Math Is Easy - Use Algebra To Solve Word Problem 9!Find three consecutive even integers such that fourtimes the first minus the third is 6 more than twice the second. (Note1)Since even consecutive integers differ by 2 (2,4,6), wecan represent them as follows:let n represent the smallest of the three consecutiveintegers; then n + 2 represents the second largest and, n + 4 represents thelargest .We can now create the following equation: 4(n) - (n + 4) =2 (n + 2) + 6Solve this equation4(n) - (n +4) =2(n + 2) + 64nn4 =2n + 4 + 6 Remove parentheses3n - 4 = 2n + 10 Add thecommon terms3n - 4 = 2n + 10 + 4 = + 4 Add 4 to each side3n + 0 = 2n + 143n = 2n + 14-2n= -2n Subtract 2n from each side1n = 0n + 14 n = 14Now we substitute the answerfor n into the original equation to see if it works.4(n) - (n + 4) =2 (n + 2) + 64(14) - (14 + 4)= 2 (14 + 2) + 64(14) - (18) = 2(16) + 656 - 18 = 32 + 638 = 38We must also see if the solution works in the originalproblem statement.Find three consecutive even integers such that fourtimes the first minus the third is 6 more than twice the second.Find three consecutive even integers (14,16,18 areconsecutive even integers) such that four times the first (4 X 14) minus thethird (18) is 6 more than twice the second (2 X 16).Find three consecutive even integers such that fourtimes the first (56) minus the third (18) is 6 more than twice the second(32).Find three consecutive even integers such that 56 minus18 is 6 more than 32.5618 = 6 + 3238 = 38We have just solved a first degree equation.(Note1)The source for the problem statement is:Intermediate AlgebraFor College Students by Jerome E. Kaufman CloseWindow