Question

How do you solve `(x+2)/6 + (x-2)/10 = 6/5`?

Answer 1

`x=4`

Explanation:

`color(blue)("3 Key points")`

A fraction's structure consists of :

`("count")/("size indicator of what you are counting") ->("numerator")/("denominator")`

You can NOT DIRECTLY add or subtract the counts unless the size indicators are the same.

Multiply by 1 and you do not change the intrinsic value. However, 1 comes in many forms so you can change the way a fraction looks without changing its intrinsic value.
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`color(blue)("Method: ")`

Make all the denominators the same but insuring that the numerators remain proportional to the denominator.

Then just solve for the numerators only. This really works!

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`color(blue)("Answering the question")`

I choose to make the denominators 60:

`color(green)([(x+2)/6color(red)(xx1)] +[(x-2)/10color(red)(xx1)] = [6/5color(red)(xx1)]`

`color(green)([(x+2)/6color(red)(xx10/10)] +[(x-2)/10color(red)(xx6/6)] = [6/5color(red)(xx12/12)]`

`color(white)(.)color(green)([(10x+20)/60] color(white)(..)+color(white)(.)[(6x-12)/60]=[72/60]`

Thus it also true that:

`" "10x+20" "+" "6x-12" "=72`

`16x+8=72`

Subtract 8 from both sides

`16x=64`

Divide both sides by 16

`x=64/16=4`

Answer 2

`x=4`

Explanation:

If you have fractions which are in an equation, you can get rid of them immediately.

Multiply each term by the LCM of the denominators. (In this case `color(blue)(30)`)

If you multiply the whole equation by `30` you do not change the value.

`(color(blue)(30)xx(x+2))/6 + (color(blue)(30)xx(x-2))/10 = (color(blue)(30xx)6)/5`

Now cancel each denominator.

`(color(blue)(cancel30)^5xx(x+2))/cancel6 + (color(blue)(cancel30)^3xx(x-2))/cancel10 = (color(blue)(cancel30^6xx)6)/cancel5`
`5(x+2)+3(x-2) = 6xx6" "larr` no more fractions!

`5x+10 +3x-6 =36" "larr` now solve for `x`

`8x = 36-4`

`8x =32`

`x=4`