Need help with Algebra QUICK

[waterboy]

I'm helping my daughter with her homework and I need the equation for the following question:

Find the length and width of the rectangle described:

The length is 5 units more than the width. The perimeter is 9 times the width.

All l really need is the equation. It's been 20 years since I've had algebra in college, and I can't remember how to form the equation.

[bigwood33]

My advice is...........drop the class. Been 28 years for me.
Seriously, Google it...you can find anything on the internet :)

[waterboy]

My advice is...........drop the class. Been 28 years for me.
This is not for me, silly! This is a question on my daughter's 9th grade Algebra I homework.

[Saggy Aggie]

I'm helping my daughter with her homework and I need the equation for the following question:

Find the length and width of the rectangle described:

The length is 5 units more than the width. The perimeter is 9 times the width.

All l really need is the equation. It's been 20 years since I've had algebra in college, and I can't remember how to form the equation.1) P=9W
2) P=2L+2W
3) L=5+W

3 equations and 3 unknowns. You can solve this by substituting equations 1 & 3 into #2.

If you do this, you should get W=2, L=7, and P=18.


Hope that helped.

[waterboy]

1) P=9W
2) P=2L+2W
3) L=5+W

3 equations and 3 unknowns. You can solve this by substituting equations 1 & 3 into #2.

If you do this, you should get W=2, L=7, and P=18.


Hope that helped.
How would I write it in a single equation? The perimeter should not have to be a separate equation from the width.

[Saggy Aggie]

How would I write it in a single equation? The perimeter should not have to be a separate equation from the width. Directly substitute p=9w and l=w+5 into the 2nd equation.


When you do that, it would look like this:

9w=2(w+5)+2w

Simplifying, you get 9w=2w+10+2w

simplifying further, you get: 9w=4w+10

Solving this for w gives you w=2. Then plugging w=2 back into equation 3 tells you l=7 and the same for equation #1 and the perimeter.


EDIT:

And there are 3 variables introduced in this question (Perimeter, Length, and Width). This means that 3 equations are required to solve it, otherwise it cant be solved. So you use the info they gave you to create 2 equations and the 3rd equation comes from general knowledge of knowing that perimeter equals the length of the four sides added up (in this case a rectangle, so 2 lengths and 2 widths)

[waterboy]

Next question:

The length is 5 units less than 2 times the width. The perimeter is 22 units more than twice the width.

formula?

[waterboy]

BTW, I'll owe you a dinner or something for this. I appreciate your help.

[waterboy]

Is the formula for this one:
2(2w-5)=w+22

I'm getting 8 for the answer.

[Saggy Aggie]

Next question:

The length is 5 units less than 2 times the width. The perimeter is 22 units more than twice the width.

formula?Exact same concept, assuming this is also a rectangle?

L=2W-5
P=2W+22
P=2W+2L

Substutting the first 2 into the 3rd, you get:

2W+22=2W+2(2W-5)

Simplifying:

2W+22=2W+4W-10

Simplifying further:

32=4W

This gives: W=8

Plugging this into eqn 1 gives L=11

Plugging l=11 and w=8 into the final equation gives: P=38


And if you want to double check then make sure that 38 = 2W+22.... and when W=8, this equation says 38=16+22 or 38=38.... so you know its correct.

[Phil C]

Directly substitute p=9w and l=w+5 into the 2nd equation.


When you do that, it would look like this:

9w=2(w+5)+2w

Simplifying, you get 9w=2w+10+2w

simplifying further, you get: 9w=4w+10

Solving this for w gives you w=2. Then plugging w=2 back into equation 3 tells you l=7 and the same for equation #1 and the perimeter.


EDIT:

And there are 3 variables introduced in this question (Perimeter, Length, and Width). This means that 3 equations are required to solve it, otherwise it cant be solved. So you use the info they gave you to create 2 equations and the 3rd equation comes from general knowledge of knowing that perimeter equals the length of the four sides added up (in this case a rectangle, so 2 lengths and 2 widths)

I got on this late but as an algebra teacher of many years ago "Well done Saggy Aggie!"

I have also got to give credit to my fantastic high school teacher Mr. Eddie Saenz who was an A&M graduate.

[waterboy]

Next question:
Central High's enrollment decreases at an average rate of 55 students per year, while Washington High's enrollment increases at an average rate of 70 students per year. Central High has 2176 students and Washington High has 1866 students. If enrollments continue to change at the same rate, when will the two schools have the same number of students?

[Saggy Aggie]

Next question:
Central High's enrollment decreases at an average rate of 55 students per year, while Washington High's enrollment increases at an average rate of 70 students per year. Central High has 2176 students and Washington Highhas 1866 students. If enrollments continue to change at the same rate, when will the two schools have the same number of students?


