Further Solving of Equations and Inequalities

1. Simultaneous Equations

Two linear equations with two unknowns (usually $x$ and $y$) that must be solved simultaneously.

Methods of Solution

• Elimination: Multiply equations so that coefficients of one variable are the same (or opposites), then add or subtract the equations to eliminate that variable.
• Substitution: Rearrange one equation to express one variable in terms of the other (e.g., $x = \dots$), then substitute this expression into the second equation.

Worked example (elimination): Solve

$$ \begin{aligned} 2x+y&=7\\ \\ 3x-y&=8 \end{aligned} $$

1. Add the two equations. The $+y$ and $-y$ cancel:

   $$ \begin{aligned} (2x+y)+(3x-y)&=7+8\\ \\ 5x&=15 \end{aligned} $$

2. Divide both sides by $5$:

   $$x=3$$

3. Substitute $x=3$ into the first equation:

   $$2(3)+y=7$$

4. Simplify and solve for $y$:

   $$ \begin{aligned} 6+y&=7\\ \\ y&=1 \end{aligned} $$

5. Check both values in the original equations:

   $$2(3)+1=7$$

   $$3(3)-1=8$$

6. Write the solution as an ordered pair:

   $$(x,y)=(3,1)$$

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2. Linear Inequalities

Inequalities describe a range of values rather than a single value.

Solving Inequalities

Solve as you would a linear equation, with one critical rule: When multiplying or dividing by a negative number, reverse the inequality sign.

• Example: $-2x < 6 \implies x > -3$

Representation

• Number Line:
  • Open circle $\circ$ for $<$ or $>$.
  • Closed circle $\bullet$ for $\le$ or $\ge$.
• Regions in a Plane:
  • Solid line for $\le$ or $\ge$.
  • Dashed line for $<$ or $>$.
  • Shade the region that satisfies the inequality.
  • To choose the side, test a point not on the boundary. For $y>2x+1$, test $(0,0)$: $0>1$ is false, so shade the side not containing $(0,0)$.
  • Read boundary lines carefully: a solid boundary means equality is included; a dashed boundary means it is excluded.

A shaded region can be described by listing every boundary inequality. Check a point inside the shaded region in every inequality, then check that points on solid boundaries satisfy equality.

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3. Quadratic Methods

Solving equations of the form $ax^2 + bx + c = 0$.

The Quadratic Formula

Use this when factorisation is difficult or impossible:

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

For $ax^2+bx+c=0$, identify $a$, $b$, and $c$ before substituting. The term $b^2-4ac$ is the discriminant:

• $b^2-4ac>0$: two distinct real roots;
• $b^2-4ac=0$: one repeated real root;
• $b^2-4ac<0$: no real roots.

Worked example: For $2x^2+3x-2=0$, use $(a,b,c)=(2,3,-2)$:

$$x=\frac{-3\pm\sqrt{3^2-4(2)(-2)}}{2(2)}=\frac{-3\pm5}{4}$$

Hence $x=\frac12$ or $x=-2$.

Completing the Square

Completing the square rewrites a quadratic expression exactly. It is useful for solving quadratic equations and finding the turning point of a quadratic graph.

A perfect square has the form of one squared bracket. For example:

$$x^2+4x+4=(x+2)^2$$

Most quadratic expressions are not perfect squares. Add and subtract the same number so part of the expression becomes a perfect square. This keeps the equality exact.

Some expressions have an extra amount:

$$x^2+4x+7=(x+2)^2+3$$

Some expressions are missing an amount:

$$x^2+6x+4=(x+3)^2-5$$

For larger coefficients, halve the coefficient of $x$ and square the result. For a quadratic with an outside factor, use the coefficient of $x$ inside the brackets.

Completing the square is useful for solving quadratic equations and finding the maximum or minimum of a quadratic graph.

The completed-square form is:

$$a(x+p)^2+q$$

For this form, the turning point is $(-p,q)$.

