Answer:
Caden had 47 gifts and Collin had 13 gifts at the start.
Step-by-step explanation:
Given:
Caden and Collin had 60 gifts altogether.
Let number of gifts of Caden had be x
also Let number of gift Collin had be y.
Hence the equation can be framed as;
[tex]x+y=60 \ \ \ \ equation \ 1[/tex]
Also given Caden gave Collin 12 of his gifts and Collin gave 10 of his gifts to Caden, Caden would have 3 times as many as Collin.
Hence the equation can be framed as;
[tex]((x-12)+10)=3((y-10)+12)[/tex]
Now Solving the above equation we get;
[tex]x-12+10=3(y-10+12)\\x-2=3(y+2)\\x-2=3y+6\\x-3y=6+2\\x-3y=8 \ \ \ \ equation \ 2[/tex]
Now Subtracting equation 2 from equation 1 we get;
[tex](x+y)-(x-3y)=60-8\\x+y-x+3y=52\\4y=52\\y = \frac{52}{4}=13[/tex]
Now Substituting the value of y in equation 1 we get;
[tex]x+y=60\\x+13=60\\x=60-13\\x=47[/tex]
Hence Caden had 47 gifts and Collin had 13 gifts at the start.
