## Solving a system of two eqn's.

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
Luke53
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### Solving a system of two eqn's.

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

FWT
Posts: 153
Joined: Sat Feb 28, 2009 8:53 pm

### Re: Solving a system of two eqn's.

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

MrAlgebra
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### Re: Solving a system of two eqn's.

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.

Matt
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### Re: Solving a system of two eqn's.

You can always check your answers using WolframAlpha:

http://goo.gl/lRNOV

This matches MrAlgebra's answer.