Solution of Literal Equations and Systems of Equations Revision Notes for VCE SSCE Mathematical Methods

Solution of Literal Equations and Systems of Equations

Understanding literal equations

A literal equation is an equation that contains multiple letters or symbols representing different values. When solving literal equations, you're asked to rearrange the equation to isolate one particular variable in terms of the others.

While linear literal equations involve straightforward rearrangement, non-linear literal equations require additional techniques. These equations may contain powers, roots, or products of variables.

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Although computer algebra systems (CAS) can solve these equations quickly, understanding the manual solving process helps build strong algebraic skills and deepens your mathematical reasoning.

Solving literal equations with power functions

When working with literal equations involving powers, you need to understand how different types of powers behave. The key distinction is between odd and even powers.

Properties of odd powers

If $n$ is an odd natural number, then the equation $b^n = a$ is equivalent to:

b = a^{\frac{1}{n}}

This means there is only one solution. For example, if $x^3 = 8$ , then $x = 8^{\frac{1}{3}} = 2$ (only one answer).

Properties of even powers

If $n$ is an even natural number, then the equation $b^n = a$ is equivalent to:

b = \pm a^{\frac{1}{n}} \text{, where } a \geq 0

This means there are two solutions: one positive and one negative. For example, if $x^2 = 4$ , then $x = \pm 2$ (two answers: $+2$ and $-2$ ).

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When dealing with even powers, you must always remember to include both the positive and negative solutions. A common error is forgetting the negative solution.

For instance, the equation $x^2 = 2$ has solutions $x = \sqrt{2}$ and $x = -\sqrt{2}$ , which we write as $x = \pm\sqrt{2}$ .

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Worked Example: Solving literal equations with the quadratic formula

Let's solve $x^2 + kx + k = 0$ for $x$ .

This is a quadratic equation in standard form, so we can apply the quadratic formula with $a = 1$ , $b = k$ , and $c = k$ :

x = \frac{-k \pm \sqrt{k^2 - 4k}}{2}

Notice that real solutions only exist when the discriminant is non-negative. This means we need:

k^2 - 4k \geq 0

k(k - 4) \geq 0

This inequality holds when $k \geq 4$ or $k \leq 0$ .

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Worked Example: Factorising cubic equations

Solve $x^3 - 3ax^2 + 2a^2x = 0$ for $x$ .

First, factor out the common term $x$ :

x(x^2 - 3ax + 2a^2) = 0

Now factorise the quadratic expression. We need two numbers that multiply to give $2a^2$ and add to give $-3a$ . These are $-a$ and $-2a$ :

x(x - a)(x - 2a) = 0

Using the null factor law, we get three solutions:

x = 0 \text{ or } x = a \text{ or } x = 2a

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Worked Example: Equations with even powers

Solve $x^4 = c$ for $x$ , where $c > 0$ .

Since we have an even power, we need both positive and negative solutions:

x = \sqrt[4]{c} \text{ or } x = -\sqrt[4]{c}

We can write this more concisely as $x = \pm\sqrt[4]{c}$ .

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Worked Example: Equations requiring multiple steps

Solve $ax^{\frac{1}{5}} = b$ for $x$ .

Divide both sides by $a$ :

x^{\frac{1}{5}} = \frac{b}{a}

Since $\frac{1}{5}$ is an odd power (the fifth root), we raise both sides to the fifth power:

x = \left(\frac{b}{a}\right)^5

This can also be written as:

x = \frac{b^5}{a^5}

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Worked Example: Cubic equations in transformed form

Solve $a(x + b)^3 = c$ for $x$ .

Divide both sides by $a$ :

(x + b)^3 = \frac{c}{a}

Since this involves an odd power (cube), take the cube root of both sides:

x + b = \left(\frac{c}{a}\right)^{\frac{1}{3}}

Subtract $b$ from both sides:

x = \left(\frac{c}{a}\right)^{\frac{1}{3}} - b

Solving simultaneous equations

When two curves intersect, their coordinates satisfy both equations at the same time. To find these points of intersection, we solve the equations simultaneously using the substitution method.

