With a distance-speed-time problem, remember the relationship:

#s=d/t " "# Let the original speed be #x# kph.

We can then write the speeds and times in terms of #x#

#"Original speed" = x color(white)(xxxxxxxxxx)"Faster speed" = x+10#

#"distance = "10kmcolor(white)(xxxxxxxxxx)" distance ="25km#

#rarr time_1 = 10/x "hours" color(white)(xxxxxxxx)rarrtime_2 = 25/(x+10)#

The total time for the ride was #3/4# hour #" "(time_1 + time_2#)

#10/x +25/(x+10) = 3/4" "larr# now solve the equation

Multiply through by the LCD which is #color(blue)(4x(x+10))#

#(color(blue)(4cancelx(x+10))xx10)/cancelx +(color(blue)(4xcancel(x+10))xx25)/(cancel(x+10) )= (3xxcolor(blue)(cancel4x(x+10)))/cancel4#

=#40(x+10) +100x = 3x(x+10)#

#40x+400 +100x = 3x^2 +30x " "larr# make = 0

#0 = 3x^2 -110x -400" "larr# find factors

#(3x+10)(x-40) = 0#

If # 3x+10 = 0 " "rarr x = -10/3# reject negative speed

if#x-40 = 0" "rarr x = 40#

The original speed was #40 # km per hour