First, convert the two mixed numbers to improper fractions:
#4 1/6 = 3 1/3c#
#4 + 1/6 = (3 + 1/3)c#
#[6/6 xx 4] + 1/6 = ([3/3 xx 3] + 1/3)c#
#24/6 + 1/6 = (9/3 + 1/3)c#
#25/6 = 10/3c#
Next, multiply each side of the equation by #color(red)(3)/color(blue)(10)# to solve for #c# while keeping the equation balanced:
#color(red)(3)/color(blue)(10) xx 25/6 = color(red)(3)/color(blue)(10) xx 10/3c#
#color(red)(3)/color(blue)(5 xx 2) xx (5 xx 5)/(3 xx 2) = cancel(color(red)(3))/cancel(color(blue)(10)) xx color(blue)(cancel(color(black)(10)))/color(red)(cancel(color(black)(3)))c#
#cancel(color(red)(3))/color(blue)(color(black)(cancel(color(blue)(5))) xx 2) xx (color(blue)(cancel(color(black)(5))) xx 5)/(color(red)(cancel(color(black)(3))) xx 2) = c#
#5/(2 xx 2) = c#
#5/4 = c#
#c = 5/4#
We can now convert this solution into a mixed number:
#c = (4 + 1)/4#
#c = 4/4 + 1/4#
#c = 1 + 1/4#
#c = 1 1/4#