Question 1.7: In this worked example, try solving the following equations ......
In this worked example, try solving the following equations yourself. The answers are given below, followed by the detailed solutions.
(a) 2x + 3 = 5x −8 (b) \frac{1}{x} + \frac{2}{x} = 5
(c) x² + 4x − 6 = 2(2x + 5) (d) (x − y) =4 (e) x³ − 2x = 0
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
\frac{3}{x} = \frac{5}{1}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.