Math Problem Statement

If xx and yy are in direct proportion and yy is 10 when xx is 2, find yy when xx is 6.

Solution

If x2x^2 and y2y^2 are in direct proportion, we can express the relationship as:

x2y2orx2y2=kx^2 \propto y^2 \quad \text{or} \quad \frac{x^2}{y^2} = k

where kk is the constant of proportionality.

We are given that y2=10y^2 = 10 when x2=2x^2 = 2. Substituting these values into the equation to find kk:

210=kk=15\frac{2}{10} = k \quad \Rightarrow \quad k = \frac{1}{5}

Now, we are asked to find y2y^2 when x2=6x^2 = 6. Using the same relationship:

6y2=15\frac{6}{y^2} = \frac{1}{5}

Solving for y2y^2:

y2=6×5=30y^2 = 6 \times 5 = 30

Thus, when x2=6x^2 = 6, y2=30y^2 = 30.

Would you like more details or further explanations?

Here are some related questions for further exploration:

  1. What is the constant of proportionality in general terms for this kind of problem?
  2. How does the solution change if the relationship were inverse instead of direct?
  3. Can you determine yy instead of y2y^2 given the same conditions?
  4. How would the equation look if we introduced a constant multiplier in x2x^2?
  5. What happens to y2y^2 if x2x^2 is doubled?

Tip: In problems involving direct proportion, once you know the proportionality constant, you can solve for unknowns by multiplying or dividing appropriately.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Direct Proportion
Algebra

Formulas

x^2 ∝ y^2 (Direct Proportion Relationship)
k = x^2 / y^2 (Proportionality Constant)

Theorems

Direct Proportionality Theorem

Suitable Grade Level

Grades 6-8