Christina X
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- Sep 22, 2011
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solve the system by elimination.
-4x - 5y - z = 18
-2x - 5y - 2z = 12
-2x + 5y + 2z = 4
-4x - 5y - z = 18
-2x - 5y - 2z = 12
-2x + 5y + 2z = 4
Last edited:
I NEED HELP with solving systems of three equations w/ elimination
- Thread starter Christina X
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heartywaffles
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- Sep 28, 2011
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Just in case you still need help after Denis's post, I'll offer further explanation of how to do this problem. Working off of Denis's post, we have one new equation. Now, we need a second. You can combine together any combination of the three equations (1&2, 1&3, and 2&3). Denis combined 2&3, so we have to combine another two. Assuming we're eliminating the z-column/variable, let's combine 1&2.
-4x - 5y - z = 18
-2x - 5y - 2z = 12
You can't just add these together, because it wouldn't eliminate a variable (in this case, z). First, you have to make the coefficients of the variable you're eliminating the same. Since we're doing z, we need both coefficients to be 2. So, you multiply the first equation by 2, making it -8x -10y -2z = 36. You then have to subtract the two equations (i.e. switch all the signs of the second equation, and add them together):
-8x -10y -2z = 36
-2x -5y -2z = 12
------------------
-6x -5y = 24
You now have a new set of equations with only two variables, as opposed to the original three. Sort of. This particular problem is special in that the combination of 2&3 eliminates both the y and z variables, so you can easily solve for x. Normally, you'd have to take the two new equations and again eliminate a variable, until you had only one, and solve from there.
Once you know the values of x and y, you just plug them into one of your original equations, and solve for z. Finally, check another of the original equations you have with your three values of x, y, and z to make sure everything checks out, and your values are right.
Hope this helps.
PS: tkhunny was right in that eliminating the y-variable first is probably easiest (no multiplication, only addition and subtraction), but I wanted to demonstrate the multiplication that's required to eliminate the z-variable, since some of your other problems may need it.
PPS: In other problems of this type, you may have coefficients that you can't easily make one another (for example, 3z and 7z). In that case, you multiply the two coefficients together (mutliply all the terms by 7 in the 3z equation, and by 3 in the 7z equation), and go from there.
-4x - 5y - z = 18
-2x - 5y - 2z = 12
You can't just add these together, because it wouldn't eliminate a variable (in this case, z). First, you have to make the coefficients of the variable you're eliminating the same. Since we're doing z, we need both coefficients to be 2. So, you multiply the first equation by 2, making it -8x -10y -2z = 36. You then have to subtract the two equations (i.e. switch all the signs of the second equation, and add them together):
-8x -10y -2z = 36
-2x -5y -2z = 12
------------------
-6x -5y = 24
You now have a new set of equations with only two variables, as opposed to the original three. Sort of. This particular problem is special in that the combination of 2&3 eliminates both the y and z variables, so you can easily solve for x. Normally, you'd have to take the two new equations and again eliminate a variable, until you had only one, and solve from there.
Once you know the values of x and y, you just plug them into one of your original equations, and solve for z. Finally, check another of the original equations you have with your three values of x, y, and z to make sure everything checks out, and your values are right.
Hope this helps.
PS: tkhunny was right in that eliminating the y-variable first is probably easiest (no multiplication, only addition and subtraction), but I wanted to demonstrate the multiplication that's required to eliminate the z-variable, since some of your other problems may need it.
PPS: In other problems of this type, you may have coefficients that you can't easily make one another (for example, 3z and 7z). In that case, you multiply the two coefficients together (mutliply all the terms by 7 in the 3z equation, and by 3 in the 7z equation), and go from there.