Equations and Inequalities

------- | ------- | ------------------------------------------- | | AQA | Paper 1 | Simultaneous equations, inequalities | |” date: 2025-06-02T16:25:28.480Z tags:

• Maths
• ALevel categories:
• Maths

flowchart TD
    A[03 Equations And Inequalities] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

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

Theorem. The system

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 $b_1$:

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).

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.

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.

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.

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.

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.

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:

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.

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:

Define the coefficient determinant:

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

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:

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

• $|f(x)| \lt a$ (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.

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 $b$:

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 $b \lt -b = |b|$:

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

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 $a + b$.

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

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

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

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

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

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

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

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

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 $q$.

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

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 $f(x)$.

Problem 12. Solve the system of equations:

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

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

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