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What is the solution of the system of equations?
{ 2x + 2y + 3z = 6
3x + 5y + 4z = 3
2x + 3y + 4z =  10}
Three equations
 one year ago
 one year ago
What is the solution of the system of equations? { 2x + 2y + 3z = 6 3x + 5y + 4z = 3 2x + 3y + 4z =  10} Three equations
 one year ago
 one year ago

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iiamentertainmentBest ResponseYou've already chosen the best response.0
So 2x + 2y + 3z = 6 2y + 3z = 2x  6 2y = 2x 3z  6 y = 2x 3z  6/2 y = x 3/2z  3 ?
 one year ago

iiamentertainmentBest ResponseYou've already chosen the best response.0
@CalebBeavers
 one year ago

iiamentertainmentBest ResponseYou've already chosen the best response.0
@abb0t please help ..
 one year ago

abb0tBest ResponseYou've already chosen the best response.2
Well, you can use 1 of 2 methods. Substitution, or Elimination. Are you familiar with either of them?
 one year ago

iiamentertainmentBest ResponseYou've already chosen the best response.0
im familar with substitution
 one year ago

abb0tBest ResponseYou've already chosen the best response.2
Ah, I almost want to suggest using gausselimination, but that's beyond algebra 2. Well, I think it might be easier to use elimination actually to eliminate one equation first and then a second to get 2variable equations. Does that make sense?
 one year ago

abb0tBest ResponseYou've already chosen the best response.2
Hint, start by eliminating the x's since equations (1) and (3) both begin with 2. Multiply the first (or second) by 1.
 one year ago

iiamentertainmentBest ResponseYou've already chosen the best response.0
Um im not that good with elimination , butf you walk me through and let me solve it i think ill be ok
 one year ago

abb0tBest ResponseYou've already chosen the best response.2
Yes. I will walk you through it, mate. Start by eliminating one variable in the system of equations. As I said before, It might be easier to eliminate the x from equations (1) and (3) since they both begin with the number 2 [NOTE: Do NOT do anything to eq (2), yet]. You can do this by multiplying either (1) or (3) by 1. Try it and let me know which one you choose and what ur answer is.
 one year ago

iiamentertainmentBest ResponseYou've already chosen the best response.0
1(2x + 2y + 3z)= 6 2x  2y  3z = 6 ?
 one year ago

abb0tBest ResponseYou've already chosen the best response.2
You're 90% correct, when you multiply (1) to the eq, you also have to multiply it to the 6 and you know (1)(6) = 6. making eq(1) 2x2y3z = 6 I'm going to use the term: eq(#) just so I type less :P But notice that eq(1) and eq(2) can be added. And the x' term cancels out nicely. So that you get a new equation with only y's and z's. When you add everything! Including the 6 and 10. Try it.
 one year ago

iiamentertainmentBest ResponseYou've already chosen the best response.0
2x + 3y + 4z =  10 2x  2y  3z = 6 y + 1 = 4 y = 5 ?
 one year ago

abb0tBest ResponseYou've already chosen the best response.2
i have to step out for a moment, but I will tell you the next few steps. Next, do the same thing to eliminate x! from eq(2)! To get a second equation with only y's and z's! Notice that now you have TWO system of equations which you can use elimination (again) or substitution (which you said you like) to solve for either y or z! Once you have your y or z value, plug it back into the original 2 system equation to solve for either y or z (which ever variable you chose to solve for). Once you solved for the 2nd variable you should have TWO variables: y AND z! Now plug in everything into ANY of the equations of the 3systems and solve for x! You can check by plugging everything in and you should get either 6, 3, or 10! If you are correct, you should get those answers!
 one year ago

iiamentertainmentBest ResponseYou've already chosen the best response.0
@blondie16 can you help , im confused
 one year ago

