(a) k = 7 and m – k = 4 - Junior Cycle Mathematics - Question 9 - 2022

To factorise, we can group the terms in pairs and factor out common factors:

$$(8ax – 14bx) + (4ay – 7by)$$

This gives us:

$$2x(4a – 7b) + y(4a – 7b)$$

Now, since (4a - 7b) is a common factor, we can combine them:

$$(2x + y)(4a – 7b)$$

To combine these fractions, we need a common denominator, which is ((2x + 1)(3x + 5)). Rewriting gives:

$$\frac{2(3x + 5) + 3(2x + 1)}{(2x + 1)(3x + 5)}$$

Expanding the numerator:

$$= \frac{(6x + 10) + (6x + 3)}{(2x + 1)(3x + 5)}$$

Combining like terms results in:

$$= \frac{12x + 13}{(2x + 1)(3x + 5)}$$

For the quadratic equation (2x² – 7x – 3 = 0), we will use the quadratic formula:

$$x = \frac{-b \pm \sqrt{b² - 4ac}}{2a}$$

Where (a = 2, b = -7, c = -3):

First, calculating the discriminant:

$$b² - 4ac = (-7)² - 4(2)(-3) = 49 + 24 = 73$$

Now substituting into the formula:

$$x = \frac{7 \pm \sqrt{73}}{4}$$

Evaluating the two possible solutions gives:

$$x = \frac{7 + \sqrt{73}}{4} \approx 3.86 \text{ and } x = \frac{7 - \sqrt{73}}{4} \approx -0.39$$

Thus, the solutions are approximately 3.86 and -0.39.