14) x=0, y=3, z=-2
Solution Set (0,3,-2)
16) x=1, y=1 and z=1
Solution set = (1,1,1)
20) x = -263/31, y=164/31 ,z=122/31
Solution set (-263/31, 164/31 ,122/31)
Step-by-step explanation:
14)

Rearranging and solving:

Eliminate y:
Adding eq(1) and eq(2)

Multiply eq(2) with 2 and add with eq(3)

Eliminate x:
Subtract eq(4) and eq(5)

So, value of z = -2
Now putting value of z in eq(2)

So, value of y = 3
Now, putting value of z and y in eq(1)

So, value of x = 0
So, x=0, y=3, z=-2
S.S(0,3,-2)
16)

Let:

Eliminating y:
Adding eq(1) and (2)

Multiply eq(1) by 2 and add with eq(3)

Now eliminating z in eq(4) and eq(5) to find value of x
Subtracting eq(4) and eq(5)

So, value of x = 1
Putting value of x in eq(4) to find value of x:

So, value of z = 1
Putting value of x and z in eq(2) to find value of y:

So, x=1, y=1 and z=1
Solution set = (1,1,1)
20)

Let:

Solving:
Eliminating z :
Adding eq(1) and eq(3)

Multiply eq(1) with 2 and eq(2) with 5 and add:

Eliminate y:
Multiply eq(4) with 18 and eq(5) with 5 and subtract:

So, value of x = -263/31
Putting value of x in eq(4)

Now putting x = -263/31 and y=164/31 in eq(1) and finding z:
We get z=122/31
So, x = -263/31, y=164/31 ,z=122/31
Solution set (-263/31, 164/31 ,122/31)
Keywords: Solving system of Equations
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