How do you solve #(2(x+7))/(x+4)-2=(2x+20)/(2x+8)#?

2 Answers
Oct 7, 2017

Answer:

#x=-4#

Explanation:

#(2(x+7)-2(x+4))/(x+4)=(2(x+10))/(2(x+4))#
#(2x+14-2x-8)/(cancel(color(red)(x+4)))=(cancel(2)(x+10))/(cancel(2)*(cancel(color(red)(x+4)))#
#cancel(color(blue)(2x-2x))+14-8=x+10#
#6=x+10#
#x=-4#

Oct 7, 2017

Answer:

A lot of detail using first principles so that you understand where everything is coming from. Shortcut methods and jumping steps is much faster.

#x=-4#

Explanation:

#color(blue)("Preamble")#

A fractions structure is such that you have:

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

You can not directly add or subtract the 'counts' unless the size indicator are the same.

So either change them to match before you add or subtract or apply some form of adjustment afterwards. The afterwards bit is what is happening in division by a fraction where you invert and multiply.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Answering your question")#

Lets see if we can recognise any common values in the denominators and take it from there.

Note that #2x+8# is the same as #2(x+4)#. This occurs on the right side so it is a good start.

#(2(x+7))/(x+4)-2=(2x+20)/(2(x+4))#

Can we 'get rid of' the 2 from #2(x+4)#

Notice that the numerator has even numbers. We can factor out the 2 from that as well.

#(2(x+7))/(x+4)-2=(2(x+10))/(2(x+4))#

Now we can cancel those 2's out

#(2(x+7))/(x+4)-2=(cancel(2)^1(x+10))/(cancel(2)^1(x+4))#

Next we can 'force' the -2 to have the same denominator as all the rest. Multiply by 1 and you do not change the value. However, 1 comes in many forms.

#color(green)([ (2(x+7))/(x+4)] -[2color(red)(xx1)]=(x+10)/(x+4))#

#color(green)([ (2(x+7))/(x+4)] -[2color(red)(xx(x+4)/(x+4))]=(x+10)/(x+4))#

#[ (2(x+7))/(x+4)]-[(2(x+4))/(x+4)]=(x+10)/(x+4)#

#[ (2x+14)/(x+4)]-[(2x+8)/(x+4)]=(x+10)/(x+4)#

All the denominators are the same. So the equation is equally true if we just totally forget about the denominators. A purist would say: multiply both sides by (x+4). This will give the same result!

#[color(red)(2x+14)] -[color(green)(2x+8)]=x+10#

#color(green)(color(red)(2x)-2xcolor(red)(+14)-8)color(white)("d.d")=x+10#

#color(white)("dddddd")6color(white)("ddddddd.d")=x+10#

Subtract 10 from both sides

#-4=x#

#x=-4#