Question 73922
Let s=smaller number, L=larger number
*[Tex \large s+L=53 \\ 3s=L-1]The word problem translates to this
Solve for L
*[Tex \large s+L=53]
*[Tex \large L=53-s]
Plug this into *[Tex \large 3s=L-1]
*[Tex \large 3s=(53-s)-1]
Now we can solve for s
*[Tex \large 4s=52]
*[Tex \large s=13]
Use this to find the other number
*[Tex \large s+L=53]
*[Tex \large 13+L=53]
*[Tex \large L=40]
So the 2 numbers are 13 and 40
<p>
Check:
*[Tex \large 13+40=53]
*[Tex \large 53=53]works
*[Tex \large 3s=L-1]
*[Tex \large 3(13)=40-1]
*[Tex \large 39=39]works