Question

Please help!! I need step by steps!! Thank you

1. There are three consecutive even numbers such that twice the first is 20 more than the second. Find the numbers.

2. 10x - 5-2 (x+7) = 2 (4-x) + 10x

3. -4x + 5 - 1 (x-4) = 5x + 5 + 4 (1 - 3x)

4. 8 - (9x+6) = 5x + 5 + 4 (1 - 3x)

Answer 1

Part 1) The numbers are 22,24 and 26

Part 2) The value of x has no solution

Part 3) The value of x has no solution

Part 4) The value of x is
`x=-3.5`

Explanation:

Part 1)

Let

x ----> the first even number

x+2 ---> the second consecutive even number

x+4 ---> the third consecutive even number

we know that

`2(x)=(x+2)+20`

Solve for x

`2x=x+22`

`2x-x=22`

`x=22`

therefore

The numbers are 22,24 and 26

Part 2) we have

`10x-5-2(x+7)=2(4-x)+10x`

Distribute in the left and the right side to remove the parenthesis

`10x-5-2x-14=8-2x+10x`

Combine like terms

`8x-19=8x+8`

`-19=8` -----> is not true

therefore

The value of x has no solution

Part 3) we have

`-4x+5-1(x-4)=5x+5+4(1-3x)`

Distribute in the left and the right side to remove the parenthesis

`-4x+5-x+4=5x+5+4-12x`

Combine like terms

`-5x+9=-7x+9`

`-5x=-7x` -----> is not true

therefore

The value of x has no solution

Part 3) we have

`8-(9x+6)=5x+5+4(1-3x)`

Distribute in the right side to remove the parenthesis

`8-9x-6=5x+5+4-12x`

Combine like terms

`2-9x=-7x+9`

`9x-7x=2-9`

`2x=-7`

`x=-3.5`