Question 33348This question is from textbook algebra for college students
: I need to solve this system of linear equations and I am having a hard time. Please help.
x-3y+2z =-11
2x-4y+3z =-15
3x-5y-4z =5
This question is from textbook algebra for college students
Found 2 solutions by mukhopadhyay, sarah_adam:
Answer by mukhopadhyay(490)   (Show Source):
You can put this solution on YOUR website! There are many ways to solve such equations. You can either use substitution, or elimination, or Cramer's Rule to solve this.
I would Use Cramer's Rule to do so:
D = Determinant of [(1,-3,2);(2,-4,3);(3,-5,-4)] = 1(31)-2(22)+3(-1) = -16;
D(x) = Determinant of [(-11,-3,2);(-15,-4,3);(5,-5,-4)]
= -11(31)+15(22)+5(-1) = -341+330-5 = -16;
D(y) = Determinant of [(1,-11,2);(2,-15,3);(3,5,-4)]
= 1(45)-2(34)+3(-3) = 45-68-9 = -32;
D(z) = Determinant of [(1,-3,-11);(2,-4,-15);(3,-5,5)]
= 1(-95)-2(-70)+3(1) = -95+140+3 = 48;
x = D(x)/D = -16/-16 = 1;
y = D(y)/D = -32/-16 = 2;
z = D(z)/D = 48/-16 = -3;
Answer: (x,y,z) = (1,2,-3)
Answer by sarah_adam(201) (Show Source):
You can put this solution on YOUR website! x-3y+2z =-11 --- eq 1
2x-4y+3z =-15 --- eq 2
3x-5y-4z =5 --- eq 3
from eq 1 we get x = 3y - 2z -11
substituting the value of x in eq 2
2(3y - 2z -11) - 4y +3z = -15
6y - 4z -22 -4y +3z = -15
2y - z = 22 - 15 = 7
2y - z = 7 ----eq 4
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substitute value of x again in eq 3
3(3y - 2z -11)-5y-4z = 5
9y - 6z -33-5y-4z = 5
4y-10z = 33 + 5 = 38
4y-10z = 38 --- eq 5
-------------------------------------------
2y - z = 7 ----eq 4
4y-10z = 38 --- eq 5
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multiply entire eq 4 with 2
4y - 2z = 14 --- eq 4 (after multipling the entire equation with 4)
subtracting eq 5 and eq 4
4y-10z = 38
4y - 2z = 14
(- ) (+ ) (-) changing signs
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-8z = 24
z = -(24/8)= -3
substituting the value of Z in eq 4
2y - z = 7
2y -(-3)=7
2y +3 = 7
2y = 7-3 = 4
y = 4/2 = 2
substitue the values of y and z in : x = 3y - 2z -11
x = 3(2)-2(-3)-11
x = 6 +6-11 = 12 -11 = 1
x = 1 ; y = 2 ; z = -3
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