system [of equations]

kevin3000 · #1 $Old$ November 5th, 2009, 04:14 AM

can someone help me solve this system [of equations].
Be very methodic.

$(1+ \frac{(x+2)}{100})\cdot Y=2350$
$(1+ \frac{(x)}{100})\cdot 2Y=3100$

Wilmer · #2 $Old$ November 5th, 2009, 08:20 AM

Quote:

Originally Posted by kevin3000 $View Post$

Be very methodic.
$(1+ \frac{(x+2)}{100})\cdot Y=2350$

Sorry; don't have time to be "very methodic".
Do 1st one like this:

Distribute left side:
y + y(x - 2) / 100 = 2350
y + (xy - 2y) / 100 = 2350
multiply by 100:
100y + xy + 2y = 235000 [1]

Handle 2nd one similarly; you'll end up with 200y + 2xy = 310000 [2]

Now combine [1] and [2]: very easy...

Raoh · #3 $Old$ November 5th, 2009, 08:23 AM

Hello

multiply ur first line with 2,and u should get.
$2350\times\frac{100}{102+x}=3100\times\frac{100}{100+x}\Leftrightarrow \frac{2350}{102+x}=\frac{3100}{100+x}.$
solve for $x$ .

I4talent · #4 $Old$ November 5th, 2009, 06:40 PM

Quote:

Originally Posted by kevin3000 $View Post$

can someone help me solve this system [of equations].
Be very methodic.

$(1+ \frac{(x+2)}{100})\cdot Y=2350$
$(1+ \frac{(x)}{100})\cdot 2Y=3100$

$(1+ \frac{(x+2)}{100})\cdot Y=2350$ $\implies y = \frac{2350}{1+ \frac{(x+2)}{100}}$
$(1+ \frac{(x)}{100})\cdot 2Y=3100 \implies 2y = \frac{3100}{1+ \frac{(x)}{100}} \implies y = \frac{3100}{\left(1+ \frac{(x)}{100}\right){2}}$

Let $\frac{2350}{1+ \frac{(x+2)}{100}} = \frac{3100}{\left(1+ \frac{(x)}{100}\right){2}}$

Then $\frac{2350}{1+ \frac{(x+2)}{100}} - \frac{3100}{\left(1+ \frac{(x)}{100}\right){2}} = 0$

Solve for $x$ .

HallsofIvy · #5 $Old$ November 6th, 2009, 04:31 AM

Dividing the first equation by the second eliminates Y and gives
$\frac{1+\frac{X+2}{100}}{1+\frac{X}{100}}= \frac{2350}{3100}$

Multiplying both numerator and denominator on the left by 100 gives
$\frac{102+ X}{100+ X}= \frac{2350}{3100}$ .
That should be easy to solve for X.

Wilmer · #6 $Old$ November 6th, 2009, 06:24 AM

Not quite Sir Halls; to do that, you need to double 1st equation;
that'll make right side 4700/3100; ends up with x = -96 1/8

dhiab · #7 $Old$ November 6th, 2009, 10:23 PM

Hello I'have a simpl solution
LOOK HERE

Wilmer · #8 $Old$ November 7th, 2009, 05:40 AM

But I see no need to do all that work Dhiab.

We start with:
y(1 + (x+2)/100) = 2350 [1]
2y(1 + x/100) = 3100 [2]

Multiply [1] by 2:
2y(1 + (x+2)/100) = 4700 [1]
2y(1 + x/100) = 3100 [2]

Divide [1] by [2] (the y's cancel out):

(1 + (x+2)/100) / (1 + x/100) = 47 / 31

Solve for x:
31(x + 102) = 47(x + 100)
16x = 1538
x = 96 1/8