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? a table titled rahul's solution with 2 columns and 5 rows. the first column, steps, has the entries 1, 2, 3, 4. the second column, resulting equations, has the entries, 2 x minus startfraction 1 over 4 endfraction minus startfraction 3 over 5 endfraction x equals startfraction 55 over 4 endfraction, startfraction 7 over 5 endfraction x minus startfraction 1 over 4 endfraction equals startfraction 55 over 4 endfraction, startfraction 7 over 5 endfraction x equals startfraction 56 over 4 endfraction, x equals 10. step 1 step 2 step 3 step 4