Problem Solving

Could not get why options A,B,C are incorrect when all are satisfying LHS = RHS.

OA E

If xy + z = x(y + z), which of the following must be true:

1. x = 0 and z = 0
2. x = 1 and y = 1
3. y = 1 and z = 0
4. x = 1 or y = 0
5. x = 1 or z = 0

xy + z = x(y +z)

xy + z = xy + xz

z = xz

z - xz = 0

z(1 - x) = 0.

For the resulting equation to be valid, either z=0 or x=1.

The correct answer is E.

A is incorrect because if z=0, then x can be ANY VALUE.
Thus, it does NOT have to be true that z=0 AND x=0.
For example, it is possible that z=0 and x=1.

B and C are incorrect because -- when the equation is simplified -- it no longer includes y.
The implication is that y can be ANY VALUE.
Thus, it does NOT have to be true that y=1.
For example:
It is possible that x=1 and y=2, proving that B does not have to be true.
It is possible that z=0 and y=2, proving that C does not have to be true.

eitijan wrote:Could not get why options A,B,C are incorrect when all are satisfying LHS = RHS.

OA E

The key word MUST means we need to find the solution that will always satisfy the equation.
Hence we need to solve xy+z = x(y+z) as

xz - z = 0
z(x-1) = 0
Hence z = 0 or x = 1 will always satisfy the equation

The algebra described in the other solutions is probably best, but if you're still stuck, you can always try the answers to see what happens.

For instance, if x = 1, we have

1*y + z = 1*(y + z)

or y + z = y + z, which is true, so x = 1 is a solution.

If z = 0, we have

xy + 0 = x*(y + 0)

or xy = xy, which is also true, so z = 0 is a solution.

Since x = 1 and z = 0 both work independently, E is the only possible answer.