Question

Can someone help me on 1, 2, 3 and 4 plz I really need help

Answer 1

Answer:

NO

YES

NO

NO

Step-by-step explanation:

For a number to be the solution of the equation, the number should satisfy it.

To check if t satisfies, plug the number and compare LHS and RHS.

1) 3z + 4 = 15

Substituting z = -5, we get 3(-5) + 4 = -15 + 4 = -11 $\ne$ 15.

Therefore, -5 is not the solution of the equation.

2) 8 - $\frac{x}{2}$ = 0

Substituting x = 16, we get 8 - $\frac{16}{2} = 8$

⇒ 8 - 8 = 0

Therefore, 16 is a solution of the equation.

3) 4(p + 5) = 60

Substitute p = 12. 4(12 + 5) = 4(17) $\ne$ 60.

Not a solution.

4) 8 - k = -48

If k = -40, then 8 - (-40) = 48

On the RHS we have -48. Not equal. So, it is not a solution.