First, multiply each side of the inequality by #color(red)(30)# to eliminate the fractions while keeping the inequality balanced. We are using #color(red)(30# because it is the Lowest Common Denominator for the three fractions:
#color(red)(30)((1 + m)/10 - (m + 2)/6) < color(red)(30) xx -m/30#
#(color(red)(30) xx (1 + m)/10) - (color(red)(30) xx (m + 2)/6) < cancel(color(red)(30)) xx -m/color(red)(cancel(color(black)(30)))#
#(cancel(color(red)(30))" " 3 xx (1 + m)/color(red)(cancel(color(black)(10)))) - (cancel(color(red)(30))" "5 xx (m + 2)/color(red)(cancel(color(black)(6)))) < -m#
#3(1 + m) - 5(m + 2) < -m#
#(3 xx 1) + (3 xx m) - (5 xx m) - (5 xx 2) < -m#
#3 + 3m - 5m - 10 < -m#
Next, we can group and combine like terms on the left side of the inequality:
#3 - 10 + 3m - 5m < -m#
#(3 - 10) + (3 - 5)m < -m#
#-7 + (-2)m < -m#
#-7 - 2m < -m#
Now, we can add #color(red)(2m)# to each side of the inequality to solve for #m# while keeping the inequality balanced:
#-7 - 2m + color(red)(2m) < -m + color(red)(2m)#
#-7 - 0 < -1m + color(red)(2m)#
#-7 < (-1 + color(red)(2))m#
#-7 < 1m#
#-7 < m#
We can reverse of "flip" the entire inequality to state the solution in terms of #m#:
#m > -7#