LogicalAppleBest ResponseYou've already chosen the best response.2
You are certainly on the right track. You seem to have forgotten the 'z' variable when adding "4z" and "3z". This gives you a z. So your equation becomes: y + z = 4
 one year ago

iiamentertainmentBest ResponseYou've already chosen the best response.0
Ok thats what i did wrong ok cool, then what
 one year ago

LogicalAppleBest ResponseYou've already chosen the best response.2
What @abb0t suggested next was to use elimination once more. This time with eq(1) and eq(2). The focus here is to, again, eliminate the 'x' variable. equation 1: 2x + 2y + 3z = 6 equation 2: 3x + 5y + 4z = 3 If you multiply eq(1) by 3/2 you will be able to then add it to eq(2) to cancel out the 'x' again. What do you obtain when you multiply eq(1) through by 3/2 ?
 one year ago

iiamentertainmentBest ResponseYou've already chosen the best response.0
3/2(2x + 2y + 3z) = 6 3x  3y  4.5z = 9 3x + 5y + 4z = 3 2y .5z = 6 ?
 one year ago

iiamentertainmentBest ResponseYou've already chosen the best response.0
@LogicalApple
 one year ago

LogicalAppleBest ResponseYou've already chosen the best response.2
The 9 + 3 should be a 12 so 2y  0.5z = 12
 one year ago

iiamentertainmentBest ResponseYou've already chosen the best response.0
i thoue were subtractingnot adding
 one year ago

LogicalAppleBest ResponseYou've already chosen the best response.2
3x + 3x = 0 3y + 5y = 2y 4.5z + 4z = 0.5z 9 + 3 = 12 You added everything else correctly but the last numbers
 one year ago

iiamentertainmentBest ResponseYou've already chosen the best response.0
Oh Ok , Next Step?
 one year ago

LogicalAppleBest ResponseYou've already chosen the best response.2
In any event you are now down to a system of 2 equations with 2 variables y + z = 4 2y  0.5z = 12 You can use substitution at this point.
 one year ago

LogicalAppleBest ResponseYou've already chosen the best response.2
Notice how the elimination method eliminates one of the variables, leaving you with only two to deal with. From here you could set y = 4  z in the first equation and substitute this value of y into the second equation and solve for z
 one year ago

iiamentertainmentBest ResponseYou've already chosen the best response.0
2x + 2(4 z) + 3z = 6 like that
 one year ago

LogicalAppleBest ResponseYou've already chosen the best response.2
No, I meant the second equation listed here: y + z = 4 2y  0.5z = 12 Not any of the original equations
 one year ago

iiamentertainmentBest ResponseYou've already chosen the best response.0
2(4 z)  0.5z = 12 8  2z  .5z = 12 8 + 3z = 12 3z = 20 z = 6.67
 one year ago

LogicalAppleBest ResponseYou've already chosen the best response.2
2z  0.5z = 2.5z So you should have 8  2.5z = 12
 one year ago

iiamentertainmentBest ResponseYou've already chosen the best response.0
8  2.5z = 12 2.5z = 20 z = 8
 one year ago

LogicalAppleBest ResponseYou've already chosen the best response.2
And we know: y + z = 4
 one year ago

iiamentertainmentBest ResponseYou've already chosen the best response.0
y + (8) = 4 y = 4
 one year ago

iiamentertainmentBest ResponseYou've already chosen the best response.0
@LogicalApple
 one year ago

LogicalAppleBest ResponseYou've already chosen the best response.2
Yes, that is also correct. Now you know what y and z are. Use these values in any one of the original three equations and you will obtain the value for x.
 one year ago

iiamentertainmentBest ResponseYou've already chosen the best response.0
2x + 2(4) + 3(8) = 6 2x + 8  24 = 6 2x  16 = 6 2x = 10 x = 5 y = 4 z = 8 @LogicalApple
 one year ago

LogicalAppleBest ResponseYou've already chosen the best response.2
Fantastic. (5, 4, 8) is the correct solution
 one year ago
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