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Question Video: Using Proportions to Find Unknowns Mathematics • Third Year of Preparatory School

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If π‘Ž/7 = 𝑏/4 = 𝑐/14 = (6π‘Ž βˆ’ 7𝑏 + 2𝑐)/3π‘₯, find the value of π‘₯.


Video Transcript

If π‘Ž over seven equals 𝑏 over four equals 𝑐 over 14, which equals six π‘Ž minus seven 𝑏 plus two 𝑐 over three π‘₯, find the value of π‘₯.

This rather strange-looking equation links the proportion between a number of variables and constants. It tells us that the proportion of π‘Ž to 𝑏 [seven] is the same as the proportion of 𝑏 to four. These in turn are the same of the proportion of 𝑐 to 14 and the proportion in our final fraction.

Now, what this also tells us is that we can extract any two expressions and set them equal to each other to solve for π‘₯. But we do have a problem. We’re certainly going to be interested in our final fraction since that contains the variable π‘₯ that we’re trying to find the value of. But it also contains three further variables. Our job will be to make the numerator in terms of just one of our variables, π‘Ž, 𝑏, or 𝑐. It doesn’t matter which we choose. We’ll get the same result either way.

We’re going to in fact choose 𝑏, and then we’re going to choose the first pair of expressions π‘Ž over seven equals 𝑏 over four. By multiplying both sides of this equation by seven, we get an equation for π‘Ž purely in terms of 𝑏. We’ll then be able to replace π‘Ž in the numerator of our final fraction and eventually 𝑐 as well to give us an expression just in terms of 𝑏. When we times by seven, we get π‘Ž equals seven 𝑏 over four.

Let’s now take the second pair of expressions: 𝑏 over four equals 𝑐 over 14. We need to find a way to replace 𝑐 with some expression in terms of 𝑏. So we’re going to multiply both sides of this equation by 14. So 𝑐 equals 14𝑏 over four. Now we’re not going to simplify any fractions just yet because we might find at the end we’re going to need to add or subtract fractions. And that’s easier if their denominators are equal.

Let’s now consider the numerator of the expression we are interested in. It’s six π‘Ž minus seven 𝑏 plus two 𝑐. Replacing π‘Ž with seven 𝑏 over four and 𝑐 with 14𝑏 over four, we find that this is equivalent to six times seven 𝑏 over four minus seven 𝑏 plus two times 14𝑏 over four. We distribute the six over the expression seven 𝑏 over four to give us 42𝑏 over four.

Next, we rewrite seven 𝑏 with a denominator of four. So it’s 28𝑏 over four. Finally, two times 14𝑏 over four is 28𝑏 over four. Then we notice that negative 28𝑏 over four plus 28𝑏 over four is zero. So the numerator of our fraction is in fact just 42𝑏 over four or equivalently 21𝑏 over two.

And so our final fraction can be expressed as 21𝑏 over two over three π‘₯. In fact, this is equivalent to multiplying 21𝑏 over two by one over three π‘₯. So it’s the same as 21𝑏 over six π‘₯.

Now remember, we can set this fraction equal to any of our other three fractions. And since we’re trying to eliminate 𝑏 so we have an equation purely in terms of π‘₯, let’s set this equal to 𝑏 over four. So 𝑏 over four equals 21𝑏 over six π‘₯. Now in fact, we know 𝑏 can’t be equal to zero since this equation would make no sense. But in fact, we also know that the coefficient of 𝑏 on the left-hand side is one-quarter and on the right is 21 over six π‘₯. So we can either divide through by 𝑏 or equivalently set the coefficients equal to zero. So one-quarter equals 21 over six π‘₯. Then we can simplify the right-hand side to one-quarter is seven over two π‘₯.

And we might then next take the reciprocal of both sides so that four equals two π‘₯ over seven. We didn’t need to perform this step, but it makes the solving process easier since our variable is on the numerator of a fraction. To solve for π‘₯, let’s multiply through by seven. So 28 equals two π‘₯. Finally, we need to divide through by two, and that gives us π‘₯ equals 14. So given the relation in our question, we have found that the value of π‘₯ is 14.

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