Equations and Inequalities

Board	Paper	Notes
AQA	Paper 1	Simultaneous equations, inequalities
Edexcel	P1	Linear and quadratic simultaneous equations
OCR (A)	Paper 1	Set notation for solutions
CIE (9709)	P1	Simultaneous equations, inequalities

We consider systems of two equations in two unknowns. The standard methods are substitution and Elimination.

Theorem. The system

$$\begin{aligned} A_1 x + b_1 y &= c_1 \\ A_2 x + b_2 y &= c_2 \end{aligned}$$

Has:

• A unique solution if $a_1 b_2 - a_2 b_1 \neq 0$ (the lines are not parallel);
• No solution if $a_1 b_2 - a_2 b_1 = 0$ and the equations are inconsistent (parallel distinct lines);
• Infinitely many solutions if $a_1 b_2 - a_2 b_1 = 0$ and the equations are consistent (coincident lines).

Proof. By elimination. Multiply the first equation by $b_2$ and the second by $b_1$:

$$\begin{aligned} A_1 b_2 x + b_1 b_2 y &= c_1 b_2 \\ A_2 b_1 x + b_1 b_2 y &= c_2 b_1 \end{aligned}$$

Subtracting: $(a_1 b_2 - a_2 b_1)x = c_1 b_2 - c_2 b_1$.

If $a_1 b_2 - a_2 b_1 \neq 0$We obtain a unique $x$. Similarly for $y$.

If $a_1 b_2 - a_2 b_1 = 0$Then either $c_1 b_2 - c_2 b_1 = 0$ (infinitely many solutions) or $c_1 b_2 - c_2 b_1 \neq 0$ (no solution). $\blacksquare$

Intuition. The quantity $a_1 b_2 - a_2 b_1$ is the determinant of the coefficient matrix. Geometrically, two lines in the plane either intersect (unique solution), are parallel but distinct (no solution), or coincide (infinitely many solutions).

Example

Solve: $$\begin{aligned} 3x + 2y &= 12 \quad \mathrm{--- (1)} \\ 5x - 3y &= 1 \quad \mathrm{--- (2)} \end{aligned}$$

Multiply (1) by 3 and (2) by 2:

$$\begin{aligned} 9x + 6y &= 36 \\ 10x - 6y &= 2 \end{aligned}$$

Add: $19x = 38$So $x = 2$.

Substitute into (1): $6 + 2y = 12$So $y = 3$.

Solution: $x = 2$, $y = 3$.

---

When one equation is linear and the other is quadratic (or of higher degree), we use substitution.

Method.

1. From the linear equation, express one variable in terms of the other.
2. Substitute into the quadratic equation.
3. Solve the resulting quadratic.
4. Back-substitute to find both variables.

The discriminant of the resulting quadratic determines the number of intersection points.

Example

Solve: $$\begin{aligned} Y &= 2x - 1 \\ X^2 + y^2 &= 25 \end{aligned}$$

Substitute $y = 2x - 1$ into $x^2 + y^2 = 25$:

$$\begin{aligned} X^2 + (2x - 1)^2 &= 25 \\ X^2 + 4x^2 - 4x + 1 &= 25 \\ 5x^2 - 4x - 24 &= 0 \end{aligned}$$

$x = \frac{4 \pm \sqrt{16 + 480}}{10} = \frac{4 \pm \sqrt{496}}{10} = \frac{4 \pm 4\sqrt{31}}{10} = \frac{2 \pm 2\sqrt{31}}{5}$

$\Delta = 496 > 0$So the line intersects the circle at two points.

Tip: Tip Quadratic in both variables, which is harder to solve.

---

The rules for manipulating inequalities are the same as for equations, with one crucial exception.

Theorem (Order-Reversing Property). If $a < b$ and $c < 0$Then $ac > bc$.

Proof. From $a < b$We have $b - a > 0$. Since $c < 0$ and $b - a > 0$: $c(b - a) < 0$ (product Of positive and negative). So $cb - ca < 0$Giving $ca > cb$. $\blacksquare$

Corollary. Multiplying or dividing both sides of an inequality by a negative number reverses the Inequality.

