Rahul solved the equation 2(x – ) – 2 left-parenthesis x minus startfraction 1 over 8 endfraction right-parenthesis minus startfraction 3 over 5 endfraction x equals startfraction 55 over 4 endfraction x = 2 left-parenthesis x minus startfraction 1 over 8 endfraction right-parenthesis minus startfraction 3 over 5 endfraction x equals startfraction 55 over 4 endfraction . in which step did he use the addition property of equality?

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answer: 16

step-by-step explanation: first you would add 6+6 which equals 12 then add 2+2 which equals 4. then add 12 + 4 which is 16.

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see explanation

step-by-step explanation:

a

(6r + 7) + (13 + 7r) ( remove brackets and collect like terms )

= 6r + 7 + 13 + 7r

= 13r + 20

b

remove brackets and collect like terms

13 - r - 1 + r

= 12 - r

c

remove brackets and collect like terms

= - 8 - r + 2r - 4

= - 12 + r

d

remove brackets and collect like terms

= 7r - - - 6r

[-   - = - - = - ]

= r -