Solving a system of two eqn's.  TOPIC_SOLVED

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Solving a system of two eqn's.

Postby Luke53 on Tue Sep 18, 2012 7:14 pm

How to solve the following system of two equations?
(13/x) - (2/y) = - 0.3
(14.6/x) - (0.4/y) = 0.04
Thanks.
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Re: Solving a system of two eqn's.

Postby FWT on Tue Sep 18, 2012 11:20 pm

Luke53 wrote:How to solve the following system of two equations?
(13/x) - (2/y) = - 0.3
(14.6/x) - (0.4/y) = 0.04

Solve it in the usual way. But you might want to times by 10 and 100 first to get

130/x - 20/y = -3
1460/x - 40/y = 4

I'd times by -2 on the first row next.

-260/x + 40/y = 6
1460/x - 40/y = 4

Add down. Etc.
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Re: Solving a system of two eqn's.  TOPIC_SOLVED

Postby MrAlgebra on Wed Sep 19, 2012 8:31 pm

Don't get too caught on up on the fact that the equations might look a little different than what you are used to. The same ideas and concepts work just like always.

(13/x) - (2/y) = - 0.3
(14.6/x) - (0.4/y) = 0.04

When your variables are the denominator of fractions, it's usually going to be easier to try to add or subtract the equations together like in the post above. Just to prove the point, let's solve this one using substitution. We'll start off by solving the first equation for x. I'll put my comments in brackets.

(13/x) - (2/y) = - 0.3
13/x = -0.3 + (2/y) [added (2/y) to each side]
1/x = -0.3/13 + (2/13y) [divided both sides by 13]
x = 1/(-0.3/13 + (2/13y)) [took the reciprocal of each side]

Let's clean up what we have for x a bit before we continue.

x = 1/(-0.3/13 + (2/13y))
x = 1/((-0.3y/13y) + (2/13y)) [multiplied -0.3/13 by y/y to get a common denominator]
x = 1/((-0.3y + 2)/13y) [added the fractions]
x = 13y/(-0.3y + 2) [took the reciprocal]

Now we'll plug in what we got for x into the second equation and solve for y.

(14.6/x) - (0.4/y) = 0.04
(14.6/(13y/(-0.3y + 2))) - (0.4/y) = 0.04 [did the substitution]
(14.6(-0.3y + 2))/13y) - (0.4/y) = 0.04 [did the division for the first set of terms]
(-4.38y + 29.2)/13y - (0.4/y) = 0.04 [multiplied out the 14.6]
(-4.38y + 29.2)/13y - (5.2/13y) = 0.04 [0.4/y by 13/13 to get a common denominator]
(-4.38y + 24)/13y = 0.04 [subtracted the fractions]
-4.38y + 24 = 0.52y [multiplied both sides by 13y]
24 = 4.9y [added 4.38y to both sides]
4.8980 = y [divided both sides by 4.9]

Using this answer and x = 13y/(-0.3y + 2), we can find the true value of x.

x = 13y/(-0.3y + 2)
x = 13(4.8980)/(-0.3(4.8980) + 2) [did the substitution for y]
x = 63.674/0.5306 [simplifying]
x = 120.0038 [simplifying]

Now let's check our answer.

(13/x) - (2/y) = - 0.3
(13/120.0038) - (2/4.8980) = -0.3
0.1083 - 0.4083 = -0.3
-0.3 = -0.3

Ta da.
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Re: Solving a system of two eqn's.

Postby Matt on Wed Sep 19, 2012 9:41 pm

You can always check your answers using WolframAlpha:

http://goo.gl/lRNOV

This matches MrAlgebra's answer.
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