Direct Variation Question 1 with Solutions

Yesterday, I went over direct variation, and I gave you three direct variation problems to try on your own. You can see that post here: Direct Variation

Today I would like to give a solution to the first of those three problems. I will give solutions to the other two problems throughout this week.

Example: If y = kx and y = 7 when x = 11, then what is y when x = 33?

Try to solve the problem yourself before checking the solutions below.

Solutions: 

(1) We are given that y = 7 when x = 11, so that 7 = k(11), or k = 7/11. Therefore y = 7x/11. When x = 33, we have y = 7(33)/11 = 21.

(2) Since y varies directly as x,  y/x  is a constant. So, we get the following ratio:  7/11 = y/33. Cross multiplying gives 231 = 11y, so that y = 21.

(3) The graph of y = f(x) is a line passing through the points (0,0) and (11, 7) The slope of this line is  (7 – 0)/(11 – 0) = 7/11. Writing the equation of the line in slope-intercept form we have y = 7/11 x. As in solution 1, when x = 33, we have y = 7(33)/11 = 21.

(4) To get from x = 11 to x = 33 we multiply x by 3. So we have to also multiply y by 3. We get 3(7) = 21.

More Problems with Explanations

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