### 3.2 Quadratic Inequalities

See Quadratics, Section 6.

When an inequality involves fractions, multiply through by the square of the denominator (which is Always non-negative, so the inequality direction is preserved) or use a sign chart.

Example

Solve $\frac{2x - 1}{x + 3} \geq 1$. $$\begin{aligned} \frac{2x - 1}{x + 3} - 1 &\geq 0 \\ \frac{2x - 1 - (x + 3)}{x + 3} &\geq 0 \\ \frac{x - 4}{x + 3} &\geq 0 \end{aligned}$$

Critical values: $x = 4$ (zero of numerator) and $x = -3$ (zero of denominator, undefined).

Sign chart:

Interval	$x - 4$	$x + 3$	Quotient
$x < -3$	$-$	$-$	$+$
$-3 < x < 4$	$-$	$+$	$-$
$x > 4$	$+$	$+$	$+$

The quotient is $\geq 0$ when $x < -3$ or $x \geq 4$.

Solution: $x \in (-\infty, -3) \cup [4, \infty)$.

---

To represent $ax + by + c \geq 0$ graphically:

1. Draw the line $ax + by + c = 0$.
2. Test a point not on the line ( the origin).
3. If the point satisfies the inequality, shade the region containing it.
4. If the point does not satisfy the inequality, shade the other region.
5. Use a solid line for $\geq$ or $\leq$And a dashed line for $>$ or $<$.

When multiple inequalities define a region, the solution is the intersection of all individual Regions.

Example

Shade the region defined by: $$\begin{aligned} X + y &\leq 6 \\ X &\geq 0 \\ Y &\geq 0 \\ Y &\geq 2x \end{aligned}$$

This defines a polygon bounded by the lines $x + y = 6$, $x = 0$, $y = 0$And $y = 2x$. The Vertices are (0, 0)$$(0, 6)And the intersection of $x + y = 6$ with $y = 2x$: $3x = 6$ x = 2$$y = 4. So the third vertex is $(2, 4)$.

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If $a < b$ and $b < c$Then $a < c$.

Proof. $b - a > 0$ and $c - b > 0$. Adding: $(c - b) + (b - a) = c - a > 0$. So $a < c$. $\blacksquare$

If $a < b$ and $c < d$Then $a + c < b + d$.

Proof. $b - a > 0$ and $d - c > 0$. Adding: $(b - a) + (d - c) = (b + d) - (a + c) > 0$. So $a + c < b + d$. $\blacksquare$

If $a < b$ and $c > 0$Then $ac < bc$.

Proof. $b - a > 0$ and $c > 0$. Product: $c(b - a) > 0$So $cb - ca > 0$Giving $ac < bc$. $\blacksquare$

If $0 < a < b$Then $\frac{1}{a} > \frac{1}{b}$.

Proof. Since $a, b > 0$ and $a < b$: $\frac{1}{a} - \frac{1}{b} = \frac{b - a}{ab}$. Since $b - a > 0$ and $ab > 0$The result is positive. So $\frac{1}{a} > \frac{1}{b}$. $\blacksquare$

Intuition. Consider a = 2$$b = 4. Then $\frac{1}{2} > \frac{1}{4}$. The smaller the positive Number, the larger its reciprocal — like how slicing a cake into more pieces makes each piece Smaller.

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Theorem (Factor Theorem). If $f(a) = 0$Then $(x - a)$ is a factor of $f(x)$.

Proof. By polynomial division, for any polynomial $f(x)$ and constant $a$There exist a quotient Polynomial $Q(x)$ and a constant remainder $R$ such that:

$f(x) = (x - a)Q(x) + R$

Setting $x = a$: $f(a) = (a - a)Q(a) + R = R$.

If $f(a) = 0$Then $R = 0$So $f(x) = (x - a)Q(x)$. Hence $(x - a)$ divides $f(x)$ exactly. $\blacksquare$

Theorem (Remainder Theorem). When a polynomial $f(x)$ is divided by $(x - a)$The remainder Equals $f(a)$.

