Math Problem Statement

Which of the following points is a solution to the inequality y + 3 > 5/2(x - 6)?

Solution

We are given the inequality:

y+3>52(x6)y + 3 > \frac{5}{2}(x - 6)

We need to determine which of the following points satisfies this inequality.

Steps:

  1. Rewrite the inequality by isolating yy: y>52(x6)3y > \frac{5}{2}(x - 6) - 3 Simplifying the right side: y>52x153y > \frac{5}{2}x - 15 - 3 y>52x18y > \frac{5}{2}x - 18

Now, we will substitute each point into this inequality to see which one satisfies it.


Option 1: (8,0)(8, 0)

Substitute x=8x = 8 and y=0y = 0 into the inequality:

0>52(8)180 > \frac{5}{2}(8) - 18 0>20180 > 20 - 18 0>20 > 2

This is false.


Option 2: (10,4)(10, 4)

Substitute x=10x = 10 and y=4y = 4 into the inequality:

4>52(10)184 > \frac{5}{2}(10) - 18 4>25184 > 25 - 18 4>74 > 7

This is false.


Option 3: (0,15)(0, -15)

Substitute x=0x = 0 and y=15y = -15 into the inequality:

15>52(0)18-15 > \frac{5}{2}(0) - 18 15>18-15 > -18

This is true.


Conclusion:

The only point that satisfies the inequality is (0,15)(0, -15).

So, the correct answer is (0,15)(0, -15).


Would you like further clarification or details? Here are 5 follow-up questions related to this problem:

  1. How do we determine if an inequality is strict or non-strict?
  2. How would the solution change if the inequality was y+352(x6)y + 3 \geq \frac{5}{2}(x - 6)?
  3. What is the geometric interpretation of this inequality on a coordinate plane?
  4. How do we handle fractional coefficients in inequalities when solving for yy?
  5. What happens to the inequality sign when multiplying or dividing by a negative number?

Tip: When solving inequalities involving fractions, it's helpful to first isolate the variable of interest (e.g., yy) and then substitute the values to check for correctness.

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Math Problem Analysis

Mathematical Concepts

Algebra
Linear Inequalities
Substitution

Formulas

y > (5/2)(x - 6) - 3

Theorems

Properties of inequalities
Substitution method for inequalities

Suitable Grade Level

Grades 9-10