(pen)o (g)Linear Tables(pen)
(pen)-> Solve for y before building a (g)table(pen).
(pen)-> x + 3y - 6 = 0 gives 3y = -x + 6.
(pen)-> Divide by 3: y = -x/3 + 2.
(pen)-> x values: -6, -3, 0, 3, 6.
(pen)-> y values: 4, 3, 2, 1, 0, respectively.
(pen)-> Plot the ordered pairs; connect with a straight line.
(pen)-> (g)Check(pen): 6x + 2y = 10 gives y = -3x + 5.
(pen)-> Pick x values that make the (g)division(pen) easy, like 0 and 6.
(pen)-> Equal x steps of 3 give equal y steps of -1: (g)slope(pen) -1/3.
(pen)o (pu)Intercepts and Slope(pen)
(pen)-> For 3x - 2y = -12, set y = 0 to find x.
(pen)-> 3x = -12 gives the x-intercept (-4, 0).
(pen)-> Set x = 0: -2y = -12 gives (0, 6).
(pen)-> Draw the line through these two (pu)intercepts(pen).
(pen)-> In y = (3/2)x - 4, (pu)slope(pen) is 3/2.
(pen)-> Start at (0, -4); move right 2 and up 3.
(pen)-> Slope means change in y divided by change in x.
(pen)-> The (pu)intercepts(pen) are (-4, 0) and (0, 6); no other crossings.
(pen)-> (pu)Slope(pen) from the intercepts is (6 - 0)/(0 - (-4)) = 3/2.
(pen)o (br)Inequality Boundaries(pen)
(pen)-> Replace the inequality sign with = for the boundary.
(pen)-> Use a (br)solid(pen) line for <= or >=; equality is included.
(pen)-> Use a (br)dashed(pen) line for < or >; equality is excluded.
(pen)-> For y < mx + b, shade below the boundary.
(pen)-> For y > mx + b, shade above the boundary.
(pen)-> Multiplying or dividing by a negative reverses the sign.
(pen)-> For x < c, the (br)vertical boundary(pen) is x = c; shade left.
(pen)-> For x > c, use a (br)dashed(pen) vertical line and shade right.
(pen)o (hp)Rearrange and Test(pen)
(pen)-> 12 - 4y > x gives -4y > x - 12.
(pen)-> Divide by -4: y < -x/4 + 3; dashed, shade below.
(pen)-> 9 + 3y <= 8x gives y <= (8/3)x - 3.
(pen)-> Use a solid boundary and shade below it.
(pen)-> At (3, 5), 9 + 3(5) = 24 and 8(3) = 24.
(pen)-> (hp)Equality(pen) holds, so (3, 5) is included on the boundary.
(pen)-> (hp)Test(pen) a point off the line to confirm the shaded side.
(pen)-> (hp)Test(pen) (0, 0): 9 <= 0 is false, so the shade is below.
(pen)-> The (hp)boundary(pen) y = -x/4 + 3 passes (0, 3), (4, 2), (8, 1).
(pen)o (y)Weighted Grade Constraints(pen)
(pen)-> Let x = test grade and y = homework grade.
(pen)-> With 0 <= x, y <= 100, (y)target(pen): 0.7x + 0.3y >= 90.
(pen)-> Multiply by 10: 7x + 3y >= 900.
(pen)-> Solve: y >= 300 - (7/3)x.
(pen)-> If y = 100, then x >= 600/7, about 85.71.
(pen)-> If x = 100, then y >= 200/3, about 66.67.
(pen)-> Shade above the solid line within the (y)grade square(pen).
(pen)-> y = 300 - (7/3)x has (y)intercept(pen) (0, 300), above the square.
(pen)-> On the (y)boundary(pen), raising x by 3 lowers required y by 7.
(pen)o (hp)Regions and Functions(pen)
(pen)-> A system's solution satisfies every inequality.
(pen)-> Keep only the (hp)overlap(pen) of all allowed regions.
(pen)-> The grade region includes (100, 100), not (0, 0).
(pen)-> Its corners are (600/7, 100), (100, 100), (100, 200/3).
(pen)-> A (hp)function(pen) assigns exactly one output to each input.
(pen)-> An inequality region generally allows many y values per x.
(pen)-> At x = 100, both y = 80 and y = 100 satisfy the (hp)target(pen).
(pen)-> At x = 600/7 (about 85.7), y = 100 is the (hp)cutoff(pen).