Question

$\sf{ \red{Solve \: the \: following \: :}}$

$\sf{i) \: \frac{1}{9} - \frac{4}{8} + \frac{3}{6} }$

$\sf{ii) \: 12 - 4\times \frac{1}{3} }$

$\sf{ \red {Solve \: the \: following \: equation \: : }}$

$\sf{a) \: 3(x - 2) + 7 = 22}$

$\sf{b) \: - 2(x - 5) + 7 = 17}$

​

Answer 1

Answer:

$a) \: 3x - 6 = 22 - 7 \\ 3x = 15 + 6\\ 3x = 21 \\ x = \frac{21}{3} \\ x = 7$

$b) - 2x + 10 + 7 = 17 \\ - 2x + 17 = 17 \\ - 2x = 17 - 17 \\ x = 0$

Answer 2

Solution:

$\sf{{i)\:\dfrac{1}{9}-\dfrac{4}{8}+\dfrac{3}{6}}}$

In order to add three improper fractions, we need to equivalize the denominators. Here the denominators are:

9, 8 and 6.

LCM of 9, 8 and 6 = 72

Therefore:

$\sf{=\dfrac{1}{9}\times\dfrac{8}{8}=\dfrac{8}{72}}$

$\sf{=\dfrac{4}{8}\times\dfrac{9}{9}=\dfrac{36}{72}}$

$\sf{=\dfrac{3}{6}\times\dfrac{12}{12}=\dfrac{36}{72}}$

So,

$\sf{=\dfrac{8}{72}-\dfrac{36}{72}+\dfrac{36}{72}}$

$\sf{=\dfrac{8-36+36}{72}}$

$\sf{=\dfrac{8}{72}}$

Simplifying the fraction:

$\sf{\dfrac{8}{72}=\dfrac{1}{9}}$

Answer: 1/9

$\sf{ii)\:12-4\times\dfrac{1}{3}}$

According to the BODMAS theorem, multiplication comes before subtraction.

Hence:

$\sf{12\times\dfrac{-4}{3}}$

Cancelling 3 and 12:

$\sf{4\times-4}$

$\sf{-16}$

Answer: -16

$\sf{a)\:3(x-2)+7=22}$

Opening the brackets:

$\sf{3x-6+7=22}$

Moving the constants to the RHS:

$\sf{3x=22+6-7}$

$\sf{3x=21}$

$\sf{x=\dfrac{21}{3}}$

$\sf{x=7}$

Answer: x = 7

$\sf{b)\:-2(x-5)+7=17}$

Opening the brackets:

$\sf{-2x+10+7=17}$

Moving the constants to the RHS:

$\sf{-2x=17-10-7}$

$\sf{-2x=0}$

$\sf{x=\dfrac{0}{2}}$

$\sf{x=0}$

Answer: x = 0

Final Answers:

→ 1/9

→ -16

→ 7

→ 0