Solve the following:
3x-5y+2z=19
5x+2y-3z=-8
-2x+3y+5z=7
To solve for systems of 3 linear equation in 3 variable, we use Substitution
Method, Elimination Method, Gaussian Reduction method, Cramer's rules.
I choose Elimination Method
equation 1 : 3x-5y+2z=19
equation 2 : 5x+2y-3z=-8
equation 3 : -2x+3y+5z=7
Let us Eliminate x in equation 1 and equation 2.
3x-5y+2z=19
5x+2y-3z=-8
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Since we cannot eliminate 3x and 5x by adding, we will find a certain
number to that will be multiplied to the two equations.
Multiply 5 to 3x-5y+2z=19 and -3 to 5x+2y-3z=-8
5(3x-5y+2z=19)
-3(5x+2y-3z=-8)
_________________
15x - 25y +10z = 95
-15x - 6y + 9z = 24 Add
____________________
- 31y + 19z = 119 Equation 4
Now eliminate x in equation 1 and equation 3. Multiply 2 to 3x-5y+2z=19
and 3 to -2x+3y+5z=7
2(3x-5y+2z=19)
3(-2x+3y+5z=7)
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6x - 10y + 4z = 38
-6x + 9y + 15z = 21
____________________
-y + 19z = 59 Equation 5
Then using equation 4 and equation 5 eliminate another variable.
I choose z then solve for y.
-31y + 19z = 119
-1( - y + 19z = 59 )
______________________
-31y + 19z = 119
y - 19z = -59
________________
-30y = 60
y = -2
The value of y will be substituted to either equation 5 or 4
I choose equation 5
-y + 19z = 59, y = -2
-(-2) + 19z = 59
19z = 59 - 2
19z = 57 Divide 19 both sides
z = 3
The Substitute y = -2, z = 3 to equation 1 or 2 or 3.
I choose equation 3.
-2x + 3y + 5z = 7, y = -2, z = 3
-2x + 3(-2) + 5(3) = 7
-2x - 6 + 15 = 7
-2x = -2
x = 1
Checking:
equation 1 : 3x-5y+2z=19, x = 1, y = -2, z = 3
3(1)-5(-2)+2(3)=19
3 + 10 + 6 = 19
19 = 19 ---------->True
equation 2 : 5x+2y-3z=-8, x = 1, y = -2, z = 3
5(1)+2(-2)-3(3)=-8
5 - 4 - 9 = -8
-8 = -8 ------------>True
equation 3 : -2x+3y+5z=7, x = 1, y = -2, z = 3
-2(1)+3(-2)+5(3)= 7
-2 -6 + 15 = 7
7 = 7 ----------> True
Therefore, the solution is x = 1, y = -2 and z = 3
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