Question

If \(13x + 1 < 2z\) and \(z + 3 = 5y^2\), then

• A. x is necessarily less than y
• B. x is necessarily greater than y
• C. x is necessarily equal to y
• D. None of the above is necessarily true

Hint:

Eliminate z to get one inequality in x and y, then test whether x can be both smaller and larger than y.

Answer 1

The Correct Option is D

Step 1: Eliminate z:
From \(z + 3 = 5y^2\), \(z = 5y^2 - 3\). Then \(13x + 1 < 2z = 10y^2 - 6\), so \(13x < 10y^2 - 7\), i.e. \(x < \frac{10y^2 - 7}{13}\).

Step 2: Can x be less than y?:
Take y = 0, so z = -3, and choose x = -1: 13(-1) + 1 = -12 < -6 = 2z. This is allowed and x < y.

Step 3: Can x be greater than y?:
Take y = -1, so z = 2, and choose x = 0: 13(0) + 1 = 1 < 4 = 2z. This is allowed and x > y.

Step 4: Conclude:
Since x < y is possible, option (2) is not necessary; since x > y is possible, option (1) is not necessary; and x = y is not forced (for example y = 0 with x = -1). So none of (1), (2), (3) is necessarily true. This is option (4).

Final Answer:
The correct option is (4): None of the above is necessarily true