I can't figure out how to write this problem out, so that I can solve it. Here it is: Three times a certain number is as much less than 55 as 4 times the number exceeds 50. What is the number? Can someone help me please? Thanks.
Let the number be $\displaystyle x$
then the amount that three times the number is less than 55 is: $\displaystyle 55 - 3x$
and the amount that 4 times the number exceeds 50 is: $\displaystyle 4x - 50$
we are told these are the same, that is: $\displaystyle 55 - 3x = 4x - 50$
now solve for $\displaystyle x$
EDIT: Ah, Colby beat me to it and did it for you
It's also interesting to note that you could interpret the 'amount less' or the 'amount exceeding' as negative values $\displaystyle \Rightarrow 3x>55$ and $\displaystyle 50>4x$ so you get
$\displaystyle 3x-55=50-4x$
But this still gets you to $\displaystyle x = 15$
Note that you cannot allow the 'amount 3x is less than 55' and the 'amount 4x exceeds 55' to alternate in sign, as this would lead to the incorrect equations:
$\displaystyle 3x-55=4x-50$ or $\displaystyle 55-3x=50-4x$
As the LHS's sign is not equal to the RHS's sign, this cannot be possible.
Hello, jesuslover!
Sometimes, "baby talk" will help . . .
Three times a certain number is as much less than 55
as 4 times the number exceeds 50.
What is the number?
Let $\displaystyle x$ = the number.
"Three times $\displaystyle x$ is as much less than 55" tells us that $\displaystyle 3x$ is less than 55.
How much less?
. . The difference is: .$\displaystyle 55-3x$
"Four times number exceeds 50" tells us that $\displaystyle 4x$ is greater than 50.
How much more?
. . The amount is: .$\displaystyle 4x - 50$
We are told that these amounts are equal.
. . There is our equation! . $\displaystyle 55-3x \:=\:4x-50$
Go for it!