One number is 7 times a first number. A third number is 100 more than the first number. If the sum of thrre numbers is 568, find the numbers

Question

Has to be in order of First, Secind, Third

Answer 1

A = 1st number, B = 2nd number, C = 3rd number

1. 7A = B
2. A+100 = C
3. A + B + C = 568

These are your three knowns. You would solve for one of the variables, then substitute in another. Then repeat with the same variable in the other equation. Until you solve for one variable. With that solved, you would solve for the second and finally the third variable.

C is already isolated in the second equation above, and B is isolated in the first. both are in terms of A so substitute for both in the third equation.

A + B + C = 568

A + (7A) + (A+100) = 568 now combine like terms

9A + 100 = 568

9A = 468 divide both sides by 9

A = 52 Use this in the first equation above to solve for B; in the second to solve for C

7A = B A+100 = C

7(52) = B 52 + 100 = C

364 = B 152 = C

A = 52, B = 364, C = 152

Do a check: 52 + 364 + 152 = 568 Correct.

Answer 2

b = 7a

c = a + 100

a + 7a + a + 100 = 568

9a + 100 = 568

9a + 100 - 100 = 568 - is 00

9a = 468

9a/9 = 468/9

a = 52

7(52) = 364

52 + 100 = 152

52, 364, 152