The equation given in the image is $2x = 7y = 28$, which can be interpreted as two separate equations:

$2x = 28 \quad \text{and} \quad 7y = 28$

1. Solve $2x = 28$: $x = \frac{28}{2} = 14$

2. Solve $7y = 28$: $y = \frac{28}{7} = 4$

Thus, the solution to the system of equations is $x = 14$ and $y = 4$. Now, we check which of the given options is not a solution:

Option (a) $(7, 2)$:

• For $x = 7$: $2x = 2(7) = 14 \quad (\text{not } 28)$ Thus, option (a) is not a solution.

Checking the other options:

Option (b) $(0, 4)$:

• For $x = 0$: $2x = 2(0) = 0 \quad (\text{not } 28)$ So, (b) is also not a solution.

Option (c) $(2, 7)$:

• For $x = 2$: $2x = 2(2) = 4 \quad (\text{not } 28)$ Thus, (c) is not a solution.

Option (d) $(14, 0)$:

• For $x = 14$: $2x = 2(14) = 28$
• For $y = 0$: $7y = 7(0) = 0 \quad (\text{not } 28)$ Thus, (d) is also not a solution.

However, (a) is the first incorrect one when compared directly to the expected solutions.

Let me know if you want to dive deeper into the explanation or solve further problems.

Follow-up questions:

1. How would you graph the equation $2x + 7y = 28$?
2. How do you test for solutions in a system of linear equations?
3. Can you generalize this method for a different equation?
4. What are the key steps to solving linear equations in two variables?
5. How would this method change if the equation involved fractions?

Tip:

When solving systems of equations, always check each possible solution by substituting the values of both variables to ensure correctness.