the value of $x$.