Proof. From the division identity $f(x) = (x - a)Q(x) + R$Substituting $x = a$ gives $f(a) = R$. $\blacksquare$

The remainder theorem provides a quick way to evaluate $f(a)$: perform polynomial division of $f(x)$ By $(x - a)$ and read off the constant remainder, avoiding full expansion.

The systematic approach:

1. Find a root $a$ by testing small integer values (try factors of the constant term).
2. Confirm $(x - a)$ is a factor via $f(a) = 0$.
3. Divide to obtain a quotient of lower degree.
4. Repeat on the quotient until fully factorised.

Example

Fully factorise $f(x) = x^3 - 6x^2 + 11x - 6$.

Test integer values of $f(x)$:

$f(1) = 1 - 6 + 11 - 6 = 0$. So $(x - 1)$ is a factor.

Divide $x^3 - 6x^2 + 11x - 6$ by $(x - 1)$:

$x^3 - 6x^2 + 11x - 6 = (x - 1)(x^2 - 5x + 6) = (x - 1)(x - 2)(x - 3)$

Example

Fully factorise $f(x) = 2x^3 + x^2 - 5x + 2$.

By the rational root theorem, possible rational roots are factors of 2 divided by factors of 2: $\pm 1, \pm 2, \pm \frac{1}{2}$.

$f(1) = 2 + 1 - 5 + 2 = 0$. So $(x - 1)$ is a factor.

Divide by $(x - 1)$:

$2x^3 + x^2 - 5x + 2 = (x - 1)(2x^2 + 3x - 2)$

Factorise the quadratic: $2x^2 + 3x - 2 = (2x - 1)(x + 2)$.

So $f(x) = (x - 1)(2x - 1)(x + 2)$.

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For a system of three equations in three unknowns, the elimination method extends :

1. Use the first equation to eliminate one variable from equations 2 and 3.
2. Use the resulting pair of equations (now in two variables) to eliminate a second variable.
3. Back-substitute to recover all three variables.

This process is known as Gaussian elimination. It can be systematised using augmented matrices And three elementary row operations: swapping rows, multiplying a row by a non-zero constant, and Adding a multiple of one row to another.

A 3x3 system may have a unique solution, no solution, or infinitely many solutions, depending on the Determinant of the coefficient matrix (analogous to the 2x2 case in Section 1).

For the system:

$$\begin{aligned} A_1 x + b_1 y + c_1 z &= d_1 \\ A_2 x + b_2 y + c_2 z &= d_2 \\ A_3 x + b_3 y + c_3 z &= d_3 \end{aligned}$$

Define the coefficient determinant:

$D = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}$

If $D \neq 0$The unique solution is:

$x = \frac{D_x}{D}, \quad y = \frac{D_y}{D}, \quad z = \frac{D_z}{D}$

Where $D_x$ is formed by replacing the first column of $D$ with (d_1, d_2, d_3)^T$$D_y by Replacing the second column, and $D_z$ by replacing the third.

The determinant of a 3x3 matrix expands along the first row as:

$\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = a_1 \begin{vmatrix} b_2 & c_2 \\ b_3 & c_3 \end{vmatrix} - b_1 \begin{vmatrix} a_2 & c_2 \\ a_3 & c_3 \end{vmatrix} + c_1 \begin{vmatrix} a_2 & b_2 \\ a_3 & b_3 \end{vmatrix}$

Each 2x2 minor evaluates as $\begin{vmatrix} p & q \\ r & s \end{vmatrix} = ps - qr$.

Worked Example

Solve: $$\begin{aligned} X + 2y - z &= 3 \quad \mathrm{--- (1)} \\ 2x - y + z &= 1 \quad \mathrm{--- (2)} \\ X + y + 2z &= 8 \quad \mathrm{--- (3)} \end{aligned}$$

Step 1: Eliminate $x$ from (2) and (3).

