Find two consecutive integers such that the sum of 2 times the first integer and 4 times second integer is 46. Define variables, set up equation and solve.
2007-01-21 16:49:59 · 4 answers · asked by sittinncat 1 in Science & Mathematics ➔ Mathematics
Let x be the first number and (x+1) be the second:
2x + 4(x+1) = 46 Then isolate x
2x + 4x +4 =46
6x = 42
x=7 and therefore x+1 = 8
Plug into equation to check your work
2(7) + 4(8) = 46
14 + 32 = 46
Numbers are 7 and 8. Just remember to let what you are trying to find be x and then use the question to create the equation!
2007-01-21 17:03:46 · answer #1 · answered by Maple Leaf 7 · 0⤊ 0⤋
Let x be one of the integers.
Since the integers are consecutive, the next one would be
x + 1.
"Two times the first integer" means 2x
"4 times the second integer" means 4(x + 1)
"The sum of" means we add these two values together.
"is 46" means we equate the sum to 46.
2x + 4(x + 1) = 46
Distribute the 4.
2x + 4x + 4 = 46
Group like terms.
6x + 4 = 46
6x = 42
Therefore,
x = 7
It follows that the next consecutive integer is 8. So the integers are 7 and 8.
2007-01-22 00:54:43 · answer #2 · answered by Puggy 7 · 0⤊ 0⤋
Try x for the first integer and (x+1) for the second
So 2*x + 4 *(x+1) = 46; distribute, combine like variables, and solve!
2007-01-22 00:55:07 · answer #3 · answered by lynn y 3 · 0⤊ 0⤋
integers are n & n+1
2n+4(n+1)=46
2n+4n+4=46
6n=42
n=7
n+1=8
the integers are 7 & 8
check
2*7+4*8=14+32=46
2007-01-22 01:07:25 · answer #4 · answered by yupchagee 7 · 0⤊ 0⤋