We are asked to solve four problems: (a) Expand and simplify the expression $6(2y-3) - 5(y+1)$. (b) Find the value of $q$ in the equation $3^q \times \frac{1}{27} = 81$. (c) Find the exact value of $8^{\frac{2}{3}} \times 49^{-\frac{1}{2}}$. (d) Factorise the expression $9x^2 - 64y^2$.

1. Problem Description

We are asked to solve four problems:

(a) Expand and simplify the expression $6(2y-3) - 5(y+1)$.

(b) Find the value of $q$ in the equation $3^q \times \frac{1}{27} = 81$.

(c) Find the exact value of $8^{\frac{2}{3}} \times 49^{-\frac{1}{2}}$.

(d) Factorise the expression $9x^2 - 64y^2$.

2. Solution Steps

(a) Expand and simplify $6(2y-3) - 5(y+1)$:

First, distribute the 6 and the -5:

$6(2y-3) = 12y - 18$

$-5(y+1) = -5y - 5$

Now, combine the terms:

$12y - 18 - 5y - 5 = (12y - 5y) + (-18 - 5) = 7y - 23$

(b) Find the value of $q$ in the equation $3^q \times \frac{1}{27} = 81$:

We can rewrite the equation as:

$3^q \times \frac{1}{3^3} = 3^4$

$3^q \times 3^{-3} = 3^4$

$3^{q-3} = 3^4$

Since the bases are equal, we can equate the exponents:

$q - 3 = 4$

$q = 4 + 3 = 7$

(c) Find the exact value of $8^{\frac{2}{3}} \times 49^{-\frac{1}{2}}$:

$8^{\frac{2}{3}} = (8^{\frac{1}{3}})^2 = (2)^2 = 4$

$49^{-\frac{1}{2}} = \frac{1}{49^{\frac{1}{2}}} = \frac{1}{\sqrt{49}} = \frac{1}{7}$

So, $8^{\frac{2}{3}} \times 49^{-\frac{1}{2}} = 4 \times \frac{1}{7} = \frac{4}{7}$

(d) Factorise $9x^2 - 64y^2$:

This is a difference of squares, so we can write it as:

$9x^2 - 64y^2 = (3x)^2 - (8y)^2$

Using the formula $a^2 - b^2 = (a+b)(a-b)$, we get:

$(3x)^2 - (8y)^2 = (3x + 8y)(3x - 8y)$

3. Final Answer

(a) $7y - 23$

(b) $7$

(c) $\frac{4}{7}$

(d) $(3x + 8y)(3x - 8y)$