(2) $-$ 2 $\times$ (1): $-5y + 3z = -5$ --- (4)

(3) $-$ (1): $-y + 3z = 5$ --- (5)

Step 2: Eliminate $y$ from (5).

(4) $-$ 5 $\times$ (5): $(-5y + 3z) - 5(-y + 3z) = -5 - 25$

$-5y + 3z + 5y - 15z = -30$

$-12z = -30$So $z = \frac{5}{2}$.

Step 3: Back-substitute into (5):

$-y + 3 \cdot \frac{5}{2} = 5 \implies -y + \frac{15}{2} = 5 \implies y = \frac{5}{2}$.

Step 4: Back-substitute into (1):

$x + 2 \cdot \frac{5}{2} - \frac{5}{2} = 3 \implies x + \frac{5}{2} = 3 \implies x = \frac{1}{2}$.

Solution: $x = \frac{1}{2}, \; y = \frac{5}{2}, \; z = \frac{5}{2}$.

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• $|f(x)| \lt a$ (with $a \gt 0$) is equivalent to $-a \lt f(x) \lt a$.
• $|f(x)| \gt a$ (with $a \gt 0$) is equivalent to $f(x) \lt -a$ or $f(x) \gt a$.
• $|f(x)| \lt g(x)$ requires $g(x) \gt 0$ and is equivalent to $-g(x) \lt f(x) \lt g(x)$.
• $|f(x)| \gt g(x)$ is equivalent to $f(x) \lt -g(x)$ or $f(x) \gt g(x)$.

Two principal approaches:

1. Case analysis: Split into $f(x) \geq 0$ and $f(x) \lt 0$Replacing $|f(x)|$ with $f(x)$ or $-f(x)$ respectively. Solve each case and take the union.
2. Squaring: Since $|f(x)|^2 = f(x)^2$The inequality $|f(x)| \lt g(x)$ becomes $f(x)^2 \lt g(x)^2$ provided $g(x) \geq 0$. This is often cleaner when both sides are non-negative.

Example

Solve $|2x - 1| \lt x + 3$.

Since $|2x - 1| \geq 0$We require $x + 3 \gt 0$I.e. $x \gt -3$.

Case 1: $2x - 1 \geq 0$I.e. $x \geq \frac{1}{2}$.

Then $|2x - 1| = 2x - 1$So $2x - 1 \lt x + 3$Giving $x \lt 4$.

Combined with $x \geq \frac{1}{2}$: $\frac{1}{2} \leq x \lt 4$.

Case 2: $2x - 1 \lt 0$I.e. $x \lt \frac{1}{2}$.

Then $|2x - 1| = 1 - 2x$So $1 - 2x \lt x + 3$Giving $-2 \lt 3x$I.e. $x \gt -\frac{2}{3}$.

Combined with $x \lt \frac{1}{2}$: $-\frac{2}{3} \lt x \lt \frac{1}{2}$.

Solution: $-\frac{2}{3} \lt x \lt 4$.

Example

Solve $|x^2 - 4| \gt 5$.

Case 1: $x^2 - 4 \geq 0$I.e. $|x| \geq 2$.

Then $x^2 - 4 \gt 5$Giving $x^2 \gt 9$So $x \gt 3$ or $x \lt -3$.

Case 2: $x^2 - 4 \lt 0$I.e. $-2 \lt x \lt 2$.

Then $-(x^2 - 4) \gt 5$Giving $4 - x^2 \gt 5$I.e. $x^2 \lt -1$.

No real solution from this case.

Solution: $x \lt -3$ or $x \gt 3$.

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Proposition. $|x|^2 = x^2$ for all real $x$.

Proof. If $x \geq 0$Then $|x| = x$So $|x|^2 = x^2$.

If $x \lt 0$Then $|x| = -x$So $|x|^2 = (-x)^2 = x^2$.

In both cases $|x|^2 = x^2$. $\blacksquare$

Theorem. $|ab| = |a||b|$ for all real $a$ and $b$.

