Two find the point in the solution set for both these functions equation the two functions (#f(x) = g(x)#) and solve for #x#.
Therefore:
#2x - 1 = -4/3x + 9#
We can now solve for #x#:
#2x - 1 + color(red)(4/3x) + color(blue)(1) = -4/3x + 9 + color(red)(4/3x) + color(blue)(1)#
#2x + color(red)(4/3x) - 1 + color(blue)(1) = -4/3x + color(red)(4/3x) + 9 + color(blue)(1)#
#2x + color(red)(4/3x) - 0 = 0 + 9 + color(blue)(1)#
#2x + color(red)(4/3x) = 10#
#(3/3 xx 2)x + color(red)(4/3x) = 10#
#6/3x + color(red)(4/3x) = 10#
#10/3x = 10#
#color(red)(3)/color(blue)(10) xx 10/3x = color(red)(3)/color(blue)(10) xx 10#
#cancel(color(red)(3))/cancel(color(blue)(10)) xx color(blue)(cancel(color(black)(10)))/color(red)cancel(color(black)(3))x = color(red)(3)/cancel(color(blue)(10)) xx color(blue)(cancel(color(black)(10)#
#x = 3# is the point which lies in the solution set for both functions.
