Inequalities and Forming Equations

Translating a real-world or geometric situation into an equation is a key exam skill. The same solving techniques apply once the equation is set up.

Worked example — perimeter:

A rectangle has length $(3x+1)$ cm and width $(x-2)$ cm. The perimeter is 30 cm. Find $x$.

$2(3x+1)+2(x-2)=30⟹6x+2+2x-4=30⟹8x-2=30⟹8x=32⟹x=4$

Check: length $=13$ cm, width $=2$ cm, perimeter $=2(13+2)=30$ cm ✓

Worked example — two unknowns (simultaneous equations):

Two numbers add to 15 and their difference is 3. Find both numbers.

Let the numbers be $a$ and $b$: $a+b=15$ and $a-b=3$. Add: $2a=18⟹a=9$; then $b=6$.

Worked example — ages:

Amy is 3 times as old as Ben. In 4 years, the sum of their ages will be 32. Find their current ages.

Let Ben's current age = $b$; Amy's = $3b$.

$(b+4)+(3b+4)=32⟹4b+8=32⟹b=6$

Ben is 6, Amy is 18. ✓

An inequality uses $<$, $>$, $\le$, $\ge$ instead of $=$. Solve with the same steps as an equation, but with one critical rule:

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

Worked examples:

$3x-5>7⟹3x>12⟹x>4$

$2-4x\le 10⟹-4x\le 8⟹x\ge -2$ (sign reverses when dividing by $-4$)

$-3<2x+1\le 9$

Solve each part: $-3<2x+1⟹-4<2x⟹-2<x$; and $2x+1\le 9⟹2x\le 8⟹x\le 4$

Combined: $-2<x\le 4$

Solutions to inequalities are represented as intervals on a number line.

Symbol	Endpoint type	Drawn as
$<$ or $>$	Open (not included)	Open circle $\circ$
$\le$ or $\ge$	Closed (included)	Filled circle $\bullet$

Examples:

$x>3$: open circle at 3, arrow pointing right

$-2<x\le 4$: open circle at $-2$, filled circle at $4$, line between them

Integer solutions: if the question asks for integers satisfying an inequality, list them explicitly.

$-2<x\le 4$, $x$ integer: $x\in \{-1,0,1,2,3,4\}$

To solve a quadratic inequality (e.g. $x^2-5x+6<0$):

1. Find the roots by setting the expression equal to zero and solving.
2. Sketch the parabola (or reason from its shape).
3. Identify which region satisfies the inequality.

Worked example — solve $x^2-5x+6<0$:

Roots: $(x-2)(x-3)=0⟹x=2$ or $x=3$

Parabola opens upward ($+x^2$). The expression is negative between the roots.

Solution: $2<x<3$

Worked example — solve $x^2-x-6\ge 0$:

Roots: $(x-3)(x+2)=0⟹x=3$ or $x=-2$

Parabola opens upward. The expression is non-negative outside the roots.

Solution: $x\le -2$ or $x\ge 3$

In set notation (Higher): $\{x:x\le -2\}\cup \{x:x\ge 3\}$

A linear inequality in two variables defines a region of the coordinate plane.

Method:

1. Draw the boundary line $y=mx+c$ (solid if $\le$ or $\ge$; dashed if $<$ or $>$).
2. Test a point not on the line (usually the origin) to determine which side is the solution region.
3. Shade the required region (or indicate with shading and labelling — check the question).

Worked example — show the region satisfying $y>2x-1$:

Draw the dashed line $y=2x-1$. Test origin: $0>2(0)-1=-1$ — TRUE. The origin is in the solution region; shade the region containing the origin.

Combined inequalities — the feasible region satisfies all given inequalities simultaneously. Shade the overlapping area and label it R (or as directed).

1. Forgetting to reverse the inequality when dividing by a negative

$-2x>6⟹x<-3$ (sign reverses). Missing this reversal is a very common error.

2. Confusing open and closed circles

$x<4$: open circle at 4 (4 is not included). $x\le 4$: filled circle at 4 (4 is included). Check the inequality symbol carefully.

3. Quadratic inequalities — wrong region

For $a>0$, the quadratic is positive outside the roots and negative between them. Sketch the parabola to confirm which region to take — guessing the wrong side loses all marks.

4. Forming equations — not defining the variable

State what your letter represents before writing the equation. "Let $x$ = number of years" avoids ambiguity and earns method marks even if arithmetic goes wrong.

Mistake	Correction
"$-3x>9⟹x>-3$"	Divide by $-3$: $x<-3$ (sign reverses)
"The solution to $x^2<9$ is $x<3$"	$x^2<9⟹-3<x<3$ (both directions)
"Open circle for $x\le 5$"	$x\le 5$ uses a filled (closed) circle at 5