Proof. Exhaustive case analysis on the signs of $a$ and $b$:

• $a \geq 0, \; b \geq 0$: $|ab| = ab = |a| \cdot |b|$.
• $a \geq 0, \; b \lt 0$: $ab \lt 0$So $|ab| = -(ab) = a(-b) = |a| \cdot |b|$.
• $a \lt 0, \; b \geq 0$: $ab \lt 0$So $|ab| = -(ab) = (-a)b = |a| \cdot |b|$.
• $a \lt 0, \; b \lt 0$: $ab \gt 0$So $|ab| = ab = (-a)(-b) = |a| \cdot |b|$.

In all four cases, $|ab| = |a||b|$. $\blacksquare$

Theorem (Triangle Inequality). For all real $a$ and $b$:

$|a + b| \leq |a| + |b|$

Proof. We split into cases based on the signs of $a$ and $b$.

Case 1: $a \geq 0, \; b \geq 0$.

Then $a + b \geq 0$So $|a + b| = a + b = |a| + |b|$. Equality holds.

Case 2: $a \geq 0, \; b \lt 0$.

Sub-case (i): $a + b \geq 0$. Then $|a + b| = a + b$. Since $b \lt 0$ implies $b \lt -b = |b|$:

$|a + b| = a + b \lt a + |b| = |a| + |b|$

Sub-case (ii): $a + b \lt 0$. Then $|a + b| = -(a + b) = -a - b$. Since $a \geq 0$ implies $-a \leq a = |a|$:

$|a + b| = -a + (-b) = -a + |b| \leq |a| + |b|$

Case 3: $a \lt 0, \; b \geq 0$. Symmetric to Case 2 (swap $a$ and $b$).

Case 4: $a \lt 0, \; b \lt 0$.

Then $a + b \lt 0$So $|a + b| = -(a + b) = (-a) + (-b) = |a| + |b|$. Equality holds.

In all cases, $|a + b| \leq |a| + |b|$. $\blacksquare$

Intuition. On the number line, going from the origin to $a + b$ directly covers at most as much Distance as going from the origin to $a$ and then from $a$ to $a + b$.

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Problem 1. Solve the simultaneous equations $3x + y = 13$ and $x^2 + y^2 = 25$.

Solution

From (1): $y = 13 - 3x$. Substitute into (2): $$\begin{aligned} X^2 + (13 - 3x)^2 &= 25 \\ X^2 + 169 - 78x + 9x^2 &= 25 \\ 10x^2 - 78x + 144 &= 0 \\ 5x^2 - 39x + 72 &= 0 \end{aligned}$$

$x = \frac{39 \pm \sqrt{1521 - 1440}}{10} = \frac{39 \pm \sqrt{81}}{10} = \frac{39 \pm 9}{10}$

$x = \frac{48}{10} = \frac{24}{5}$: $y = 13 - \frac{72}{5} = \frac{65 - 72}{5} = -\frac{7}{5}$.

$x = \frac{30}{10} = 3$: $y = 13 - 9 = 4$.

Solutions: $(3, 4)$ and $\left(\frac{24}{5}, -\frac{7}{5}\right)$.

If you get this wrong, revise: [Linear-quadratic simultaneous equations](#2-linear-quadratic-simultaneous-equations)

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Problem 2. Solve $\frac{3}{x - 1} > \frac{2}{x + 1}$.

Solution

$$\frac{3}{x - 1} - \frac{2}{x + 1} > 0$$

$\frac{3(x + 1) - 2(x - 1)}{(x - 1)(x + 1)} > 0$

$\frac{3x + 3 - 2x + 2}{(x - 1)(x + 1)} > 0$

$\frac{x + 5}{(x - 1)(x + 1)} > 0$

Critical values: $x = -5, -1, 1$.

Sign chart:

Interval	$x + 5$	$x - 1$	$x + 1$	Quotient
$x < -5$	$-$	$-$	$-$	$-$
$-5 < x < -1$	$+$	$-$	$-$	$+$
$-1 < x < 1$	$+$	$-$	$+$	$-$
$x > 1$	$+$	$+$	$+$	$+$

Solution: $-5 < x < -1$ or $x > 1$.