Completing the square where $a\neq1$

If the coefficient of $x^2$ is not $1$, first factor it out from the $x^2$ and $x$ terms. Then use the coefficient of $x$ inside the brackets.

For example:

$$2x^2+8x+10=2\left(x^2+4x+5\right)$$

The coefficient to halve is now $4$, not the original $8$.

Step-by-step example: Solve:

$$2x^2+8x-10=0$$

1. Factor out $2$ from the quadratic and linear terms:

   $$2(x^2+4x-5)=0$$

2. Halve the coefficient of $x$ inside the brackets and square the result:

   $$\frac{4}{2}=2, \qquad 2^2=4$$

3. Add and subtract $4$ inside the brackets. This keeps the expression equal:

   $$ \begin{aligned} 2(x^2+4x-5)&=2\left(x^2+4x+4-4-5\right)\\ \\ &=2\left(x^2+4x+4-9\right)\\ \\ &=2\left((x+2)^2-9\right) \end{aligned} $$

4. Distribute the outside factor:

   $$2\left((x+2)^2-9\right)=2(x+2)^2-18$$

5. Solve the equation:

   $$ \begin{aligned} 2(x+2)^2-18&=0\\ \\ 2(x+2)^2&=18\\ \\ (x+2)^2&=9\\ \\ x+2&=\pm\sqrt{9}=\pm3 \end{aligned} $$

6. Find both solutions:

   $$x+2=3 \quad\Rightarrow\quad x=1$$

   $$x+2=-3 \quad\Rightarrow\quad x=-5$$

7. Check both solutions in the original equation:

   $$2(1)^2+8(1)-10=0$$

   $$2(-5)^2+8(-5)-10=0$$

Therefore, the roots, or $x$-intercepts, are $x=1$ and $x=-5$. For the graph $y=2x^2+8x-10$, the turning point is $(-2,-18)$.

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4. Quadratic Factorisation

Factorising $ax^2 + bx + c$ where $a \neq 1$.

Method: Split the Middle Term

1. Find two numbers that multiply to $ac$ and add to $b$.
2. Rewrite $bx$ using these two numbers.
3. Factorise by grouping.

Example 1: Factorise and solve:

$$2x^2+7x+3=0$$

1. Find two numbers that multiply to $ac$ and add to $b$:

   $$ac=2(3)=6, \qquad b=7$$

   The numbers are $6$ and $1$ because:

   $$6(1)=6, \qquad 6+1=7$$

2. Split the middle term:

   $$2x^2+7x+3=2x^2+6x+x+3$$

3. Factorise by grouping:

   $$2x^2+6x+x+3=2x(x+3)+1(x+3)$$

4. Take out the common bracket:

   $$2x^2+7x+3=(2x+1)(x+3)$$

5. Set each factor equal to zero:

   $$2x+1=0 \quad\Rightarrow\quad x=-\frac{1}{2}$$

   $$x+3=0 \quad\Rightarrow\quad x=-3$$

Therefore, the solutions are:

$$x=-\frac{1}{2} \quad\text{or}\quad x=-3$$

Example 2: Factorise and solve $6x^2-x-2=0$.

1. Find two numbers that multiply to $ac$ and add to $b$:

   $$ac=6(-2)=-12, \qquad b=-1$$

   The numbers are $3$ and $-4$ because:

   $$3(-4)=-12, \qquad 3+(-4)=-1$$

2. Split the middle term:

   $$6x^2-x-2=6x^2+3x-4x-2$$

3. Factorise by grouping:

   $$6x^2+3x-4x-2=3x(2x+1)-2(2x+1)$$

4. Take out the common bracket:

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

5. Solve by setting each factor equal to zero:

   $$3x-2=0 \quad\Rightarrow\quad x=\frac{2}{3}$$

   $$2x+1=0 \quad\Rightarrow\quad x=-\frac{1}{2}$$

Therefore, the solutions are:

$$x=\frac{2}{3} \quad\text{or}\quad x=-\frac{1}{2}$$