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56. CHALLENGE Let a/b and c/d be two distinct rational numbers. Find the rational number that lies exactly halfway between a/b and c/d on a number line. (This algebra 2)

User Dmehro
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Answer: midpoint = (ad + bc)/2bd

Explanation:

The number that lies at the midpoint between these two points will be an average of the two numbers.

To get an average, we add the two numbers and then multiply by one-half (or divide by 2, same thing).

x = (1/2) * ( a/b + c/d)

To add a/b and c/d, we must use the least common denominator (LCD). Since we don't have numerical values, we can only multiply the two denominators together and not worry about whether to 'reduce' the result. So the LCD = b*d

To give a/b a denominator of bd, we must multiple top and bottom both by 'd' and we get a/b ---> ad/bd

To give c/d a denominator of bd, we must multiple top and bottom both by 'b' and we get c/d ---> bc/bd

This gives us x = (1/2)( ad/bd + bc/bd )

Since the two fractions in the second set of parentheses now have the same denominator, we can add them.

x = (1/2) * [ (ad + bc)/bd ]

Now we have two fractions to multiply.

The numerators ---> 1 * (ad + bc) = ad + bc

The denominators ---> 2 * bd = 2bd

So, we get a final answer of

x = (ad + bc)/2bd

midpoint = (ad + bc)/2bd

I hope this helps.

User Aforankur
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