Finding Solutions to Systems of Equations
Date: 10/18/96 at 22:34:58 From: Christian Sheehy Subject: Finding Solutions to Systems of Equations Dear Dr. Math, I need help solving systems of equations! My friends keep telling me what to do, but I just can't get the hang of it :( Examples: Find the solution set of each equation: 3m+5n=21 4m+10n=38 -2x+9y=5 8x-36y=-20 x+5y-2z=5 2x+y-z=4 Thanks for your help! :)
Date: 10/19/96 at 11:5:54 From: Doctor Anthony Subject: Re: Finding Solutions to Systems of Equations Try to get the number in front of either m or n the same in both equations. If we multiply the first equation throughout by 2, we shall get 10n in both equations. When we subtract the equations, the term in n will disappear. > 3m+5n=21 multiply along by 2 6m+10n = 42 > 4m+10n=38 4m+10n = 38 ------------- subtract 2m = 4 divide by 2 m = 2 Return to one of the original equations and plug in m = 2 3 x 2 + 5n = 21 so 5n = 15 and n = 3 So m=2 and n=3. Try these values in the second equation: 4 x 2 + 10 x 3 = 38 38 = 38 Yep! So these values are correct. > > -2x+9y=5 multiply by 4 -8x+36y = 20 > 8x-36y=-20 8x-36y = -20 If we add these equations both x and y terms go out. If we look closely, we can see that in fact the two equations are really the same equation. If you multiply the second equation by -1 throughout, you get the first equation. In this case there is no single point (x,y) satisfying the equations. We have a 'solution set' which is all points on the line -2x+9y=5. So there are an infinite number of solutions. > x+5y-2z = 5 > 2x+y-z = 4 These are the equations of two planes, and the solution set is the line of intersection of the planes. If you are familiar with 3 dimensional work, we first write the equations in terms of any common point that we like to choose. For example the point (1,0,-2) lies on both planes, so we could write the equations: (x-1) + 5(y) - 2(z+2) = 0 2(x-1) + y - (z+2) = 0 Now using determinants the equation of the line can be written (x-1) -y (z+2) ----- = ----- = ----- |5 -2| |1 -2| |1 5| |1 -1| |2 -1| |2 1| (x-1) -y (z+2) ----- = ----- = ------ -3 3 -9 (x-1) y (z+2) ----- = ----- = ------ = k 1 1 3 So x = 1 + k y = k z = -2 + 3k We can check for any other point on the common line of the two planes. Try k = 1. This gives x=2, y=1, z=1. If you put these values in the equations of the planes you have x+5y-2z = 5 2+5-2 = 5 Yes OK 2x+y- z = 4 4+1-1 = 4 Yes OK So the expressions for x, y, z in terms of k (any value you like) will give the solution set for these equations. Again there are an infinite number of points that lie on the line of intersection of the planes. This last question was a lot more advanced than the earlier ones, so I expect they just wanted you to realize that two equations with three unknowns, (x,y,z) will not give you a single solution. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Date: 10/18/2002 at 13:48:19 From: Judy Foster Subject: Solutions to Systems of Equations Dear Dr. Math: In your answer for 3m + 5n = 21 4m + 10n = 38 you suggested using 2 in the first equation. Should it have been -2 in order to cancel out 10n? I'm in college and learning myself and wanted to be sure I was thinking correctly. Thank you.
Date: 10/18/2002 at 23:28:04 From: Doctor Peterson Subject: Re: Solutions to Systems of Equations Hi, Judy. In this example, you are told to multiply the first equation by 2, and then to SUBTRACT the equations. That is the same as multiplying by -2 and ADDING, which is what you want to do. So your thinking is fine. In fact, when I work these out on paper I usually indicate each step I do by showing the numbers I multiply by before adding; this can be taken instead as the positive multiplier together with a sign indicating whether to add or subtract: 3m + 5n = 21 \ 2 \ 2m = 4 m = 2 / /-1 4m + 10n = 38 Here I chose to multiply the first by 2 and the second by -1 (that is, to subtract the second) so I would get a positive coefficient of m; but you could just as well have used -2 and 1 to get -2m=-4. I drew a line from each equation to the place where I planned to put the new combined equation, and wrote the multipliers next to them to remind me what to multiply by; then I multiplied and subtracted in one step to save writing. But that trick is just a little extra for you. The answer to your question is, you're doing fine! If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math
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