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The general approach for solving simultaneous equations:

Express one variable in terms of the other using one equation
Substitute this expression into the second equation
Solve for the remaining variable
Substitute back to find the other coordinate
State the intersection points as coordinate pairs

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Worked Example: Parabola and straight line intersection

Find where the parabola $y = x^2 - 2x - 2$ intersects the straight line $y = x + 4$ .

Since both expressions equal $y$ , we can equate them:

x^2 - 2x - 2 = x + 4

Rearrange to standard form:

x^2 - 3x - 6 = 0

This doesn't factorise easily, so we use the quadratic formula with $a = 1$ , $b = -3$ , and $c = -6$ :

x = \frac{3 \pm \sqrt{9 - 4 \times 1 \times (-6)}}{2}

x = \frac{3 \pm \sqrt{9 + 24}}{2}

x = \frac{3 \pm \sqrt{33}}{2}

Now we find the corresponding $y$ -coordinates using $y = x + 4$ :

When $x = \frac{3 - \sqrt{33}}{2}$ :

y = \frac{3 - \sqrt{33}}{2} + 4 = \frac{3 - \sqrt{33} + 8}{2} = \frac{11 - \sqrt{33}}{2}

When $x = \frac{3 + \sqrt{33}}{2}$ :

y = \frac{3 + \sqrt{33}}{2} + 4 = \frac{11 + \sqrt{33}}{2}

Therefore, the points of intersection are:

\left(\frac{3 - \sqrt{33}}{2}, \frac{11 - \sqrt{33}}{2}\right) \text{ and } \left(\frac{3 + \sqrt{33}}{2}, \frac{11 + \sqrt{33}}{2}\right)

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Worked Example: Circle and line intersection

Find where the circle $(x - 4)^2 + y^2 = 16$ intersects the line $x - y = 0$ .

First, rearrange the line equation to make $y$ the subject:

y = x

Substitute this into the circle equation:

(x - 4)^2 + x^2 = 16

Expand the bracket:

x^2 - 8x + 16 + x^2 = 16

Simplify:

2x^2 - 8x = 0

Factor:

2x(x - 4) = 0

Therefore:

x = 0 \text{ or } x = 4

Since $y = x$ , the points of intersection are $(0, 0)$ and $(4, 4)$ .

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Worked Example: Hyperbola and line tangency

Find where the line $\frac{1}{9}x + y = \frac{2}{3}$ meets the curve $xy = 1$ .

Rearrange the line equation:

y = -\frac{1}{9}x + \frac{2}{3}

And the curve equation:

y = \frac{1}{x}

Equate the expressions for $y$ :

-\frac{1}{9}x + \frac{2}{3} = \frac{1}{x}

Multiply through by $9x$ to clear fractions:

-x^2 + 6x = 9

Rearrange:

x^2 - 6x + 9 = 0

This is a perfect square:

(x - 3)^2 = 0

Therefore $x = 3$ (a repeated root, indicating the line is tangent to the curve).

When $x = 3$ :

y = \frac{1}{3}

The point of contact is $\left(3, \frac{1}{3}\right)$ .

Using technology to verify solutions

Computer algebra systems and graphing calculators can quickly solve systems of equations. On a TI-Nspire or Casio ClassPad, you can use the "Solve System of Equations" function to verify your algebraic work.

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While these tools are helpful for checking answers, understanding the manual solving process is essential for developing algebraic reasoning skills.

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Key Points to Remember:

Odd powers give one solution: When solving $x^n = a$ where $n$ is odd, there is only one real solution: $x = a^{\frac{1}{n}}$
Even powers give two solutions: When solving $x^n = a$ where $n$ is even and $a \geq 0$ , there are two solutions: $x = \pm a^{\frac{1}{n}}$ . Never forget the $\pm$ symbol!
Substitution is key for simultaneous equations: To find intersection points, express one variable in terms of the other, then substitute into the second equation
Check your discriminant: For quadratic literal equations, real solutions only exist when $b^2 - 4ac \geq 0$
Repeated roots indicate tangency: When a system of equations produces a repeated solution (perfect square), the curves touch at exactly one point rather than crossing

Solution of Literal Equations and Systems of Equations (VCE SSCE Mathematical Methods): Revision Notes