First, convert the mixed number to an improper fraction:
#2 2/19 = 2 + 2/19 = (19/19 xx 2) + 2/19 = 38/19 + 2/19 = (38 + 2)/19 = 40/19#
Next rewrite the equation as:
#3.8/(4c + 3) = 2/(40/19)#
#3.8/(4c + 3) = (2/1)/(40/19)#
Now, use this rule for dividing equations to evaluate the expression on the right side of the equation:
#(color(red)(a)/color(blue)(b))/(color(green)(c)/color(purple)(d)) = (color(red)(a) xx color(purple)(d))/(color(blue)(b) xx color(green)(c))#
#(color(red)(2)/color(blue)(1))/(color(green)(40)/color(purple)(19)) = (color(red)(2) xx color(purple)(19))/(color(blue)(1) xx color(green)(40)) = (color(green)(cancel(color(red)(2))) xx color(purple)(19))/(color(blue)(1) xx color(red)(cancel(color(green)(40)))color(green)(20)) = 19/20#
We can again rewrite the equation as:
#3.8/(4c + 3) = 19/20#
Because both sides of the equation are pure fractions we can flip the fractions giving:
#(4c + 3)/3.8 = 20/19#
Next, multiply each side of the equation by #color(red)(3.8)# to eliminate the fractions while keeping the equation balanced:
#color(red)(3.8) xx (4c + 3)/3.8 = color(red)(3.8) xx 20/19#
#cancel(color(red)(3.8)) xx (4c + 3)/color(red)(cancel(color(black)(3.8))) = cancel(color(red)(3.8))color(red)(0.2) xx 20/color(red)(cancel(color(black)(19)))#
#4c + 3 = 0.2 xx 20#
#4c + 3 = 4#
Then, subtract #color(red)(3)# from each side of the equation to isolate the #c# term while keeping the equation balanced:
#4c + 3 - color(red)(3) = 4 - color(red)(3)#
#4c + 0 = 1#
#4c = 1#
Now, divide each side of the equation by #color(red)(4)# to solve for #c# while keeping the equation balanced:
#(4c)/color(red)(4) = 1/color(red)(4)#
#(color(red)(cancel(color(black)(4)))c)/cancel(color(red)(4)) = 1/4#
#c = 1/4#
