Solve the system of equations: 2x - 5y + 3z = -1 x + 4y - 2z = 9 x - 2y - 4z = -5
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We have to solve the following equations by elimination
2x-5y+3z=-1 ...(1)
x+4y-2z=9 ...(2)
x-2y-4z=-5 ...(3)
Let's eliminate x and create 2 equations with y and z.
(2) - (3)
=> 4y - 2z + 2y + 4z = 9+5
=> 6y + 2z = 14
=> 3y + z = 7 ...(4)
(1) - 2*(2)
=> 2x - 5y + 3z - 2x - 8y + 4z = -1 - 18
=> -13y + 7z = -19 ...(5)
Now 7*(4) - (5)
=> 21y + 7z + 13y - 7z = 49 + 19
=> 34y = 68
=> y = 2
substitute in (4)
3y + z = 7
=> 6 + z = 7
=> z = 1
substitute y and z in (2)
=> x + 8 - 2 = 9
=> x = 3
Therefore x = 3, y= 2 and z = 1.
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To solve the system of equations:
2x - 5y + 3z = -1...(1)
x + 4y - 2z = 9......(2)
x - 2y - 4z = -5......(3).
(1)-(2)*2 = 2x-5y+3z-2(x+4y-2z) = -1-2*9 = -19.
=> -13y+7z = -19....(4).
(2)-(3): (x+4y-2z) -(x-2y-4z) = 9-(-5) = 14
=> 6y+2z = 14. We divide both sides by (2):
3y+z = 7...(5):
5*7-(4) = 7(3y+z)-(-13y+7z) = 7*7 -(-19) = 68
34y = 68. So y = 68/34 = 2.
Substitute y = 2 in (5): 3y+z= 7. Or z = 7-3y = 7-3*2 = 1. So z = 1.
We put y = 2, z= 1 in (1): 2x-5y+3z = 1. Or 2x-5*2+3*1 = -1.
2x = -1+10-3 = 6. Or 2x= 6, So x = 6/2 = 3.
So x= 3, y = 2 and z = 1.
It would be much more easier using matrix.
Till then, we'll apply elimination method.
We'll eliminate the unknown x from the 2nd and 3rd equations.
(2) - (3):
x+4y-2z-x+2y+4z=9+5
We'll combine and eliminate like terms:
6y + 2z = 14
We'll divide by 2:
3y + z = 7 (4)
We'll eliminate the unknown x from the 1st and 5th equations. For this reason, we'll multiply (3) by -2:
-2x + 4y + 8z = 10 (5)
(1) + (5):
2x - 5y + 3z - 2x + 4y + 8z = -1 + 10
We'll combine and eliminate like terms:
-y + 11z = 9 (6)
3y + z = 7 (4)
We'll multiply (6) by 3:
-3y + 33z = 27 (7)
(7) + (4):
-3y + 33z + 3y + z = 27 + 7
34z = 34
z = 1
3y + z = 7 <=> 3y + 1 = 7
3y = 7 - 1
3y = 6
y = 2
We'll substitute y and z in the 2nd equation:
x+ 8 - 2 = 9
x + 6 = 9
x = 3
The solution of the system, solved by elimination method is (3 ; 2 ; 1).
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