Question

Three times the square of a number is greater than a second number. The square of the second number increased by 5 is greater than the first number. Find inequality equation.

Answer 1

The inequalities representing the relationship between the two numbers are 3x² > y and y² + 5 > x, forming a system of inequalities.

Step-by-step explanation:

The problem describes two inequalities involving the squares of two numbers. Let's denote the first number as x and the second number as y. The two inequalities are described as follows:

1. Three times the square of the first number (x) is greater than the second number (y), which can be written as 3x² > y.
2. The square of the second number (y) increased by 5 is greater than the first number (x), which looks like y² + 5 > x.

Together, these form a system of inequalities that defines the relationship between x and y. These inequalities must be true at the same time for the given numbers.

To solve this problem, we can translate the given information into mathematical expressions and inequalities. Let's assume that the first number is 'x' and the second number is 'y'.

According to the problem statement, we have two inequalities:

3x² > y

y² + 5 > x

To simplify the first inequality, we can rewrite it as:

3x² - y > 0

Now, we can create an inequality equation by multiplying the second inequality by 3:

3(y² + 5) > 3x

Simplifying further, we get :

3y² + 15 > 3x

So, the inequality equation is:

3x² - y > 0

3y² + 15 > 3x