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