Question 714329
THE FIFTH GRADER WAY:
That number could be in between 40 and 70,
or could be larger than 70.
(It cannot be less than 40 because that would put it close to 40 than to 70).
If it's in between, it should be {{{2/3}}} of the way from 40 to 70,
so you have gone 2 thirds of the way, but you have 1 third of the way to go.
Since that way is {{{70-40=30}}} , {{{1/3}}} of that way is {{{10}}}, and
{{{2/3}}} of that is {{{20}}} .
The number between 40 and 70, that is 10 away from 70, and
{{{20}}} away from {{{40}}} is
{{{40+20=highlight(60)}}}.
If the number is larger than 70, and
going from 40 to that number is twice as far as
going from 70 to that number,
then 70 is half way there, and the distance from 40 to 70,
{{{70=40=30}}} ,
is the same as from {{{70}}} further to the number,
so the number is {{{30}}} past {{{70}}} .
It is {{{70+30=highlight(100)}}}
 
TWO WAYS THAT LOOK MORE LIKE ALGEBRA:
A number {{{x}}} is
{{{abs(x-40)}}} from {{{40}}} and
{{{abs(x-70)}}} from {{{70}}} .
We know that
{{{abs(x-40)=2*abs(x-70)}}}
At this point,
we can just square both sides of the equal sign and worry about eliminating any possible extraneous solution later.
Or we could examine what those absolute values could be and work for each case.
 
Squaring both sides we get
{{{(x-40)^2=2^2*(x-70)^2}}} --> {{{x^2-80x+1600=4(x^2-140x+4900)}}} --> {{{x^2-80x+1600=4x^2-560x+19600}}} --> {{{4x^2-x^2-560x+80x+19600-1600=0}}}  --> {{{3x^2-480+18000=0}}}
Dividing both sides by 3 we get
{{{x^2-160+6000=0}}}
If we are good at factoring, we factor to get
{{{(x-60)(x-100)=0}}} with solutions
{{{highlight(x=60)}}} and {{{highlight(x=100)}}}
If we cannot factor, we apply the quadratic formula:
{{{x = (-(-160)+- sqrt((-160)^2-4*1*6000 ))/(2*1) }}}
{{{x = (160+- sqrt(25600-24000 ))/2 }}}
{{{x = (160+- sqrt(1600 ))/2 }}} --> {{{x = (160+- 40)/2}}}
with solutions
{{{x=(160+40)/2}}} --> {{{x=200/2}}} --> {{{highlight(x=100)}}} and
{{{x=(160-40)/2}}} --> {{{x=120/2}}} --> {{{highlight(x=60)}}} .
 
Examining case by case the absolute values:
For {{{x>70}}} ,
{{{abs(x-40)>0}}} so {{{abs(x-40)=x-40}}} and
{{{abs(x-70)>0}}} so {{{abs(x-70)=x-70}}}
{{{abs(x-40)=2*abs(x-70)}}} transforms into
{{{x-40=2*(x-70)}}} --> {{{x-40=2x-140}}} --> {{{x-40+140-x=2x-140+140-x}}} --> {{{-40+140=2x-x}}} --> {{{highlight(x=100)}}}
For {{{x<40}}} ,
{{{abs(x-40)<0}}} so {{{abs(x-40)=-(x-40)}}} and
{{{abs(x-70)<0}}} so {{{abs(x-70)=-(x-70)}}}
{{{abs(x-40)=2*abs(x-70)}}} transforms into
{{{-(x-40)=2*(-(x-70))}}} and multiplying times {{{(-1)}}} both sides of the equal sign we get
{{{x-40=2*(x-70)}}} as above,
which gives us the solution {{{x=100}}} as above,
but as we had started with {{{x<40}}},
we did not find a solution that fit the case {{{x<40}}}.
For {{{x}}} such that {{{40<x<70}}} ,
{{{abs(x-40)>0}}} so {{{abs(x-40)=x-40}}} and
{{{abs(x-70)<0}}} so {{{abs(x-70)=-(x-70)}}}
{{{abs(x-40)=2*abs(x-70)}}} transforms into
{{{x-40=2*(-(x-70))}}} --> {{{x-40=2(-x+70))}}} --> {{{x-40=-2x+140))}}} --> {{{x-40+2x+40=-2x+140+2x+40))}}} --> {{{x+2x=140+40}}} --> {{{3x=180}}} --> {{{3x/3=180/3}}} --> {{{3x=180}}} --> {{{highlight(x=60)}}} .