How do you solve #\frac { 7} { 5} x - 5\geq \frac { 1} { 6} x + 1#?

2 Answers
Jul 20, 2017

We need to isolate the variable, #x#, but first, let's get the constants on one side

#7/5x - 5 >= 1/6x + 1#

add #5# to both sides

#7/5x >= 1/6x + 6#

subtract #1/6x# on both sides

#7/5x - 1/6x >= 6#

#6/6 xx 7/5x - 1/6x xx 5/5 >= 6#

#42/30x - 5/30x >= 6#

#37/30x >= 6#

divide by #37/30# on both sides

#x >= 6 -: 37/30#

multiply by the reciprocal

#x >= 6/1 xx 30/37#

#x >= 180/37 ~~ 4.86#

#* * * * * * * * * * * * * * * * *#

Let's check our work by solving the original equation, using #180/37# instead of #x#. If we were right, both sides of the equation should look the same.

#7/5x - 5 >= 1/6x + 1#

#7/5 (180/37) - 5 >= 1/6 (180/37) + 1#

#252/37 - 5 >= 30/37 +1#

#67/37 >= 67/37#

We were right!

#x# is #180/37#

Jul 20, 2017

#x >=4 37/180#

Explanation:

Treat an inequality in the same way as an equation, unless you multiply or divide by a negative number, (then the sign changes around.)

#7/5x-5 >= 1/6x+1" "# multiply by 30 to cancel the denominators

#(color(blue)(cancel30^6xx)7)/cancel5x-color(blue)(30xx)5 >=(color(blue)(cancel30^5xx) 1)/cancel6x+color(blue)(30xx)1#

#42x-150 >= 5x+30" "larr# re-arrange the terms

#42x -5x >= 30 +150#

#37x >= 180#

#x >= 180/37#

#x >= 4 37/180#