1.
1x2+2y5x−6y2 = 1x2+2y5x−6y2 ⋅x2y2x2y2 = y2+2x2y5xy2−6x2 and y ≠ 0
2.
To graph y≥−12x+212 , graph the line y=−12x+212 then lightly shade the portion above the line.
To graph y<15x+6 , graph the line y=15x+6 but make it a dotted line. Then lightly shade the portion below it.
The solution to the system of inequalities is the portion of the graph that is shaded twice.
See here: https://www.desmos.com/calculator/hzox2qggig
According to the graph, the point (0, 4) should satisfy both inequalities.
Is it true that 4≥−12(0)+212 ? Yes it is true that 4≥212
Is it true that 4<15(0)+6 ? Yes it is true that 4<6
So (0, 4) does satisfy both inequalities.
1.
1x2+2y5x−6y2 = 1x2+2y5x−6y2 ⋅x2y2x2y2 = y2+2x2y5xy2−6x2 and y ≠ 0
2.
To graph y≥−12x+212 , graph the line y=−12x+212 then lightly shade the portion above the line.
To graph y<15x+6 , graph the line y=15x+6 but make it a dotted line. Then lightly shade the portion below it.
The solution to the system of inequalities is the portion of the graph that is shaded twice.
See here: https://www.desmos.com/calculator/hzox2qggig
According to the graph, the point (0, 4) should satisfy both inequalities.
Is it true that 4≥−12(0)+212 ? Yes it is true that 4≥212
Is it true that 4<15(0)+6 ? Yes it is true that 4<6
So (0, 4) does satisfy both inequalities.