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Given that π₯ plus π¦ over 14 is equal to π¦ plus π over 10, which is equal to π plus π₯ over four, find the value of the missing denominator in the equation π₯ minus π over four is equal to π¦ minus π₯ over what.
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Weβre given three rational expressions containing various combinations of the unknowns π₯, π¦, and π and told that the three expressions are equal.
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We can express these in terms of proportions and say that π₯ plus π¦ is to 14 as π¦ plus π is to 10.
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And π¦ plus π is to 10 as π plus π₯ is to four.
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And what this tells us is that for some constant π, π₯ plus π¦ multiplied by π is 14, π¦ plus π multiplied by π is 10, and π plus π₯ multiplied by π is four.
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And letβs label these equations one, two, and three.
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Now, given the information we have, remember weβre looking for the missing denominator in the equation π₯ minus π over four is equal to π¦ minus π₯ over an unknown.
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So we want to find a value such that the proportion of π₯ minus π to four is equal to that of π¦ minus π₯ to the unknown denominator.
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Now, as things stand, none of our three expressions contain either π₯ minus π or π¦ minus π₯, which are the two numerators in the equation weβre trying to solve.
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But going back to our three equations, if we subtract equation two from equation one, we have π multiplied by π₯ plus π¦ minus π multiplied by π¦ plus π is equal to 14 minus 10.
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And now distributing our parentheses, we have ππ₯ plus ππ¦ minus ππ¦ minus ππ is four.
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ππ¦ minus ππ¦ is equal to zero.
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And so we have ππ₯ minus ππ is equal to four.
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Taking out the common factor of π, this gives us π multiplied by π₯ minus π is equal to four.
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And so we now have one of the terms in the given equation.
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Now making some space, if we look back at our three equations again, subtracting equation three from equation two, we have π multiplied by π¦ plus π minus π multiplied by π plus π₯ is equal to 10 minus four.
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And distributing our parentheses, this gives us ππ¦ plus ππ minus ππ minus ππ₯ is equal to six.
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Subtracting ππ from itself gives us zero.
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And so weβre left with ππ¦ minus ππ₯ is equal to six.
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Now, taking the common factor of π outside some parentheses, we have π multiplied by π¦ minus π₯ is equal to six.
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So now we notice that π¦ minus π₯ is the numerator on the right-hand side of the given equation.
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Now making some room so that we can compare, in terms of proportions, what we found tells us that π₯ minus π is to four as π¦ minus π₯ is to six.
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The missing denominator in the equation π₯ minus π over four is equal to π¦ minus π₯ over what is therefore six.