Total number of kids = 2176 -55(X number of years) for Central

Total number of kids = 1866 +70(X number of years) for washington

You want these total number of kids to be equal for both schools(because the question asks when will the enrollments be the same) so you just set the 2 equations we created equal to each other. Thus, 2176-55x=1866+70x and solve for x.


Doing this, gives.... x=2.48 years?

IDK if this is correct, but that's how I'd do it. Anyway you can check the answer in the back of the book?

[Yoe_09]

As long as it is odd.

[waterboy]

Total number of kids = 2176 -55(X number of years) for Central

Total number of kids = 1866 +70(X number of years) for washington

You want these total number of kids to be equal for both schools(because the question asks when will the enrollments be the same) so you just set the 2 equations we created equal to each other. Thus, 2176-55x=1866+70x and solve for x.


Doing this, gives.... x=2.48 years?

IDK if this is correct, but that's how I'd do it. Anyway you can check the answer in the back of the book?
That's exactly what I got. The only problem is these problems are on a worksheet, not in the back of a textbook. I hate it when she gets this much homework every night, especially on church night. It wears me out. Word problems were tough for me in college, too. I ended up acing the class, but that was twenty years ago.

I appreciate your help!

[RoyceTTU]

That's exactly what I got. The only problem is these problems are on a worksheet, not in the back of a textbook. I hate it when she gets this much homework every night, especially on church night. It wears me out. Word problems were tough for me in college, too. I ended up acing the class, but that was twenty years ago.

I appreciate your help!

Iunderstand totally. My daughter is in 8th grade now and it isn't too bad, but last year her 7th grade teacher would send her home with 2 or 3 of those worksheets. It would take her 1-2 hours to finish and take me 30 minutes to grade and help her find her mistakes. It was a bit overboard. I feel your pain.

[Cam]

Exact same concept, assuming this is also a rectangle?

L=2W-5
P=2W+22
P=2W+2L

Substutting the first 2 into the 3rd, you get:

2W+22=2W+2(2W-5)

Simplifying:

2W+22=2W+4W-10

Simplifying further:

32=4W

This gives: W=8

Plugging this into eqn 1 gives L=11

Plugging l=11 and w=8 into the final equation gives: P=38


And if you want to double check then make sure that 38 = 2W+22.... and when W=8, this equation says 38=16+22 or 38=38.... so you know its correct.

What he said!

[Saggy Aggie]

That's exactly what I got. The only problem is these problems are on a worksheet, not in the back of a textbook. I hate it when she gets this much homework every night, especially on church night. It wears me out. Word problems were tough for me in college, too. I ended up acing the class, but that was twenty years ago.

I appreciate your help! no problem. Glad I could help.

[garciap77]

Total number of kids = 2176 -55(X number of years) for Central

Total number of kids = 1866 +70(X number of years) for washington

You want these total number of kids to be equal for both schools(because the question asks when will the enrollments be the same) so you just set the 2 equations we created equal to each other. Thus, 2176-55x=1866+70x and solve for x.


Doing this, gives.... x=2.48 years?

IDK if this is correct, but that's how I'd do it. Anyway you can check the answer in the back of the book?

http://i220.photobucket.com/albums/dd185/garciap77/Smilies/hmm.gif At Wylie we work it like this ....... nm :D

[Cam]

Next question:
Central High's enrollment decreases at an average rate of 55 students per year, while Washington High's enrollment increases at an average rate of 70 students per year. Central High has 2176 students and Washington High has 1866 students. If enrollments continue to change at the same rate, when will the two schools have the same number of students?

So, Central High = 5A and Washington High = 4A, but the real questions should be:
What classification will both schools be in 2-1/2 years?
And, with decreasing enrollment numbers, how long will it take for Central High to drop down to 3A and compete with the mighty Gilmer Buckeyes?
Why are families moving their kids from Central? What the hell is going on there?...is everybody moving to Celina & Brownwood? ...wait a minute, are these schools for real??......you guys!!!.....:)

[garciap77]

So, Central High = 5A and Washington High = 4A, but the real questions should be:
What classification will both schools be in 2-1/2 years?
And, with decreasing enrollment numbers, how long will it take for Central High to drop down to 3A and compete with the mighty Gilmer Buckeyes?
Why are families moving their kids from Central? What the hell is going on there?...is everybody moving to Celina & Brownwood? ...wait a minute, are these schools for real??......you guys!!!.....:)
:evillol:

[regaleagle]

Please compute for me the difference between the means and the extremes of the season offensive totals of the Wylie Bulldogs vs. Lucas Lovejoy. This should give us some idea of what kind of offensive numbers we might expect tonite, right???????