Study guide: Algebra1: Systems and Feasible Linear Models
A mathematical plan can require simultaneous equalities or inequalities. The feasible set contains inputs satisfying all constraints, and a linear objective asks which allowed choice gives the greatest or least modeled value.
Ideas to explain in your own words
- A system can be solved by adding equations whose opposite variable coefficients cancel. The resulting one-variable equation must still be paired with an original equation to recover the full solution. The operation applies to right-side constants as well as coefficients.
- A feasible point satisfies all listed restrictions at once. Nonnegative coordinates and an inclusive sum cap can be checked directly without drawing the entire feasible region. Passing one inequality does not make a point feasible.
- A strict linear inequality excludes its equality line, represented by a dashed boundary. Output values below the line satisfy a less-than comparison when the inequality is solved for y. A solid line includes equality, so it would misrepresent a strict sign.
- An objective assigns a modeled value to each feasible choice. When one resource can be allocated to variables with unequal positive contributions, the larger contribution can determine the optimal corner. An optimum must satisfy the original constraints, not merely make the expression large.
Useful vocabulary
- feasible set
- All choices satisfying every constraint.
- objective
- A quantity being maximized or minimized.
- boundary line
- The equality line associated with a linear inequality.
Watch your reasoning
- The operation applies to right-side constants as well as coefficients. A solved x value alone is incomplete; find y and verify the ordered pair in every original relationship.
- Passing one inequality does not make a point feasible. Test each constraint separately, retaining equality for an at-most cap; sign restrictions often rule out otherwise plausible points.
- A solid line includes equality, so it would misrepresent a strict sign. The shading depends on the solved comparison; use an off-boundary test point when unsure rather than relying on a sketch.
A worked review example
Complete supplied guide examples: Use an Eliminating Linear Combination — Solve 2x+y=7 and x−y=-1. Candidate answers: (A) (2,3) (B) (3,2) (C) (4,2) (D) (7,-1). Add the equations: 3x=6. Thus x=2, and y=x−(-1)=3. Verification in the first equation gives 7. Test a Point against Every Constraint — Which point satisfies x≥0, y≥0 and x+y≤4? Candidate answers: (A) (2,2) (B) (2,−1) (C) (3,2) (D) (−1,2). The point (2,2) has nonnegative coordinates and sum 4, equal to the inclusive cap. The first distractor exceeds the cap; the others use a negative coordinate. Choose the Boundary Style and Shaded Side — How should y<2x+2 be represented on a coordinate plane? Candidate answers: (A) Dashed boundary line; shade above it (B) Dashed boundary line; shade below it (C) Solid boundary line; shade above it (D) Solid boundary line; shade below it. The strict sign excludes equality, so the boundary is dashed. Smaller y-values lie below the line y=2x+2. Test any convenient off-line point to check the side. Optimize within a Simple Feasible Region — For x≥0, y≥0, x+y≤5, maximize P=2x+3y. Candidate answers: (A) 14 (B) 15 (C) 16 (D) 30. Each unit assigned to y contributes 3 rather than 2 to P, so the maximum uses x=0,y=5. P=15. Equivalently P=2(x+y)+y≤2(5)+5.
Make a practice plan
- Study the four methods in the listed order; attempt each independent check before opening its explanation.
- Complete each lesson’s three changed-input questions and record the operation or reasoning step behind any correction.
- Take the four-question unit review, then solve the transfer task with written justifications.
- After correcting errors, attempt the separate unit exam and revisit the linked lesson for any unresolved method.