We are given three problems: a) Given two sets $M$ and $N$, find their union and intersection. $M$ is the set of prime integers between 1 and 11, i.e., $M = \{2, 3, 5, 7\}$. $N$ is the set of factors of 12, i.e., $N = \{1, 2, 3, 4, 6, 12\}$. We need to find $M \cup N$ and $M \cap N$. b) Simplify the expression $45 \div 3 + 2 \times 8 - 12 + 42$. c) We are given that angles $x$, $y$, and $z$ are on a straight line. Also, the ratio of angles $x$ and $z$ is $x:z = 2:3$, and the value of angle $y$ is $y = 80^\circ$. We need to find the value of $x$.

1. Problem Description

We are given three problems:

a) Given two sets $M$ and $N$, find their union and intersection.

$M$ is the set of prime integers between 1 and 11, i.e., $M = \{2, 3, 5, 7\}$.

$N$ is the set of factors of 12, i.e., $N = \{1, 2, 3, 4, 6, 12\}$.

We need to find $M \cup N$ and $M \cap N$.

b) Simplify the expression $45 \div 3 + 2 \times 8 - 12 + 42$.

c) We are given that angles $x$, $y$, and $z$ are on a straight line.

Also, the ratio of angles $x$ and $z$ is $x:z = 2:3$, and the value of angle $y$ is $y = 80^\circ$. We need to find the value of $x$.

2. Solution Steps

a)

(i) $M \cup N$ is the union of the sets $M$ and $N$, which includes all the elements in either $M$ or $N$ or both.

$M = \{2, 3, 5, 7\}$ and $N = \{1, 2, 3, 4, 6, 12\}$.

$M \cup N = \{1, 2, 3, 4, 5, 6, 7, 12\}$.

(ii) $M \cap N$ is the intersection of the sets $M$ and $N$, which includes all the elements that are in both $M$ and $N$.

$M = \{2, 3, 5, 7\}$ and $N = \{1, 2, 3, 4, 6, 12\}$.

$M \cap N = \{2, 3\}$.

b) To simplify $45 \div 3 + 2 \times 8 - 12 + 42$, we follow the order of operations (PEMDAS/BODMAS).

First, perform division and multiplication from left to right:

$45 \div 3 = 15$

$2 \times 8 = 16$

So the expression becomes $15 + 16 - 12 + 42$.

Next, perform addition and subtraction from left to right:

$15 + 16 = 31$

$31 - 12 = 19$

$19 + 42 = 61$

c) Since angles $x$, $y$, and $z$ are on a straight line, we have:

$x + y + z = 180^\circ$.

We are given that $y = 80^\circ$, so we have:

$x + 80 + z = 180$

$x + z = 180 - 80 = 100$.

We are also given that $x:z = 2:3$. Let $x = 2k$ and $z = 3k$ for some constant $k$.

Then $2k + 3k = 100$, which gives $5k = 100$.

Solving for $k$, we get $k = 100/5 = 20$.

Therefore, $x = 2k = 2(20) = 40$.

3. Final Answer

a)

(i) $M \cup N = \{1, 2, 3, 4, 5, 6, 7, 12\}$

(ii) $M \cap N = \{2, 3\}$

b) $45 \div 3 + 2 \times 8 - 12 + 42 = 61$

c) $x = 40$