Simplify the following. h 2 - h - 20 h 2 - 25

We have, h 2 - h - 20 h 2 - 25 We can factorize both the numerator and the denominator. h 2 - 5 h + 4 h - 20 h 2 - 5 2 [using a 2 - b 2 = a - b a + b ] h h - 5 + 4 h - 5 h - 5 h + 5 h - 5 h + 4 h - 5 h + 5 Now we can see that h - 5 is a factor of both the numerator and the denominator, so we can cancel this common factor. Then, h + 4 h + 5 Therefore, the simplest form of h 2 - h - 20 h 2 - 25 is h + 4 h + 5 .

Simplify. x 2 + 6 x - 7 3 x + 21

Given: x 2 + 6 x - 7 3 x + 21 We have to simplify the given fraction to its lowest term. Then we can factorize both the numerator and the denominator. x 2 + 7 x - x - 7 3 x + 21 x x + 7 - 1 x + 7 3 x + 7 x - 1 x + 7 3 x + 7 Now we can see that x + 7 is a factor of both the numerator and the denominator, so we can cancel this common factor. x - 1 3 Therefore, the simplest form of x 2 + 6 x - 7 3 x + 21 is x - 1 3 .

Write as a single fraction in its simplest form. 3 2 x - 1 - 1 x + 2

Given: 3 2 x - 1 - 1 x + 2 The lowest common multiple of 2 x - 1 and x + 2 is 2 x - 1 x + 2 . This will be the common denominator. Then, 3 x + 2 - 1 2 x - 1 2 x - 1 x + 2 3 x + 6 - 2 x + 1 2 x - 1 x + 2 x + 7 2 x - 1 x + 2 Therefore, as a single fraction in its simplest form is x + 7 2 x - 1 x + 2 .

Simplify. 4 x 2 - 16 2 x 2 + 6 x - 56

We have, 4 x 2 - 16 2 x 2 + 6 x - 56 We can factorize both the numerator and the denominator. 2 x 2 - 4 2 2 x 2 + 14 x - 8 x - 56 [using a 2 - b 2 = a - b a + b and middle term splitting] 2 x - 4 2 x + 4 2 x x + 7 - 8 x + 7 2 x - 4 2 x + 4 2 x - 8 x + 7 4 x - 2 x + 2 2 x - 4 x + 7 Now we can see that 2 is a factor of both the numerator and the denominator, so we can cancel this common factor. Then, 2 x - 2 x + 2 x - 4 x + 7 Therefore, the simplest form of 4 x 2 - 16 2 x 2 + 6 x - 56 is 2 x - 2 x + 2 x - 4 x + 7 .

Karen Morrison and Nick Hamshaw Solutions for Exercise 15: Examination practice

Author:Karen Morrison & Nick Hamshaw

Karen Morrison Mathematics Solutions for Exercise - Karen Morrison and Nick Hamshaw Solutions for Exercise 15: Examination practice

Attempt the practice questions from Exercise 15: Examination practice with hints and solutions to strengthen your understanding. Cambridge IGCSE® Mathematics Core and Extended Coursebook Second Edition solutions are prepared by Experienced Embibe Experts.

Questions from Karen Morrison and Nick Hamshaw Solutions for Exercise 15: Examination practice with Hints & Solutions

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Upper Secondary: IGCSE

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Cambridge IGCSE® Mathematics Core and Extended Coursebook Second Edition>Chapter 14 - Further Solving of Equations and Inequalities>Examination practice>Q 5

Jay makes wooden boxes in two sizes. He makes $x$ small boxes and $y$ large boxes.
He makes at least $5$ small boxes.
The greatest number of large boxes he can make is $8$ .
The greatest total number of boxes is $14$ .
The number of large boxes is at least half the number of small boxes.

The price of the small box is $$ 20$ and the price of the large box is $$ 45$ .

What is the greatest amount of money he receives when he sells all the boxes he has made?