If you get this wrong, revise: [Inequalities involving fractions](#33-inequalities-involving-fractions)

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Problem 3. Show that the simultaneous equations $x + 2y = 1$ and $2x + 4y = 3$ have no solution.

Solution

From (1): $x = 1 - 2y$. Substitute into (2):

$2(1 - 2y) + 4y = 3 \implies 2 - 4y + 4y = 3 \implies 2 = 3$

This is a contradiction, so there is no solution.

Alternatively: $a_1 b_2 - a_2 b_1 = 1 \times 4 - 2 \times 2 = 0$So the lines are parallel. Since $c_1 \cdot 2 \neq c_2 \cdot 1$ ($2 \neq 3$), they are distinct parallel lines.

If you get this wrong, revise: [Linear simultaneous equations](#1-linear-simultaneous-equations)

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Problem 4. Solve the inequality $x^2 - 2x - 15 \leq 0$.

Solution

$(x - 5)(x + 3) \leq 0$.

The parabola opens upwards. It is $\leq 0$ between and including the roots:

$-3 \leq x \leq 5$

If you get this wrong, revise: [Quadratic inequalities](02-quadratics.mdx)

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Problem 5. Solve the inequality $\frac{1}{x} \leq \frac{1}{x - 2}$.

Solution

$$\frac{1}{x} - \frac{1}{x - 2} \leq 0$$

$\frac{(x - 2) - x}{x(x - 2)} \leq 0$

$\frac{-2}{x(x - 2)} \leq 0$

$\frac{2}{x(x - 2)} \geq 0$

Critical values: $x = 0$, $x = 2$.

Sign chart for $x(x - 2)$:

Interval	$x$	$x - 2$	Product
$x < 0$	$-$	$-$	$+$
$0 < x < 2$	$+$	$-$	$-$
$x > 2$	$+$	$+$	$+$

So $\frac{2}{x(x-2)} \geq 0$ when $x < 0$ or $x > 2$.

Solution: $x \in (-\infty, 0) \cup (2, \infty)$.

If you get this wrong, revise: [Reciprocals reverse order](#54-reciprocals-reverse-order-for-positive-numbers)

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Problem 6. Find the vertices of the region defined by x \geq 0$$y \geq 0$$2x + y \leq 8 And $x + 2y \leq 8$.

Solution

Intersection of $2x + y = 8$ and $x + 2y = 8$:

Multiply first by 2: $4x + 2y = 16$. Subtract: 3x = 8$$x = \frac{8}{3}.

$y = 8 - 2 \cdot \frac{8}{3} = \frac{24 - 16}{3} = \frac{8}{3}$.

Vertices: (0, 0)$$(4, 0)$$(0, 4)And $\left(\frac{8}{3}, \frac{8}{3}\right)$.

If you get this wrong, revise: [Graphical inequalities](#4-graphical-inequalities)

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Problem 7. Prove that if $a > b > 0$Then $a^2 > b^2$.

Solution

Since $a > b > 0$We have $a - b > 0$ and $a + b > 0$.

$a^2 - b^2 = (a - b)(a + b)$.

Both factors are positive, so their product is positive: $a^2 - b^2 > 0$Hence $a^2 > b^2$. $\blacksquare$

If you get this wrong, revise: [Rigorous treatment](#5-rigorous-treatment-of-inequality-manipulation)

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Problem 8. Solve the inequality $|2x - 3| \leq 5$.

Solution

$|2x - 3| \leq 5$ means $-5 \leq 2x - 3 \leq 5$.

Adding 3: $-2 \leq 2x \leq 8$.

Dividing by 2: $-1 \leq x \leq 4$.

If you get this wrong, revise: [Modulus function](./05-functions.md)

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Problem 9. Given that $x^2 + px + q = 0$ has roots $\alpha$ and $\beta$And $\alpha + \beta = 6$ and $\alpha\beta = 8$Find $p$ and $q$.

Solution

By Viète's formulas (sum and product of roots): $-p = 6$ and $q = 8$.

So p = -6$$q = 8.

