Question 1.7: In this worked example, try solving the following equations ......

Now check your answers:

(a)   x = 11/3          (b)  x = 0.6             (c)  x = 4,    x = −4

(d)   Infinitely many solutions for which x = y +4    (e)  x = 0,   x = $\sqrt{2}$   x = –$\sqrt{2}$

Suggested solutions to Worked Example 1.7

(a) 2x + 3 = 5x − 8

3 + 8 = 5x − 2x

11 = 3x

$\frac{11}{3}$ = x

(b)  $\frac{1}{x} + \frac{2}{x}$ = 5

$\frac{1 + 2}{x}$ = 5

3 = 5x           multiply each side by x

5x = 3           swap sides

x = $\frac{3}{5}$ = 0.6      dividing each side by 5

(c)  x² + 4x − 6 = 2(2x + 5)

This time, simplify first by multiplying out the brackets and collecting similar terms:

x² + 4x − 6 = 4x + 10

x² + 4x − 4x = 10 + 6       bring all x-terms to one side and numbers to the other side

x² = 16

x = ±4

(d)  (x − y) = 4: Here we have one equation in two unknowns, so it is not possible to find a unique solution. The equation may be rearranged as

x = y + 4

This equation now states that for any given value of y (there are infinitely many values), x is equal to that value plus 4. So, there are infinitely many solutions.

(e) x is common to both terms, therefore we can separate or factor x from each term as follows:

x³ − 2x = 0

x(x² − 2) = 0

↓

x = 0 and/or (x² − 2) = 0 the product x(x² − 2) is zero if one or both terms are zero

x = 0 and/or x² = 2

Solution: x = 0,    x = ±$\sqrt{2}$

An exercise for the reader: check that the solutions satisfy the equation.