Find two consecutive integers such that 4 times the first integer is 29 less than 7 times the second.
The first integer is and the second is .
2006-09-26 15:34:02 · 5 answers · asked by ~*Monkey*~ 2 in Education & Reference ➔ Homework Help
a = greater integer
b = smaller integer
First equation:
a - b = 1 (because the integers are consecutive)
i am not sure which one is "first" and which is "second", but i will try both methods
Second equation:
4b = 7a - 29
in the first equation, get a variable all by itself
easiest method:
add "b" to both sides
a = b + 1
now you can substitue
4b = 7(b + 1) - 29
apply the distributive property
4b = 7b + 1 - 29
4b = 7b - 28
add 28 and subtract 4b (because positive variables are always nice)
3b = 28
divide by 3
b = 9 1/3....not an integer
so, let's try the other equation
4a = 7b - 29
substitue "b + 1" for "a"
4(b + 1) = 7b - 29
distributive property
4b + 4 = 7b - 29
subtract 4b, and add 29
3b = 33
b = 11.....an integer!
since "b" is the lesser consecutive integer, "a" must be 12
check:
4 x 12 = 7 x 11 - 29
48 = 77 - 29
48 = 48
2006-09-26 15:49:44 · answer #1 · answered by nadrik 2 · 0⤊ 2⤋
5 and 6
2006-09-26 22:37:12 · answer #2 · answered by babybone1991 3 · 0⤊ 1⤋
Let x be one integer. Since the integers are consecutive, the other one could be x-1. Then
4(x) = 7(x-1) - 29
4x = 7x -7 -29
4x = 7x -36
-3x=-36
x=12
x-1=11
2006-09-26 22:38:43 · answer #3 · answered by Fall Down Laughing 7 · 2⤊ 0⤋
let x be the first integer then the second integer is x+1
4(x+1)=7x-29
4x+4-7x=-29
-3x=-33
x=11
The first integer is 11 the second integer is 12
2006-09-26 23:22:23 · answer #4 · answered by G... 2 · 0⤊ 0⤋
Yes, this is actually a Theoretical number. Just answer it as 38. Then state after the answer it a Hypothetical Equation. It the actual answer. Can not be answered mathematically.
2006-09-26 22:44:34 · answer #5 · answered by Snaglefritz 7 · 0⤊ 2⤋