Math problems?

Question

I surely appreciate the great answers i recieved from everyone, therefore I am voting you all as giving the best answer , Too bad it doesnt give any points though. anyway I need help again .

" I will choose one of you for the points later"

The problem is as follows :
(1) A rectangle is 4 times as long as it is wide . A second rectangle is 5 centimeters longer and 2 centimeters wider than the first. The ares of the second rectangle is 270 square centimeters greater than the first. What are the dimensions of the
origional rectangle ?

(2) Solve and check useing the Substitution method :
3x + y= 5
-2x + 3y= 4 Please explain steps.

(3) One third of a number is 5 < than half of the same number .What is the number

(14) When the digites of a two digit # are reversed ,the new # is 9 more than origional # , and the sum of the digitsof the origional # is
11 . What is the origional number ?

Answer 1

Ah, Grasshopper, the answer lies within. You just haven't searched deep enough.

Answer 2

First of all in terms of x, I would solve the area of each rectangle.
Rectangle #1 L=four times the width or 4x
W=x
so the area of #1=4x squared
Rectangle #2 L=five longer than the length of rect.#1 or 4x+5
W=two longer than the width of rect#1 or x+2
so the area of #2=(4x+5)(x+2) or 4xsquared+13x+10
then if the area of rectangle #2 is 270 greater than the area of rectangle #1 I would write the following equation and solve for x
4x squared +13x+10 = 270 + 4x squared
then you would have left 13x + 10=270 so then
13x=260 therefore x=20
so back to the rectangle #1: L=4x which is 4(20)= 80
w=x which is 20
so the area of rectangle #1 is 1600 and the area of rectangle #2 is 1870 because in rect #2: L=4x +5 or 4(20)+5=85
W=x+2 or 20 + 2 = 22
and (85)(22)=1870

Answer 3

1) Let's set this up with figures, rectangle R1 with length n; width w, and area A; rectangle R2 has length g; width d and area B.
R1: n = 4w; A = 4w^2
R2: g = 5+n; d=2+w; B = A + 270
Now, that's a whole lot of variables. Instead, lets get it down to just working with one. The easiest that I can see to work with is going to be w. To do this, start with the formula for area of a rectangle, area=length * width, substituing in our values for R2
g*d = B
(5+n)*(2+w) = A+270
Now, since we know that both n and A can be expressed in terms of w, we can substitute those values in as well.
(4w+5)(w+2) = 4w^2 + 270
Now, multiply the right side out, giving us
4w^2 + 13w + 10 = 4w^2 + 270
Get the ws alone on one side by subtracting both 10 and 4w^2 from both sides of the equation (4w^2 to get rid of it on the right, 10 to get rid of the 10 on the left).
13w = 260
w = 20
Since the length was 20 times the width, our dimensions for R1 are going to be
80cm x 20cm
I'll leave the checks to you.

2) Substitution method is what we did in the last problem, expressing all the other variables in terms of one of the variables, then plugging those values in.... That wasn't very clear, was it. Well, on to your problem...
3x + y = 5; -2x + 3y = 4
Well, I can see it will be very easy to get (y) by itself in the first equation, so lets do that by subtracting 3x from each side.
3x + y -3x = 5-3x
y = 5-3x
Simple enough, right? ok, next, let's plug that in to the second equation where we find the y. Remember, that 5-3x is a substitue for y, and since y has a coefficient in the second equation, we'll need to preserve it...
-2x + 3y = 4
-2x + 3(5-3x) = 4
Multiply the three through the expression...
-2x + 15 - 9x = 4
And combine like terms
15-11x = 4
-11x = -11
Then divide by the coefficient of x
x = 1
And substitute the x back in to your original equation
3(1) + y = 5
3+y = 5
y = 2
So, you have the ordered pair (1, 2). Put that into your second equation as a check.
-2(1) + 3(2) = 4
-2 + 6 = 4
6-2 = 4
True
So your answer checks as x=1, y=2

3) Gods, I love this type of problem. So, I read it once, then go back and substitute in numbers for the words. One third of a number (1/3*n) is (=) five less (-5) than half of the same number (+[1/2*n])
n/3 = (n/2)-5
The easiest way to do this is going to be multiply by the entire equation by 6. Doing that will get rid of the 1/2 (6*1/2 = 3) and the 1/3 (6*1/3 = 2). I can do this because if a=b, then ca=cb. Since I really don't like working with fractions, that is what I will do.
6(1/3n) = 6([n/2]-5)
2n = 3n-30
Subtract 2n from each side...
0 = n-30
n = 30
Then check your answer (again, I leave to you).

14)OK, this is a tricky one... But it can be done mathematically, if you work at it. Because of the way our base 10 number system works, a two digit number is equal to 10 times the first digit plus the second digit. Amazing how that works out, right? Anyway, using that fact, I give you these equations...
a+b=11
10a+b+9 = 10b+a
Remember, the second number (10b+a) is 9 more than the first, so we must add 9 to the first number to get the right answer. Let's boil down the second equation before we do anything else, by subtracting both 10a and b from each side.
9 = 9b-9a
b-a=1
Let's get b by it's lonesome
b = a+1
and substitute it back into the first equation.
a+(a+1)=11
2a+1 = 11
2a = 10
a = 5
b = 6
Since in the original number, a came first, our number is 56.

Answer 4

Helping we enjoy, but doing all of your homework is bogus.

Answer 5

Do your own homework.

Answer 6

4y=x
(x+5)(y+2)=xy+270
sustitutingx=4y
(4y+y)(y+2)=4y*y+270
5y(y+2)=4y^2+270
5y^2-4y^2+10y-270=0
y^2+10y-270=0

2)3x+y=5
-2x+3y=4
y=5-3x
by sustituting
-2x+(5-3x)=4
-5x=-1
x=1/5
y=5-3/5=22/5
3)1/3x-5=1/2x
1/3x-1/2x=-5
1/6x=5
x=30