Could you guys please answer this.
1. From the sum of (98-5) and (4a - 10), subtract (a-8).
2. Find three consecutive integers such that sum of the first and the third is 100.
1. From the sum of (98-5) and (4a - 10), subtract (a-8).
2. Find three consecutive integers such that sum of the first and the third is 100.
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Answer:
1.) 3a+101
2.) 49, 50, 51
Step-by-step explanation:
1.) (98-5) + (4a - 10)
= (103) + (4a - 10)
= 4a + 103 - 10
= 4a + (103-10)
= 4a+93 is the sum
Subtracted by a-8:
= (4a+93) - (a-8)
= (4a+93) + (-a+8)
= (4a-a) + (93+8)
= 3a+101
2.) 1,2, and 3 are three consecutive integers.
Put "x" is the variable. (x+1), (x+2), and (x+3).
Sum of 1st and 3rd is 100.
(x+1) + (x+3) = 100
2x + 4 = 100
2x = 96
x = 48
(x+1) = 48+1 = 49
(x+2) = 48+2 = 50
(x+3) = 48+3 = 51