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I need two examples of Solve a proportion with a mixed number in one of its numerators. SHOW ALL WORK!!!!!!!!!!!!

User Amir Kirsh
by
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1 Answer

5 votes
5 votes

Answer:

A proportion equation is something like:


(A)/(B) = (x)/(C)

Where A, B, and C are known numbers, and we want to find the value of x.

Now we want two cases where in one of the numerators we have a mixed number, where a mixed number is something like:

1 and 1/3

which actually should be written as:

1 + 1/3

1) a random problem can be:


(1 + 1/3)/(4) = (x)/(5)

We can see that the numerator on the left is a mixed number.

First, let's rewrite the numerator then:

1 + 1/3

we need to have the same denominator in both numbers, so we can multiply and divide by 3 the number 1:

(3/3)*1 + 1/3

3/3 + 1/3 = 4/3

now we can rewrite our equation as:


(4/3)/(4) = (x)/(5)

now we can solve this:


(4/3)/(4) = (4)/(3*4) = (x)/(5) \\\\(1)/(3) = (x)/(5)

now we can multiply both sides by 5 to get:


(5)/(3) = x

Now let's look at another example, this time we will have the variable x in the denominator:


(7)/(12) = (3 + 4/7)/(x)

We can see that we have a mixed number in one numerator.

Let's rewrite that number as a fraction:

3 + 4/7

let's multiply and divide the 3 by 7.

(7/7)*3 + 4/7

21/7 + 4/7

25/7

Then we can rewrite our equation as


(7)/(12) = (25/7)/(x)

Now we can multiply both sides by x to get:


(7)/(12)*x = (25)/(7)

Now we need to multiply both sides by (12/7) to get:


x = (25)/(7)*(12)/(7) = 300/49

User Cjo
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