Which of the following ordered pairs printPoint( POINT_1 )
and printPoint( POINT_2 )
are solutions to the inequality Y_COEFy + X_COEFx COMP + RIGHT_INT
?
printPoint( POINT_1 )
printPoint( POINT_2 )
printPoint( POINT_1 )
printPoint( POINT_2 )
printPoint( POINT_1 )
and printPoint( POINT_2 )
To graph the inequality, we need to first put it in slope-intercept form (yCOMPmx + b
).
Add -X_COEFx
to both sides of the inequality.
Y_COEFy + X_COEFx +-X_COEFxCOMP + RIGHT_INTy +-X_COEFx
Y_COEFyCOMP + (-X_COEF)x + RIGHT_INT
When multiplying or dividing both sides of an inequality by a negative number you have to flip the inequality. Therefore COMP
becomes COMP_SOLUTION
.
Multiply both sides of the inequality by ONE_OVER_Y_COEF
.
(Y_COEFy) \cdot (ONE_OVER_Y_COEF)COMP_SOLUTION(-X_COEFx + RIGHT_INT) \cdot (ONE_OVER_Y_COEF)
y COMP_SOLUTION + M_FRACx + B_FRAC
Graph the line.
Since our inequality has a LESS_THAN ? "less-than" : "greater-than"INCLUSIVE ? " or equal to" : "" sign, that means that any point LESS_THAN ? "below" : "above" the line is a solution to the inequality.
Note that since the sign is LESS_THAN ? "less-than" : "greater-than" or equal to, any point on the line is also a solution.
Note that since the sign is LESS_THAN ? "less-than" : "greater-than" (and not equal to), any point on the line is not part of the solution.
After plotting the points, we can see that printPoint( POINT_1 ) is the only solution printPoint( POINT_2 ) is the only solution printPoint( POINT_1 ) andprintPoint( POINT_2 ) are both solutions there are no solutions to the inequality.