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If (x) varies directly as (y) and (x = 150) when (y = 5), find:

a) (xy = 750)
b) (x = 30y)
c) (x + y = 155)
d) (x/y = 30)

User Eliasdx
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1 Answer

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Final answer:

In direct variation where x = 30y, solutions for the conditions xy = 750, x + y = 155, and ratios x = 30y or x/y = 30 affirm the direct relationship, yielding x = 150 and y = 5 whenever these conditions are applied.

Step-by-step explanation:

When x varies directly as y, there exists a constant k such that x = ky. Given that x = 150 when y = 5, we find the constant using the equation 150 = k × 5. Solving for k, we get k = 30. This constant allows us to solve for other relations involving x and y.

  • a) To find x when the product xy = 750, we substitute k into the direct variation equation x = 30y and set it equal to 750: 30y × y = 750. Solving for y, we find that y = 5 and x = 150.
  • b) When x = 30y, this is simply the direct variation equation we've already established, meaning this relationship is always true for any values of x and y that maintain the variation.
  • c) For x + y = 155, substitute x = 30y into the equation: 30y + y = 155. This simplifies to 31y = 155. Dividing both sides by 31, we find that y = 5 and therefore x = 150.
  • d) The ratio x/y = 30 reflects the constant of variation; thus, this equation is also always true for any set of values x and y that maintain the direct variation.

User Gabe
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