Helpppp!!!!!!!

**amy07** · October 14th 2007, 08:21 AM

Solve For b: ax(squared) + bx+c=y.

Three times the larger of 2 consecutive even intergers, decreased by the smaller is 58. Find the numbers.

Find 3 consecutive intergers such that the sum of the first two is 74 more than the third.

A pipe is 48 inches long. David wants to cut the pipe into three pieces so that the second piece is 1 inch less than the length of the third. The first piece is 5 times as long as the third. How long is each piece?

Ron runs a ski train. One day he noticed that the train contained 13 more women than men *including himself*. If there were a total of 165 people on the train, how many of them were men?

**Jhevon** · October 14th 2007, 08:47 AM

Originally Posted by amy07

Solve For b: ax(squared) + bx+c=y.

this means that we want to get b on one side of the equation by itself. so get rid of everything on the same side with b.

$ax^2 + bx + c = y$ .......subtract $ax^2 + c$ from both sides

$\Rightarrow bx = y - ax^2 - c$ .........now if $x \ne 0$ we can divide by it.

$\Rightarrow b = \frac yx - ax - \frac cx$

Three times the larger of 2 consecutive even intergers, decreased by the smaller is 58. Find the numbers.

Let $n$ be the smallest even integer.
Then $n + 2$ is the next even integer.

we are told that 3 times $(n + 2)$ , subtracting the smaller is 58. thus,

$3(n + 2) - n = 58$

now solve for $n$ so you can find the integers.

Find 3 consecutive intergers such that the sum of the first two is 74 more than the third.

Let the first integer be $n$
then the next is $n + 1$
and the largest is $n + 2$

the sum of the first two is 74 + the third. thus,

$n + (n + 1) = 74 + (n + 2)$

now solve for $n$ to find the integers

A pipe is 48 inches long. David wants to cut the pipe into three pieces so that the second piece is 1 inch less than the length of the third. The first piece is 5 times as long as the third. How long is each piece?

Let $a$ be the length of the first piece,
Let $b$ be the length of the second piece,
Let $c$ be the length of the third piece,

then we have $a + b + c = 48$ .................(1)

since the second piece is 1 inch less than the length of the third, we have:

$b = c - 1$

$\Rightarrow b - c = -1$ .....................(2)

since the first piece is 5 times as long as the third, we have:

$a = 5c$

$a - 5c = 0$ ........................(3)

thus we obtain the system:

$a + b + c = 48$ ......................(1)
$b - c = -1$ ............................(2)
$a - 5c = 0$ .............................(3)

now solve this system to get $a,b, \mbox{ and } c$

if you do not know how to solve simultaneous equations, you can do this in one variable.

Let the length of the third be $a$

then the length of the second is $a - 1$

and the length of the first is $5a$

the pieces add up to 48, so:

$a + (a - 1) + 5a = 48$

now just find $a$ , and then substitute its values into the equation to find the other pieces

Ron runs a ski train. One day he noticed that the train contained 13 more women than men *including himself*. If there were a total of 165 people on the train, how many of them were men?

Let's keep this in one variable:

Let the number of men be $m$ .
since the train contained 13 more women than men, the number of women is $m + 13$ .

these must add up to 165 people, thus:

$m + (m + 13) = 165$

now solve for $m$

**amy07** · October 14th 2007, 10:03 AM

i dont know where i go to see where people responded =\

Okay, I found the response. thank you so much for the help!!!!!!!!!!!! i really really appreciate it!

Thread: Helpppp!!!!!!!

LinkBack

Thread Tools

Display

Helpppp!!!!!!!

i dont know where to...

Similar Math Help Forum Discussions

Helpppp!!!!

[SOLVED] helpppp please. nobody can solve it??

Graphing helpppp...

Graphing helpppp...

helpppp

Search Tags