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May 20th, 2014, 06:05 PM  #1 
Banned Camp Joined: May 2014 From: london Posts: 21 Thanks: 0  Simultaneous equations
solve this set of simultaneous equations. 7x  4y = 37 6x + 3y = 51 thank you for all of the help given pre hand. Last edited by skipjack; May 24th, 2014 at 08:32 PM. 
May 20th, 2014, 06:43 PM  #2 
Senior Member Joined: Sep 2012 From: British Columbia, Canada Posts: 764 Thanks: 53 
Well you can start like this: $$7x4y=37\Rightarrow 21x12y=111$$ $$6x+3y=51\Rightarrow 24x+12y=204$$ Then just add the equations and solve for x. Can you continue then solve for y? 
May 20th, 2014, 09:29 PM  #3 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038  
May 21st, 2014, 10:26 AM  #4 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 
Well, if you're not aware of the basics, then it becomes almost impossible to conduct a classroom here... Try here: Two equations in two unknowns  Math Central You can get other sites by googling "2 equations, 2 unknowns" Come back if you have questions... 
May 21st, 2014, 02:20 PM  #5  
Senior Member Joined: Jul 2013 From: United Kingdom Posts: 471 Thanks: 40  Quote:
You should find plenty of videos that cover this subject matter. Those who seek shall find. Last edited by skipjack; May 24th, 2014 at 08:35 PM.  
May 22nd, 2014, 02:33 AM  #6 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,156 Thanks: 731 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
There are two ways to solve simultaneous equations: Method 1: Elimination: This method involves adding or taking away the equations in such a way so that the y or x is eliminated. Then you have an equation with just an x or just a y, so it can be solved. This is the method that eddybob123 posted: the steps he conducted were as follows: 1. multiply the first equation by 3 2. multiply the second equation by 4 3. look at the y terms... one says $\displaystyle 12y$ and the other says $\displaystyle +12y$. $\displaystyle 21x  12y = 111$ $\displaystyle 24x+12y = 204$ 4. add both equations together (you can separate out the x bits, y bits and numbers if you like, but I normally just write it out underneath), so $\displaystyle 21x  12y = 111$ $\displaystyle 24x+12y = 204$  $\displaystyle 45x + 0y = 315$ If the above is confusing, you might want to try to separate out the bits and see if it helps you, as I've done below. If not, don't bother with it. y bit: $\displaystyle 12y + +12y = 12y + 12y = 0$ so you're left with x bit: $\displaystyle 21x + 24x = 45x$ numbers bit: $\displaystyle 111 + 204 = 315$ so $\displaystyle 45x = 315$ Dividing both sides by 45 gives $\displaystyle x = \frac{315}{45} = 7$ Now we know that x = 7, we substitute this back into any one of the equations we started with to get y. Taking the second one (it looks nice! You can use the first one if you fancy.): $\displaystyle 6x+3y = 51$ $\displaystyle 6\times7 +3y = 51$ $\displaystyle 42+3y = 51$ $\displaystyle 3y = 51  42 = 9$ $\displaystyle y = \frac{9}{3} = 3$ So the answer is x=7, y = 3. You might be thinking "I added the equation above... why was that?" It's because the signs were different. My teacher always said "remember SSAD... subtract is same, add if different" So... we added the equations above because the signs on the 12y were different (+ and ). If they are the same (+ and + or  and ), you subtract the two equations instead. Method 2: Substitution. Rearrange one equation for x or y, then substitute this into the second. This is the superior method for more difficult simultaneous equations. Take first equation and rearrange for y (or x, if you fancy. I'm going to pick y though): $\displaystyle 7x  4y = 37$ $\displaystyle 7x = 37 + 4y$ $\displaystyle 4y = 7x  37$ $\displaystyle y = \frac{7}{4}x  \frac{37}{4}$ then substitute this into the other equation. $\displaystyle 6x + 3y = 51$ $\displaystyle 6x + 3\left(\frac{7}{4}x  \frac{37}{4}\right) = 51$ $\displaystyle 6x + \frac{21}{4}x  \frac{111}{4} = 51$ $\displaystyle \frac{24}{4}x + \frac{21}{4}x  \frac{111}{4} = \frac{204}{4}$ $\displaystyle \frac{45}{4}x = \frac{305}{4}$ $\displaystyle x = \frac{305}{4} \times \frac{4}{45} = \frac{305}{45} = 7$ Then resubstitute the answer for x back into the equation as in method 1 to get y (y=3). As you can see, method 1 is much simpler, so for simultaneous equations like the ones you're getting, only use method 1 for your work at the moment. Don't bother with the other one, but remember that it exists if someone surprises you in the future with a crazy simultaneous equation 

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