(1) Let the smaller integer be x. Then the larger is x + 2.
Equation: x + 4 = 2(x + 2)
x + 4 = 2x + 4
x = 0
Smaller integer: 0
(2) Let the smaller integer be x. Then the larger is x + 2.
Equation: x + 3 = (x + 2) + 22
x + 3 = x + 24
3 = 24 → Contradiction. No solution exists.
(3) Let the smaller integer be x. Then the larger is x + 2.
Equation: 3x = 2(x + 2) + 142
3x = 2x + 4 + 142
3x = 2x + 146
x = 146
Smaller integer: 146
(4) Let the smaller integer be x. Then the middle is x + 2 and the largest is x + 4.
Equation: 4x = (x + 4) + 17
4x = x + 21
3x = 21
x = 7
Smaller integer: 7
(5) Let the smallest integer be x. Then the middle is x + 2 and the largest is x + 4.
Equation: x + 2(x + 4) = 3(x + 2) - 30
x + 2x + 8 = 3x + 6 - 30
3x + 8 = 3x - 24
8 = -24 → Contradiction. No solution exists.
(6) Let the smallest integer be x. Then the middle is x + 2 and the largest is x + 4.
Equation: x + 2(x + 4) = 3(x + 2) - 30
x + 2x + 8 = 3x + 6 - 30
3x + 8 = 3x - 24
8 = -24 → Contradiction. No solution exists.
(7) Let dimes = d, nickels = n.
d + n = 6.50 / 0.10 = 65 total coins? Wait, $6.50 in dimes and nickels, 6 coins altogether? That’s impossible since 6 coins max value is 6*0.10 = $0.60. Problem likely misstated. Assuming typo: “She has 6 more dimes than nickels” or similar. But as written: no solution.
(8) Let dimes = d, nickels = n.
d + n = 6.70 / 0.10? Wait, $6.70 in dimes and nickels, 6 coins? Max value 6*0.10 = $0.60. Impossible. Problem misstated. No solution.
(9) Let dimes = d, nickels = n.
Total value: 0.10d + 0.05n = 3.15
Also: d = n + 2
Substitute:
0.10(n + 2) + 0.05n = 3.15
0.10n + 0.20 + 0.05n = 3.15
0.15n = 2.95
n = 2.95 / 0.15 ≈ 19.666… Not integer. No solution.
Parent Tip: Review the logic above to help your child master the concept of integer word problem worksheet.
