How do you solve #6x + 4y > 12#?

1 Answer
Jun 12, 2018

#" "#
#y>3-(3/2)x#

Explanation:

#" "#
We have the inequality #color(red)(6x+4y>12#

Subtract #color(blue)(6x# from both sides of the inequality

#6x+4y-color(blue)(6x)> 12 - color(blue)(6x#

#cancel(6x)+4y-color(blue)(cancel (6x))> 12 - color(blue)(6x#

#4y>12-6x#

We want just #color(red)(y# on the left-hand side of the inequality.

Divide both sides of the inequality by #color(red)(4#

#(4y)/color(red)(4) >(12-6x)/color(red)(4#

#(cancel 4/cancel color(red)(4))y>(cancel 12^color(red)(3)/color(red)(cancel 4^color(red)(1)))-(cancel 6^color(red)(3)/color(red)(cancel 4^color(red)(2)))x#

#y>3-(3/2)x#

We can verify this solution using an inequality graph:

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The dotted line in the graph indicates the solution that is NOT the part of the solution.

Hope it helps.