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

Jun 12, 2018

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$y > 3 - \left(\frac{3}{2}\right) x$

#### Explanation:

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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

$4 y > 12 - 6 x$

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

$\left(\frac{\cancel{4}}{\cancel{\textcolor{red}{4}}}\right) y > \left({\cancel{12}}^{\textcolor{red}{3}} / \textcolor{red}{{\cancel{4}}^{\textcolor{red}{1}}}\right) - \left({\cancel{6}}^{\textcolor{red}{3}} / \textcolor{red}{{\cancel{4}}^{\textcolor{red}{2}}}\right) x$

$y > 3 - \left(\frac{3}{2}\right) x$

We can verify this solution using an inequality graph:

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

Hope it helps.