How do you solve #-3x + 32< 47#?

2 Answers
Elizabeth F. · EZ as pi · Jumbotron
Mar 13, 2018

#x > -5#

Explanation:

You transpose the constant to the their side so the number with the variable will be left.

#-3x<47-32#

When you transpose, always change the sign of the constant that you transposed. Do the subtracting and after that, divided by both sides by #-3# so #x# will be left.

#(-3x)/-3>15/-3#

Cancel #-3# on the left because they are equal to #1#. Divide #15# by #-3# so you get #-5.#

However, when you multiply or divide an inequality by a negative number, the inequality sign changes around

#cancel((-3)x)/cancel(-3)> cancel15^-5/cancel(-3)^1#

#x > -5#

When you divide a negative number, make sure to flip the less than sign to a greater than sign.

#x > -5# is the answer

Angel · Jumbotron
Mar 13, 2018

#x> - 5#

Explanation:

#-3x+32<47#

#=>-3x<47-32#

#=> -3x<15#
#=>x>15/-3# {because if we send the (-) sign by multiplication or by division to the other side the sign changes to (+).....}

#=>x> -5#

hence the value of #x# is #> -5#

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