Question

Please solve this
(Rational numbers 8th grade)

Answer 1

Question # 1

Answer:

`(-2)/(3)* (3)/(5)+(5)/(2)* (3)/(5)* (1)/(6)=-(3)/(20)`

Explanation:

Given the expression

`(-2)/(3)* (3)/(5)+(5)/(2)* (3)/(5)* (1)/(6)`

`=-(2)/(5)+(5)/(2)* (3)/(5)* (1)/(6)` ∵
`(-2)/(3)* (3)/(5)=-(2)/(5)`

`=-(2)/(5)+(1)/(4)` ∵
`(5)/(2)* (3)/(5)* (1)/(6)=(1)/(4)`

`\mathrm{Least\:Common\:Multiplier\:of\:}5,\:4:\quad 20`

`=-(8)/(20)+(5)/(20)`

`\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad (a)/(c)\pm (b)/(c)=(a\pm \:b)/(c)`

`=(-8+5)/(20)`

`\mathrm{Add/Subtract\:the\:numbers:}\:-8+5=-3`

`=(-3)/(20)`

`\mathrm{Apply\:the\:fraction\:rule}:\quad (-a)/(b)=-(a)/(b)`

`=-(3)/(20)`

Therefore,

`(-2)/(3)* (3)/(5)+(5)/(2)* (3)/(5)* (1)/(6)=-(3)/(20)`

Question # 2

Answer:

`(2)/(5)* (-3)/(7)-(1)/(6)* (3)/(2)* (1)/(14)* (2)/(5)=-(5)/(28)`

Explanation:

Given

`(2)/(5)* (-3)/(7)-(1)/(6)* (3)/(2)* (1)/(14)* (2)/(5)`

`=-(6)/(35)-(1)/(6)* (3)/(2)* (2)/(5)* (1)/(14)` ∵
`(2)/(5)* (-3)/(7)=-(6)/(35)`

`=-(6)/(35)-(1)/(140)` ∵
`(1)/(6)* (3)/(2)* (1)/(14)* (2)/(5)=(1)/(140)`

`\mathrm{Least\:Common\:Multiplier\:of\:}35,\:140:\quad 140`

`=-(24)/(140)-(1)/(140)`

`\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad (a)/(c)\pm (b)/(c)=(a\pm \:b)/(c)`

`=(-24-1)/(140)`

`\mathrm{Subtract\:the\:numbers:}\:-24-1=-25`

`=(-25)/(140)`

`\mathrm{Apply\:the\:fraction\:rule}:\quad (-a)/(b)=-(a)/(b)`

`=-(25)/(140)`

`\mathrm{Cancel\:the\:common\:factor:}\:5`

`=-(5)/(28)`

Therefore,

`(2)/(5)* (-3)/(7)-(1)/(6)* (3)/(2)* (1)/(14)* (2)/(5)=-(5)/(28)`