Math  /  Algebra

QuestionQuestion 1, 11.1.45
Solve the given system of equations. If the system has no solution, say that it is inconsistent. {x2y+3z=72x+y+z=13x+2y2z=3\left\{\begin{array}{rr} x-2 y+3 z= & 7 \\ 2 x+y+z= & -1 \\ -3 x+2 y-2 z= & -3 \end{array}\right.

Studdy Solution
Check the solution (x,y,z)=(1,1,2)(x, y, z) = (-1, -1, 2) by substituting back into the original equations:
1. x2y+3z=7x - 2y + 3z = 7
12(1)+3(2)=7-1 - 2(-1) + 3(2) = 7
1+2+6=7-1 + 2 + 6 = 7
7=77 = 7 (True)
2. 2x+y+z=12x + y + z = -1
2(1)+(1)+2=12(-1) + (-1) + 2 = -1
21+2=1-2 - 1 + 2 = -1
1=1-1 = -1 (True)
3. 3x+2y2z=3-3x + 2y - 2z = -3
3(1)+2(1)2(2)=3-3(-1) + 2(-1) - 2(2) = -3
324=33 - 2 - 4 = -3
3=3-3 = -3 (True)
All equations are satisfied.
The solution to the system of equations is:
(x,y,z)=(1,1,2) (x, y, z) = (-1, -1, 2)

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