My Math Forum  

Go Back   My Math Forum > High School Math Forum > Algebra

Algebra Pre-Algebra and Basic Algebra Math Forum

Thanks Tree1Thanks
LinkBack Thread Tools Display Modes
January 4th, 2016, 02:28 AM   #11
Senior Member
Joined: Apr 2014
From: Glasgow

Posts: 2,161
Thanks: 734

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 x-terms 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

$\displaystyle x + 3y = 5$


$\displaystyle 2x + 6y = 10$

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:

x-terms: $\displaystyle 2x - 2x = 0$
y-terms: $\displaystyle 6y - y = 5y$
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 x-value. 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
Benit13 is offline  
January 5th, 2016, 01:17 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.
manishsqrt is offline  
January 5th, 2016, 06:37 AM   #13
Math Team
Joined: Oct 2011
From: Ottawa Ontario, Canada

Posts: 14,597
Thanks: 1038 far as I'm concerned, all that was needed is:


Then this thread would have 2 posts instead of 13

Last edited by Denis; January 5th, 2016 at 06:40 AM.
Denis is offline  

  My Math Forum > High School Math Forum > Algebra

confused, equations, extremely, solve, variables

Thread Tools
Display Modes

Similar Threads
Thread Thread Starter Forum Replies Last Post
statics class, ended up with 2 equations and 2 variables, rules to isolate variables fixxit Algebra 4 September 10th, 2014 11:29 AM
Marginal cost with two variables, confused. nakota2k Calculus 3 February 1st, 2014 04:18 AM
Please help. So confused with these equations alexandra2727 Algebra 5 January 16th, 2014 10:38 AM
I'm confused with how to solve the limit of this series Crystal-Field Calculus 3 April 29th, 2009 06:12 PM
Solve for 'x' algebraically, extremely difficult (advanced)? IsenseHelp Algebra 2 January 11th, 2009 04:10 PM

Copyright © 2019 My Math Forum. All rights reserved.