How do you solve #4t-13=57#?

To solve the problem, we want to end up with #t# by itself on one side. To do that, we need to perform actions on both sides of the equals sign that are the same.

In this case, we should add #13# first:

#4t-13=57#

#4t-13color(blue)+color(blue)13=57color(blue)+color(blue)13#

#4tcolor(red)cancelcolor(black)(color(black)-13color(blue)+color(blue)13)=57color(blue)+color(blue)13#

#4t=57color(blue)+color(blue)13#

#4t=70#

Now, we can divide both sides by #4# to get #t# by itself:

#color(blue)(color(black)(4t)/4)=color(blue)(color(black)70/4)#

#color(blue)(color(black)(color(red)cancelcolor(black)4t)/color(red)cancelcolor(blue)4)=color(blue)(color(black)70/4)#

#t=color(blue)(color(black)70/4)#

#t=17.5#

That's the solution to the problem. Hope this helped!

First, in order to get t by itself you have to add #13# to both sides, so it will cancel out on the side it was previously on, after you do so your new equation will be #4t=70#. Then again, to get #t# by itself you will divide #70# by #4# so you can remove it from the other side, this will leave you equation as #t=17.5#.

You need to isolate the term with the variable,

Add #13# to both sides first.

#4t -13 color(blue)(+13) = 57#

#4tcolor(white)(xx.xxxx) =70 color(blue)(+13)" "larr (-13+13=0)#

You now need to isolate #t#.
Divide both sides by #4#

#(cancel4t)/cancelcolor(red)(4)color(white)(xx.xxxx) =70/color(red)(4)#

#" "tcolor(white)(xx.xxxx) =35/2 = 17 1/2#