SOLUTION:
Let whole number = x
x + 2x^2 = 21
2x^2 + x - 21 = 0
2x^2 + 7x - 6x - 21 = 0
x ( 2x + 7 ) - 3 ( 2x + 7 ) = 0
( 2x + 7 ) ( x - 3 ) = 0
2x + 7 = 0
2x = - 7
x = - 7 / 2
OR
x - 3 = 0
x = 3
ANSWER:
Therefore, as 3 is a whole number while - 7 / 2 isn't a whole number, the whole number must be 3.
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The first one I think not sure
Answer:
Given System of equation:
x-y =6 .....,[1]
2x-3z = 16 ......[2]
2y+z = 4 .......[3]
Rewrite the equation [1] as
y = x - 6 .......[4]
Substitute the value of [4] in [3], we get

Using distributive property on LHS ( i.e,
)
then, we have
2x - 12 +z =4
Add 12 to both sides of an equation:
2x-12+z+12=4+12
Simplify:
2x +z = 16 .......[5]
On substituting equation [2] in [5] we get;
2x+z=2x -3z
or
z = -3z
Add 3z both sides of an equation:
z+3z = -3z+3z
4z = 0
Simplify:
z = 0
Substitute the value of z = 0 in [2] to solve for x;

or
2x = 16
Divide by 2 both sides of an equation:

Simplify:
x= 8
Substitute the value of x =8 in equation [4] to solve for y;
y = 8-6 = 2
or
y = 2
Therefore, the solution for the given system of equation is; x = 8 , y = 2 and z =0