Graph the following system of inequalities:
expr([ "+", [ "*", A_1, "x" ], [ "*", B_1, "y" ] ]) STD_FORM_COMP_1 C_1
y COMP_1 PRETTY_SLOPE_1 x + YINT_1
expr([ "+", [ "*", A_2, "x" ], [ "*", B_2, "y" ] ]) STD_FORM_COMP_2 C_2
y COMP_2 PRETTY_SLOPE_2 x + YINT_2
Convert the first inequality, expr([ "+", [ "*", A_1, "x" ], [ "*", B_1, "y" ] ]) STD_FORM_COMP_1 C_1,
to slope-intercept form by solving for y.
A_1 < 0 ? "Add" : "Subtract" abs( A_1 )x A_1 < 0 ? "to" : "from" both sides:
\qquad expr( [ "*", B_1, "y" ] ) STD_FORM_COMP_1 expr([ "+", [ "*", -A_1, "x" ], C_1 ])
Divide both sides by B_1. Since you're multiplying or dividing by a negative number, don't forget to flip the inequality sign:
\qquad y COMP_1 expr([ "+", "\\dfrac{" + expr([ "*", -A_1, "x" ]) + "}{" + B_1 + "}", "\\dfrac{" + C_1 + "}{" + B_1 + "}" ])
\qquad y COMP_1 \color{purple}{PRETTY_SLOPE_1} x \color{gray}{+ YINT_1}
The y-intercept is YINT_1 and the slope is decimalFraction( SLOPE_1, true, true ).
Since the y-intercept is YINT_1, the line must pass through the point (0, YINT_1).
The slope is decimalFraction( SLOPE_1, true, true ). Remember that
the slope tells you rise over run. So in this case for every
abs( SLOPE_FRAC_1[0] )
"position" + ( abs( SLOPE_FRAC_1[0] ) !== 1 ? "s" : "" )
you move
down (because it's negative)
up
you must also move
SLOPE_FRAC_1[1]
"position" + ( abs( SLOPE_FRAC_1[1] ) !== 1 ? "s" : "" )
to the right. So the line must also pass through (SLOPE_FRAC_1[1], YINT_1 + SLOPE_FRAC_1[0])
Since our inequality has a LESS_THAN_1 ? "less-than" : "greater-than"INCLUSIVE_1 ? " or equal to" : "" sign, that means that any point LESS_THAN_1 ? "below" : "above" the line is a solution to the inequality, so the area LESS_THAN_1 ? "below" : "above" the line should be shaded.
Note that since the sign is LESS_THAN_1 ? "less-than" : "greater-than" or equal to, any point on the line is also a solution, so the line should be solid.
Note that since the sign is LESS_THAN_1 ? "less-than" : "greater-than" (and not equal to), any point on the line is not part of the solution, so the line should be dashed to indicate this.
Convert the second inequality, expr([ "+", [ "*", A_2, "x" ], [ "*", B_2, "y" ] ]) STD_FORM_COMP_2 C_2,
to slope-intercept form by solving for y.
A_2 < 0 ? "Add" : "Subtract" abs( A_2 )x A_2 < 0 ? "to" : "from" both sides:
\qquad expr( [ "*", B_2, "y" ] ) STD_FORM_COMP_2 expr([ "+", [ "*", -A_2, "x" ], C_2 ])
Divide both sides by B_2. Since you're multiplying or dividing by a negative number, don't forget to flip the inequality sign:
\qquad y COMP_2 expr([ "+", "\\dfrac{" + expr([ "*", -A_2, "x" ]) + "}{" + B_2 + "}", "\\dfrac{" + C_2 + "}{" + B_2 + "}" ])
\qquad y COMP_2 \color{purple}{PRETTY_SLOPE_2} x \color{gray}{+ YINT_2}
The y-intercept is YINT_2 and the slope is decimalFraction( SLOPE_2, true, true ).
Since the y-intercept is YINT_2, the line must pass through the point (0, YINT_2).
The slope is decimalFraction( SLOPE_2, true, true ). Remember that
the slope tells you rise over run. So in this case for every
abs( SLOPE_FRAC_2[0] )
"position" + ( abs( SLOPE_FRAC_2[0] ) !== 1 ? "s" : "" )
you move
down (because it's negative)
up
you must also move
SLOPE_FRAC_2[1]
"position" + ( abs( SLOPE_FRAC_2[1] ) !== 1 ? "s" : "" )
to the right. So the line must also pass through (SLOPE_FRAC_2[1], YINT_2 + SLOPE_FRAC_2[0])
Since our inequality has a LESS_THAN_2 ? "less-than" : "greater-than"INCLUSIVE_2 ? " or equal to" : "" sign, that means that any point LESS_THAN_2 ? "below" : "above" the line is a solution to the inequality, so the area LESS_THAN_2 ? "below" : "above" the line should be shaded.
Note that since the sign is LESS_THAN_2 ? "less-than" : "greater-than" or equal to, any point on the line is also a solution, so the line should be solid.
Note that since the sign is LESS_THAN_2 ? "less-than" : "greater-than" (and not equal to), any point on the line is not part of the solution, so the line should be dashed to indicate this.