Question 1.6: (a) Given the equation x + 3 = 2x − 6 + 5x, solve for x. (b)......

(a)     x + 3 = 2x − 6 + 5x

 x + 3 = 7x − 6                 adding the x-terms on the RHS

x + 3 + 6 = 7x − 6 + 6    to cancel the −6 on the RHS, add +6 to both sides

(Hence the frequently
quoted rule: ‘bring the
−6 to the other side
and change the sign
to +6’)

x + 3 + 6 = 7x                   bringing over −6 to the other side

x + 9 = 7x

9 = 7x − x                          bringing x over to the other side

9 = 6x

\frac {9}{6} = x                   dividing both sides by 6

1.5 = x

(b)   (x + 3)(x − 6) = 0

The LHS of this equation consists of the product of two terms (x + 3) and (x − 6). A product is equal to 0, if either or both terms in the product (x + 3), (x + 6) are 0. Here there are two solutions: (x + 3) = 0, hence x = −3; and (x − 6) = 0, hence x = 6. Each of these solutions can be confirmed by checking that they satisfy the original equation. Check the solutions:

Substitute x = −3 into (x + 3)(x − 6) = 0: (−3 + 3)(−3 − 6) = 0: (0)(−9) = 0: 0 = 0. True.
Substitute x = 6 into (x + 3)(x − 6) = 0: (6 + 3)(6 − 6) = 0: (9)(0) = 0: 0 = 0. True.

(c)  This is similar to part (b), a product on the LHS, zero on the right, hence the equation (x – 3)(x + 3) = 0 has two solutions: x = 3 from the first bracket and x = –3 from the second bracket. Alternatively, on multiplying out the brackets, this equation simplifies as follows:

(x − 3)(x + 3) = 0

x(x + 3) − 3(x + 3) = 0

x(x) + x(3) − 3(x) − 3(3) = 0

x² + 3x − 3x − 9 = 0

x² − 9 = 0

x² = 9        simplified equation

x = + \sqrt{9}

x =3 or x = −3

The reason for two solutions is that x may be positive or negative; both satisfy the simplified equation x² = 9.
If x = 3   →  x² = 9, so satisfying the equation.
If x=−3  →  x² = (−3)² = 9, also satisfying the equation.
Therefore, when solving equations of the form x² =number, there are always two solutions:

x = +\sqrt{number}       or        x =  –\sqrt{number}