find four consecutive integer such that 5 time the first is increased by the fourth is 11 less than 7 times the second

Question

Include a key and only algebraic

Answer 1

4 consecutive integers = x, x+1, x+2, x+3

5 times 1st + 4th = 11 less than 7 times 2nd

5x+ x+3 = 7(x+1) -11

6x + 3 = 7x -4

x = 7

integers are 7,8,9,10

7x5 +10 = 45

7x8-11 = 45

87 = 105-11 =94 whoops