Question Video: Applying the Properties of Proportions to Complete Equations Mathematics • 7th Grade

Given that (π‘₯ + 𝑦)/14 = (𝑦 + 𝑒)/10 = (𝑒 + π‘₯)/4, find the value of the missing denominator in the equation (π‘₯ βˆ’ 𝑒)/4 = (𝑦 βˆ’ π‘₯)/οΌΏ.

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Video Transcript

Given that π‘₯ plus 𝑦 over 14 is equal to 𝑦 plus 𝑒 over 10, which is equal to 𝑒 plus π‘₯ over four, find the value of the missing denominator in the equation π‘₯ minus 𝑒 over four is equal to 𝑦 minus π‘₯ over what.

We’re given three rational expressions containing various combinations of the unknowns π‘₯, 𝑦, and 𝑒 and told that the three expressions are equal. We can express these in terms of proportions and say that π‘₯ plus 𝑦 is to 14 as 𝑦 plus 𝑒 is to 10. And 𝑦 plus 𝑒 is to 10 as 𝑒 plus π‘₯ is to four. And what this tells us is that for some constant 𝑐, π‘₯ plus 𝑦 multiplied by 𝑐 is 14, 𝑦 plus 𝑒 multiplied by 𝑐 is 10, and 𝑒 plus π‘₯ multiplied by 𝑐 is four. And let’s label these equations one, two, and three. Now, given the information we have, remember we’re looking for the missing denominator in the equation π‘₯ minus 𝑒 over four is equal to 𝑦 minus π‘₯ over an unknown.

So we want to find a value such that the proportion of π‘₯ minus 𝑒 to four is equal to that of 𝑦 minus π‘₯ to the unknown denominator. Now, as things stand, none of our three expressions contain either π‘₯ minus 𝑒 or 𝑦 minus π‘₯, which are the two numerators in the equation we’re trying to solve. But going back to our three equations, if we subtract equation two from equation one, we have 𝑐 multiplied by π‘₯ plus 𝑦 minus 𝑐 multiplied by 𝑦 plus 𝑒 is equal to 14 minus 10. And now distributing our parentheses, we have 𝑐π‘₯ plus 𝑐𝑦 minus 𝑐𝑦 minus 𝑐𝑒 is four. 𝑐𝑦 minus 𝑐𝑦 is equal to zero. And so we have 𝑐π‘₯ minus 𝑐𝑒 is equal to four. Taking out the common factor of 𝑐, this gives us 𝑐 multiplied by π‘₯ minus 𝑒 is equal to four. And so we now have one of the terms in the given equation.

Now making some space, if we look back at our three equations again, subtracting equation three from equation two, we have 𝑐 multiplied by 𝑦 plus 𝑒 minus 𝑐 multiplied by 𝑒 plus π‘₯ is equal to 10 minus four. And distributing our parentheses, this gives us 𝑐𝑦 plus 𝑐𝑒 minus 𝑐𝑒 minus 𝑐π‘₯ is equal to six. Subtracting 𝑐𝑒 from itself gives us zero. And so we’re left with 𝑐𝑦 minus 𝑐π‘₯ is equal to six. Now, taking the common factor of 𝑐 outside some parentheses, we have 𝑐 multiplied by 𝑦 minus π‘₯ is equal to six.

So now we notice that 𝑦 minus π‘₯ is the numerator on the right-hand side of the given equation. Now making some room so that we can compare, in terms of proportions, what we found tells us that π‘₯ minus 𝑒 is to four as 𝑦 minus π‘₯ is to six. The missing denominator in the equation π‘₯ minus 𝑒 over four is equal to 𝑦 minus π‘₯ over what is therefore six.

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