Question 151016
Let "x"= 1st #, & "y"= 2nd #
1st condition: Average of 2 numbers is 15,
{{{(x+y)/2=15}}}
2nd condition: 4 times the diff. is 40,
{{{4(x-y)=40}}}
.
In 1st condtn. we get {{{x+y=30}}} --> REMEMBER THIS EQ'N,
 {{{x=30-y}}} and substitute this "x" value in 2nd condtn.
{{{4(30-y-y)=40}}}
{{{120-4y-4y=40}}}
{{{120-40=4y+4y}}} ----> {{{80=8y}}}
{{{y=10}}}, and REMEMBER EQ'N {{{x+10=30}}} ---> {{{x=30-10}}} --> {{{x=20}}}
To check,
Go back 1st condtn.; {{{(20+10)/2=15}}} ----> {{{15=15}}}
For 2nd condtn.; {{{4(20-10)=40}}} ------> {{{40=40}}}
</pre><font size=4="indigo"><b>Thank you
Jojo</pre>