Question
Find the two consecutive odd integers such that 5 times the first integer is 12 more than 3 times the second.
can some explain to me how to get this
i was thinking the formula would be '
5x+12=3x
No, that's not it. The second integer is x+2, since it is "consecutive" and also odd
5x = 3(x+2) + 12
2x = 18
x = 9
you have to realize that consecutive odd numbers are two apart.
So if you call the first one x, the next one would be x+2
Now "translate" our English to Math
"5 times the first integer is 12 more than 3 times the second. "
..... 5x = 3(x+2) + 12
this solves to get x=9, and the first number is 9 and the second is 11
check: 5*9=45
3*11=33
45 is greater than 33 by 12
can some explain to me how to get this
i was thinking the formula would be '
5x+12=3x
No, that's not it. The second integer is x+2, since it is "consecutive" and also odd
5x = 3(x+2) + 12
2x = 18
x = 9
you have to realize that consecutive odd numbers are two apart.
So if you call the first one x, the next one would be x+2
Now "translate" our English to Math
"5 times the first integer is 12 more than 3 times the second. "
..... 5x = 3(x+2) + 12
this solves to get x=9, and the first number is 9 and the second is 11
check: 5*9=45
3*11=33
45 is greater than 33 by 12