Answer :
The solution of the linear system of three simultaneous equations is presented as follows; x = 2, y = 1 and z = 4
What is a set of simultaneous equation?
Simultaneous system of equations consists of a finite set of equations for which a solution to the equation system is required.
The linear system of three equations can be presented as follows;
13•x - 17•y + 16•z = 73...(1)
-11•x + 15•y + 17•z = 61...(2)
46•x + 10•y - 30•z = -18...(3)
The above system of equations can be solved using common multiples of the coefficients as follows;
Multiply equation (2) by 2 and equation (3) by 3 to get;
2 × (-11•x + 15•y + 17•z) = 2 × 61 = 122
- -22•x + 30•y + 34•z = 122...(4)
3 × (46•x + 10•y - 30•z) = 3 × (-18) = -54
- 138•x + 30•y - 90•y = -54...(5)
Subtracting equation (4) from equation (5) gives;
138•x + 30•y - 90•z - (-22•x + 30•y + 34•z) = -54 - 122 = -176
138•x - (-22•x) + 30•y - 30•y - 90•z - 34•z = -176
160•x - 124•z = -176
40•x - 31•z = 44
[tex] \displaystyle {z = \frac{(44 + 40\cdot x)}{31}}[/tex]
Plugging in the value of z in equation (1) and (2) gives;
1043•x - 527•y + 704 = 73 × 31 = 2236...(6)
Which gives;
[tex] \displaystyle {y = \frac{(1043\cdot x - 1559)}{527}}[/tex]
339•x + 465•y + 748 = 61 × 31 = 1891...(7)
Which gives; [tex] \displaystyle {y = \frac{(381 - 113\cdot x )}{155}}[/tex] which gives;
[tex] \displaystyle { \frac{(1043\cdot x - 1559)}{527}= \frac{(381 - 113\cdot x )}{155}}[/tex]
Therefore; 221216•x - 442432 = 0
x = 442432 ÷ 221216 = 2
x = 2
y = (1043×2 - 1559)÷527 = 1
y = 1
z = (44 + 40×2) ÷ 31 = 4
z = 4
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