mossypne Posted June 9, 2011 Share Posted June 9, 2011 For my maths Homework i have to complete some simultaneous equations. I have come across some negative ones and we haven't done those. Is there anyway you could explain to me how to do them. The question is: (Please note: im not asking you to do it) 3a + 2b = 1 15a + 3b = -51 and 5m + 6n = -33 10m + 5n = -45 Link to comment Share on other sites More sharing options...

Shadow Posted June 9, 2011 Share Posted June 9, 2011 There are many ways to solve simultaneous equations; my favorite is the substitution method. Link to comment Share on other sites More sharing options...

imatfaal Posted June 9, 2011 Share Posted June 9, 2011 OK - there are a few pointers. 1. You can add and subtract simultaneous equations from each other. Your equations are made up of three terms (using the first question) an 'a' term, a 'b' term, and a constant - ie it is 3 times 'a' plus 2 times 'b' equals 1. When you add or subtract a pair of equations you get the 'like terms' together. You cannot subtract 15a from 2b and get a simpler answer, but if you subtract 3a from 15a you get a simpler answer 12a. 2. An equation remains the same if you multiply all of it by a number If you multiply each part of 3a+2b =1 by a number, say for example 4, you get the same equation but with a fixed multiple 3a+2b=1 says the same thing as 12a+8b=4 Play with these two rules and you will find that multiplying equation one (remember every single term) by a number and then taking it away from equation two (gather 'like terms') will give you a simpler equation where one of the terms (the 'a' term or the 'b' term disappear). Try it and put up your ideas If I was helpful, let me know by clicking the [+] sign -> 1 Link to comment Share on other sites More sharing options...

mossypne Posted June 9, 2011 Author Share Posted June 9, 2011 No i know how to do them it's just im not sure about the negative ones. Look closely at the = on the equations above. Link to comment Share on other sites More sharing options...

imatfaal Posted June 9, 2011 Share Posted June 9, 2011 Mossy Put down what you have done to the two questions and we will push you in the right direction. There is really no difference at all - ie the fact that the constant is positive or negative makes no odds If I was helpful, let me know by clicking the [+] sign -> 1 Link to comment Share on other sites More sharing options...

mossypne Posted June 9, 2011 Author Share Posted June 9, 2011 For the 1st question i have: 3a + 2b = 1 (x5) = 15a + 10b = 5 15a + 3 = -51 (15a + 10b = 5) - (15a + 3 = -51) = 13b = -46 Not sure where to go from there and for the second i have: 5m + 6n = -33 (x2) = 10m + 12n = -66 10m + 5n = -4 (10m + 2n = -66) - (10m + 5n = -45) = 7n = -21 Same im not sure where to go from here. Link to comment Share on other sites More sharing options...

Cap'n Refsmmat Posted June 9, 2011 Share Posted June 9, 2011 For the 1st question i have: 3a + 2b = 1 (x5) = 15a + 10b = 5 15a + 3b = -51 (15a + 10b = 5) - (15a + 3b = -51) = 13b = -46 If [math]13b = -46[/math], you just divide by 13 to get [imath]b=\frac{-46}{13}[/imath]. However, you have made an error in this step: [math] (15a + 10b = 5) - (15a + 3b = -51)[/math] 10b - 3b = 7b, not 13b, and similarly for the others. You subtracted 15a but added the rest, which is incorrect -- every term must be subtracted. You'll get: [math]7b=56[/math] And you can divide by 7 to get b. 5m + 6n = -33 (x2) = 10m + 12n = -66 10m + 5n = -4 (10m + 2n = -66) - (10m + 5n = -45) = 7n = -21 You've made the same error here. 2n - 5n = -3n, not 7n. Once you have solved for b and n, you can use that value to substitute into the other equation and solve for the other variable. Link to comment Share on other sites More sharing options...

imatfaal Posted June 9, 2011 Share Posted June 9, 2011 (edited) Ok Mossy Now we are cooking with gas. First equation firstly put stuff in nice neat columns if I was you - its helps keep things together secondly do your basic arithmetics more carefully 15a-15a= 0 10b-3b =?? 5 - (-51) =?? do these sums and it will work out a bit nicer edit the captain is damn quick with his [maths] markup Edited June 9, 2011 by imatfaal Link to comment Share on other sites More sharing options...

mossypne Posted June 9, 2011 Author Share Posted June 9, 2011 Oh the mistake is me just rushing it slightly. Im not that bad at maths (honest ). Link to comment Share on other sites More sharing options...

imatfaal Posted June 9, 2011 Share Posted June 9, 2011 Cool - glad you have got your head round it Link to comment Share on other sites More sharing options...

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