How do you graph $3 x + 2 y < 6$ $3x + 2y<6$ ?

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Answer 1 · 2018-06-03T19:54:55

See a solution process below:

First, solve for two points as an equation instead of an inequality to find the boundary line for the inequality.

For: $x = 0$ $x = 0$

$(3 \cdot 0) + 2 y = 6$ $(3 * 0) + 2y = 6$

$0 + 2 y = 6$ $0 + 2y = 6$

$2 y = 6$ $2y = 6$

$\frac{2 y}{2} = \frac{6}{2}$ $(2y)/color(red)(2) = 6/color(red)(2)$

$y = 3$ $y = 3$ or $(0, 3)$ $(0, 3)$

For: $x = 2$ $x = 2$

$(3 \cdot 2) + 2 y = 6$ $(3 * 2) + 2y = 6$

$6 + 2 y = 6$ $6 + 2y = 6$

$- 6 + 6 + 2 y = - 6 + 6$ $-color(red)(6) + 6 + 2y = -color(red)(6) + 6$

$0 + 2 y = 0$ $0 + 2y = 0$

$2 y = 0$ $2y = 0$

$\frac{2 y}{2} = \frac{0}{2}$ $(2y)/color(red)(2) = 0/color(red)(2)$

$y = 0$ $y = 0$ or $(2, 0)$ $(2, 0)$

We can then graph the two points on the coordinate plane and draw a line through the points to mark the boundary of the inequality.

graph{(x^2+(y-3)^2-0.035)((x-2)^2+y^2-0.035)(3x+2y-6)=0 [-10, 10, -5, 5]}

Now, we can shade the left side of the line.

We also need to change the boundary line to a dashed line because the inequality operator does not contain an "or equal to" clause.

graph{(3x+2y-6) < 0 [-10, 10, -5, 5]}

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