Verification: $x^2 - 6x + 8 = (x - 2)(x - 4) = 0$Giving roots $2$ and $4$ with sum $6$ and product $8$. ✓

If you get this wrong, revise: [Quadratics](02-quadratics.mdx)

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Problem 10. Solve $x^4 - 5x^2 + 4 = 0$ by treating it as a quadratic in $x^2$.

Solution

Let $u = x^2$. Then $u^2 - 5u + 4 = 0$.

$(u - 1)(u - 4) = 0$

$u = 1$ or $u = 4$.

$x^2 = 1 \implies x = \pm 1$.

$x^2 = 4 \implies x = \pm 2$.

Solutions: $x = -2, -1, 1, 2$.

If you get this wrong, revise: [Quadratic formula](02-quadratics.mdx)

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Problem 11. Given that $(x - 2)$ is a factor of $f(x) = x^3 + ax^2 + bx - 12$And $f(1) = -6$ Find $a$ and $b$. Hence fully factorise $f(x)$.

Solution

Since $(x - 2)$ is a factor, $f(2) = 0$ by the factor theorem:

$f(2) = 8 + 4a + 2b - 12 = 4a + 2b - 4 = 0 \implies 2a + b = 2 \quad \mathrm{--- (i)}$

Also $f(1) = -6$:

$f(1) = 1 + a + b - 12 = a + b - 11 = -6 \implies a + b = 5 \quad \mathrm{--- (ii)}$

Subtracting (i) from (ii): $-a = 3$So $a = -3$.

From (ii): $b = 5 - (-3) = 8$.

So $f(x) = x^3 - 3x^2 + 8x - 12$.

Divide by $(x - 2)$: $f(x) = (x - 2)(x^2 - x + 6)$.

The discriminant of $x^2 - x + 6$ is $\Delta = 1 - 24 = -23 \lt 0$So no further real factorisation Is possible.

$f(x) = (x - 2)(x^2 - x + 6)$

If you get this wrong, revise: [Polynomial equations](#6-polynomial-equations)

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Problem 12. Solve the system of equations:

$$\begin{aligned} X + y + z &= 6 \\ 2x - y + z &= 3 \\ X + 2y - z &= 5 \end{aligned}$$

Solution

**Step 1:** Eliminate $z$.

(2) $-$ (1): $x - 2y = -3$ --- (4)

(1) $+$ (3): $2x + 3y = 11$ --- (5)

Step 2: Solve (4) and (5) simultaneously.

From (4): $x = 2y - 3$. Substitute into (5):

$2(2y - 3) + 3y = 11 \implies 7y - 6 = 11 \implies y = \frac{17}{7}$

$x = 2 \cdot \frac{17}{7} - 3 = \frac{34 - 21}{7} = \frac{13}{7}$

Step 3: Find $z$ from (1):

$z = 6 - x - y = 6 - \frac{13}{7} - \frac{17}{7} = \frac{42 - 30}{7} = \frac{12}{7}$

Solution: $x = \frac{13}{7}, \; y = \frac{17}{7}, \; z = \frac{12}{7}$.

If you get this wrong, revise: [Systems of three linear equations](#7-systems-of-three-linear-equations)

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Problem 13. Solve $|x^2 - 3x + 1| \geq 2$.

Solution

**Case 1:** $x^2 - 3x + 1 \geq 0$I.e. $x \leq \frac{3 - \sqrt{5}}{2}$ or $x \geq \frac{3 + \sqrt{5}}{2}$.

Then $x^2 - 3x + 1 \geq 2$Giving $x^2 - 3x - 1 \geq 0$.

Roots: $x = \frac{3 \pm \sqrt{13}}{2}$.

So $x \leq \frac{3 - \sqrt{13}}{2}$ or $x \geq \frac{3 + \sqrt{13}}{2}$.

Since $\sqrt{13} \gt \sqrt{5}$The condition $x^2 - 3x + 1 \geq 0$ is automatically satisfied by These ranges.

