How do you graph #10x + 20y \geq - 9#?

1 Answer
smendyka
Nov 25, 2017

See a solution process below:

Explanation:

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

For: #x = 0#

#(10 * 0) + 20y = -9#

#0 + 20y = -9#

#20y = -9#

#(20y)/color(red)(20) = -9/color(red)(20)#

#y = -9/20# or #(0, -9/20)#

For: #x = -1#

#(10 * -1) + 20y = -9#

#-10 + 20y = -9#

#color(red)(10) - 10 + 20y = color(red)(10) - 9#

#0 + 20y = 1#

#20y = 1#

#(20y)/color(red)(20) = 1/color(red)(20)#

#y = 1/20# or #(-1, 1/20)#

We can now graph the two points on the coordinate plane and draw a line through the points to mark the boundary of the inequality.
The boundary line will be solid because the inequality operator contains an "or equal to" clause.

graph{(x^2+(y+(9/20))^2-0.00125)((x+1)^2+(y-(1/20))^2-0.00125)(10x+20y+9)=0 [-2, 2, -1, 1]}

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

graph{(10x + 20y + 9) >= 0 [-2, 2, -1, 1]}

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