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

A proportion equation is something like:

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:

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:

now we can solve this:

now we can multiply both sides by 5 to get:

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

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

Now we can multiply both sides by x to get:

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

what? ?

what type of question is thiss

step-by-step explanation:

the 8 in 8,420 is in the thousands place so it's value is 8 thousand.

the 8 in 6,870 is in the hundreds place so it's value is 8 hundred.

divide 8000 by 800 to find the answer:

8000 / 800 = 10

the 8 in 8,420 is 10 times greater than the 8 in 6,870