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

2 Answers
May 23, 2017

Answer:

#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.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#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!

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#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#

May 23, 2017

Answer:

#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#