We can write and solve the relationship below where #g# is the number of gallons Brian needs to drive 506 miles assuming the car averages it's EPA rating"
#(22"mi")/"gal" = (506"mi")/g#
First, multiply each side of the equation by #color(red)(g)color(blue)("gal")# to eliminate the fraction while keeping the equation balanced:
#color(red)(g)color(blue)("gal") xx (22"mi")/"gal" = color(red)(g)color(blue)("gal") xx (506"mi")/g#
#color(red)(g)cancel(color(blue)("gal")) xx (22"mi")/color(blue)(cancel(color(black)("gal"))) = cancel(color(red)(g))color(blue)("gal") xx (506"mi")/color(red)(cancel(color(black)(g)))#
#g xx 22"mi" = color(blue)("gal") xx 506"mi"#
#22g"mi" = 506"mi"" gal"#
Now, divide each side of the equation by #color(red)(22)color(red)("mi")# tp solve for #g# while keeping the equation balanced:
#(22g"mi")/(color(red)(22)color(red)("mi")) = (506"mi"" gal")/(color(red)(22)color(red)("mi"))#
#(color(red)(cancel(color(black)(22)))gcolor(red)(cancel(color(black)("mi"))))/(cancel(color(red)(22))cancel(color(red)("mi"))) = (color(red)(cancel(color(black)(506)))23color(red)(cancel(color(black)("mi")))" gal")/(cancel(color(red)(22))cancel(color(red)("mi")))#
#g = 23"gal"#
Brian would required 23 gallons to drive the 506 miles.