Case 2: $x^2 - 3x + 1 \lt 0$I.e. $\frac{3 - \sqrt{5}}{2} \lt x \lt \frac{3 + \sqrt{5}}{2}$.

Then $-(x^2 - 3x + 1) \geq 2$Giving $x^2 - 3x + 3 \leq 0$.

Discriminant: $\Delta = 9 - 12 = -3 \lt 0$. Since the parabola opens upward, $x^2 - 3x + 3 \gt 0$ For all real $x$. No solution from this case.

Solution: $x \leq \frac{3 - \sqrt{13}}{2}$ or $x \geq \frac{3 + \sqrt{13}}{2}$.

If you get this wrong, revise: [Modulus inequalities](#8-modulus-inequalities)

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Problem 14. Find the area of the region defined by x \geq 0$$y \geq 0$$3x + 2y \leq 12 And $x + y \geq 3$.

Solution

The region is bounded by four lines. Find the vertices:

• $(0, 3)$: intersection of $x = 0$ and $x + y = 3$.
• $(0, 6)$: intersection of $x = 0$ and $3x + 2y = 12$.
• $(4, 0)$: intersection of $y = 0$ and $3x + 2y = 12$.
• $(3, 0)$: intersection of $y = 0$ and $x + y = 3$.

The region is a trapezoid. Using the shoelace formula with vertices $(0, 3), (0, 6), (4, 0), (3, 0)$ In order:

$\mathrm{Area} = \frac{1}{2}\left| \sum_{i} x_i y_{i+1} - \sum_{i} y_i x_{i+1} \right|$

$= \frac{1}{2}\left| (0 \cdot 6 + 0 \cdot 0 + 4 \cdot 0 + 3 \cdot 3) - (3 \cdot 0 + 6 \cdot 4 + 0 \cdot 3 + 0 \cdot 0) \right|$

$= \frac{1}{2}\left| 9 - 24 \right| = \frac{15}{2}$

If you get this wrong, revise: [Graphical inequalities](#4-graphical-inequalities)

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Problem 15. Solve the simultaneous equations $x^2 + xy = 10$ and $y^2 + xy = 15$.

Solution

**Key insight:** add and subtract the equations.

Adding: $x^2 + 2xy + y^2 = 25$So $(x + y)^2 = 25$.

This gives $x + y = 5$ or $x + y = -5$.

Subtracting: $y^2 - x^2 = 5$So $(y - x)(y + x) = 5$.

Case 1: $x + y = 5$.

Then $(y - x)(5) = 5$Giving $y - x = 1$.

From $x + y = 5$ and $y - x = 1$: adding gives $2y = 6$So $y = 3$, $x = 2$.

Case 2: $x + y = -5$.

Then $(y - x)(-5) = 5$Giving $y - x = -1$.

From $x + y = -5$ and $y - x = -1$: adding gives $2y = -6$So $y = -3$, $x = -2$.

Verification: $(2, 3)$: $4 + 6 = 10$ ✓ and $9 + 6 = 15$ ✓.

$(-2, -3)$: $4 + 6 = 10$ ✓ and $9 + 6 = 15$ ✓.

Solutions: $(2, 3)$ and $(-2, -3)$.

If you get this wrong, revise: [Simultaneous equations](#1-linear-simultaneous-equations)

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## Common Pitfalls

1. Confusing the domain and range of functions, or not considering restrictions (e.g., denominator cannot be zero).

2. Forgetting the $+c$ constant of integration in indefinite integrals, or misusing boundary conditions in definite integrals.

3. Rounding too early in multi-step calculations. Carry full precision through and round only the final answer.

4. Incorrectly applying integration by parts by choosing $u$ and $\frac{dv}{dx}$ the wrong way around.

• Algebraic Expressions — Factorisation and the factor theorem are used to solve polynomial equations in this topic.
• Quadratics — Quadratic inequalities and the discriminant extend the quadratic equations covered there.
• Coordinates and Geometry — Graphical inequalities and intersection of lines and circles apply coordinate geometry techniques.
• Proof — The rigorous proofs of inequality properties connect to the proof methods developed in the proof topic.