Question

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Answer 1

Question 7: Distributive

Question 5:
`1.2x+12.3y`

Question 3:
`$1(3)/(5)x -(3)/(10)y $`

Question 2:
`$1(7)/(15) x$`

Question 1:
`-2x+3y`

Step-by-step solution:

Question 7

`4(x-5)=4x-20`

Distributive property,
`a(b + c) = ab + ac`. It says that multiplying the sum of two or more addends (numbers or terms added together) by a number (a) will give the same result as multiplying each addend individually. It is the same as the distributive property of multiplication over subtraction, except it will give the difference.

Question 5

`2.4(3x + 2y) - 1.5(4x - 5y)`

Use the distributive property:

`7.2x+4.8y-6x+7.5y`

`1.2x+12.3y`

Question 3

`$(2)/(5) \left(4x-(3)/(4)y \right)$`

Use the distributive property again:

`$(8)/(5)x -(6)/(20)y $`

`$(8)/(5)x -(3)/(10)y $`

Now you have to convert
`$(8)/(5) $` to mixed form:

It will give 1 with remainder 3, therefore

`$(8)/(5) = 1(3)/(5) $`

`$1(3)/(5)x -(3)/(10)y $`

Question 2

`$(4)/(5) x +(2)/(3) x$`

`$(4(3))/(5(3)) x +(2(5))/(3(5)) x$`

`$(12)/(15) x +(10)/(15) x$`

`$(22)/(15) x$`

Mixed form again:

`$1(7)/(15) x$`

Question 1

`$(3)/(4) (8x-4y) -(2)/(3) (12x - 9y)$`

`$6x-3y -8x +6y$`

`-2x+3y`

Answer 2

Question 7: Distributive

You multiple the outside element by each element inside the parenthesis

Question 5: 1.2x + 12.3y

2.4(3x + 2y) - 1.5(4x - 5y) = 7.2x + 4.8y - 6x + 7.5y = 1.2x + 12.3y

Question 3: 1 (3/5)x - (3/10)y

(2/5)(4x - 3y/4) = (8/5)x - 6y/20 = 1 (3/5)x - (3/10)y

Question 2: 1 (7/15)x

(4/5)x + (2/3)x = (12/15)x + (10/15)x = (22/15)x = 1 (7/15)x

Question 1: -2x + 3y

(3/4)(8x-4y) - (2/3)(12x - 9y) = 6x - 3y - 8x + 6y = -2x + 3y

Hope this helps!