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January 4th, 2016, 03:28 AM  #11 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,132 Thanks: 717 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
Just for the benefit of the OP... There are two basic ways of solving simultaneous equations: 1. By elimination 2. By substitution Basically, the idea is that by using the above two, you can reduce the number of unknown variables. Let's take a simpler example: $\displaystyle 2x + y = 3$ $\displaystyle x + 3y = 5$ So, we have two unknown variables and two unique equations. That means we have enough information to find a solution. Let's try both methods:  1. By elimination: We can choose whether to eliminate the x or the y first. Let's eliminate the x. In order to eliminate x, we need to make the xterms on both equations the same magnitude. At the moment, we have an x and a 2x, so they are not the same. We can make them the same by multiplying or dividing one of the equations. In our case, we want to multiply both sides of the second equation by 2 so that the x becomes a 2x. So $\displaystyle x + 3y = 5$ becomes $\displaystyle 2x + 6y = 10$ Now we are working with 2x + y = 3 2x + 6y = 10 So far so good. Now let's subtract the first equation from the second. We do this term by term: xterms: $\displaystyle 2x  2x = 0$ yterms: $\displaystyle 6y  y = 5y$ constants: $\displaystyle 10   3 = 10 + 3 = 13$ This gives a final equation: $\displaystyle 5y = 13$ The x has been eliminated and we are left with an equation purely in terms of y. We can just divide both sides by 5 now to give $\displaystyle y = \frac{13}{5}$ We now substitute this into one of the equations above to get the xvalue. So, for example, let's use the first equation: $\displaystyle 2x + y = 3$ $\displaystyle 2x + \frac{13}{5} = 3$ $\displaystyle 2x = 3  \frac{13}{5}$ $\displaystyle 2x = \frac{28}{5}$ $\displaystyle x = \frac{14}{5}$ So the solution is $\displaystyle x = \frac{14}{5}, y = \frac{13}{5}$  2. By substitution: $\displaystyle 2x + y = 3$ $\displaystyle x + 3y = 5$ We make x or y the subject in any one of the formulae above. Let's choose to make x the subject in the second equation: $\displaystyle x + 3y = 5$ $\displaystyle x = 5  3y$ We now substitute this equation into the first one: $\displaystyle 2x + y = 3$ $\displaystyle 2(5  3y) + y = 3$ Let's neaten this up: $\displaystyle 10  6y + y = 3$ $\displaystyle 10  5y = 3$ $\displaystyle 10 + 3 = 5y$ $\displaystyle 13 = 5y$ $\displaystyle y = \frac{13}{5}$ Now substitute this back into one of the equations (let's pick the first one): $\displaystyle 2x + y = 3$ $\displaystyle 2x + \frac{13}{5} = 3$ $\displaystyle 2x = 3  \frac{13}{5}$ $\displaystyle 2x = \frac{28}{5}$ $\displaystyle x = \frac{14}{5}$ So the solution is $\displaystyle x = \frac{14}{5}, y = \frac{13}{5}$ These techniques can apply for any set of simultaneous equations, no matter how complicated those equations are or how many you have in your set of simultaneous equations. However, as the problems get more difficult, you will need to rely more on substitution than elimination. You should practice both because both methods are very useful 
January 5th, 2016, 02:17 AM  #12 
Senior Member Joined: May 2015 From: Varanasi Posts: 110 Thanks: 5 Math Focus: Calculus 
I don't think there should be a lot of problem. Start with the third equation as that is expressed in terms of two variables only then put the value of y or z obtained from their to the first two equations, this will leave you with two equations with two variables and in this way you may solve it. I am not solving it here because i believe practice makes a man perfect and a true mathematics lover will just guide you the path and leave you alone to walk on it.

January 5th, 2016, 07:37 AM  #13 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,607 Thanks: 954 
Hmmm...as far as I'm concerned, all that was needed is: GO LEARN: https://www.google.ca/?gws_rd=ssl#q=...by+elimination Then this thread would have 2 posts instead of 13 Last edited by Denis; January 5th, 2016 at 07:40 AM. 

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