Mathematics Grade 10 Pair of Linear Equations Graphical and Algebraic

Grade 10 Mathematics Study Guide: Pair of Linear Equations – Graphical and Algebraic

1. The Driving Question

You’re running a lemonade stand with two recipes: one uses 3 lemons and 2 cups of sugar per pitcher, the other uses 1 lemon and 4 cups of sugar. You have exactly 12 lemons and 16 cups of sugar left. How many pitchers of each recipe can you make without running out of ingredients—and why does drawing lines on a graph actually solve this problem, not just describe it?

2. The Core Idea – Built, Not Listed

Imagine you’re at a county fair with two game booths. At Booth A, you pay $2 to play and win $5 if you hit the target. At Booth B, you pay $3 to play and win $7 if you hit. You have $10 to spend and want to play exactly 4 games total. How many times should you play each booth to break even (no profit, no loss)?

This is a system of linear equations: two rules (equations) that must both be true at the same time. The "solution" is the exact number of plays at each booth where both rules are satisfied—like finding the one spot on a map where two roads cross.

• System of linear equations: Two or more linear equations with the same variables. The solution is the point (or points) where all equations are true at once.
  Example: If you track how many steps you take per minute while walking vs. jogging, the system could tell you when you’ll cover 1 mile in exactly 15 minutes.

• Graphical solution: Plotting both equations on the same coordinate plane. The intersection point is the solution.
  Example: If you graph how many hours you study for math vs. science, the intersection shows when your total study time is 10 hours and you’ve spent twice as long on math as science.

• Algebraic solution (substitution/elimination): Solving the equations using algebra to find the exact values of the variables.
  Example: If you’re mixing two types of trail mix to get a specific ratio of nuts to raisins, substitution lets you find how much of each to use without guessing.

• Consistent/inconsistent systems: A system is consistent if it has at least one solution (the lines intersect). It’s inconsistent if there’s no solution (the lines are parallel).
  Example: If two friends are saving money for a concert, but one saves $10/week and the other $5/week, they’ll never have the same amount at the same time—parallel lines, no solution.
  Grade 10+ note: In college linear algebra, systems can have infinitely many solutions (coincident lines) or no solution, and the concept extends to higher dimensions (planes, hyperplanes).

3. Assessment Translation

How this appears on assessments:
- State standardized tests (e.g., PARCC, SBAC): Multiple-choice questions asking to identify the solution from a graph or solve algebraically. Short-answer questions may ask to explain why a system has no solution or infinitely many solutions.
- Classroom formative assessments: Exit tickets with a system to solve (e.g., "Solve by substitution: y = 2x + 1 and 3x + y = 9") or a real-world scenario to model with equations.
- SAT/ACT: Systems appear in the math sections, often as word problems (e.g., "A café sells coffee and tea. On Monday, they sold 50 drinks total and made $150. Coffee costs $3, tea costs $2. How many of each did they sell?"). The SAT may ask for the number of solutions a system has.

What a "proficient" response looks like vs. "developing":
- Proficient: Solves the system correctly, shows work (e.g., substitution steps or graph), and explains the solution in context. For example: "The solution (2, 5) means you can make 2 pitchers of Recipe A and 5 pitchers of Recipe B. I found this by graphing both lines and seeing where they intersect." - Developing: Solves the system but doesn’t show work, or graphs incorrectly (e.g., misplotting points), or gives a solution without context (e.g., just writes "x = 2, y = 5" without explaining what it means).

Model proficient response (short answer):
Prompt: A gym offers two membership plans. Plan A costs $30/month plus $5 per class. Plan B costs $50/month plus $2 per class. After how many classes do the plans cost the same? Response: 1. Write the equations:
- Plan A: C = 30 + 5x
- Plan B: C = 50 + 2x 2. Set them equal: 30 + 5x = 50 + 2x 3. Solve for x:
- 5x – 2x = 50 – 30
- 3x = 20 → x = 20/3 ≈ 6.67 classes 4. Interpretation: The plans cost the same after about 7 classes (since you can’t attend a fraction of a class, the gym would charge for 7).

4. Mistake Taxonomy

Mistake 1: Misreading the question format (graphical solution)
- Prompt: The graph shows two lines: y = 2x + 1 and y = -x + 4. What is the solution to the system? - Common wrong response: "The solution is y = 2x + 1 and y = -x + 4." - Why it loses credit: The question asks for the solution (the intersection point), not the equations. The student restated the problem.
- Correct approach: 1. Find the intersection by setting the equations equal: 2x + 1 = -x + 4.
2. Solve for x: 3x = 3 → x = 1.
3. Plug x = 1 into either equation to find y: y = 2(1) + 1 = 3.
4. Write the solution as (1, 3).

Mistake 2: Algebraic error in substitution (sign mistakes)
- Prompt: Solve the system: y = 3x – 2 and 2x + y = 4.
- Common wrong response: - Substitute y: 2x + (3x – 2) = 4 → 5x – 2 = 4 → 5x = 6 → x = 6/5.
- Then y = 3(6/5) – 2 = 18/5 – 2 = 18/5 – 10/5 = 8/5.
- But the student writes: "The solution is (6/5, 8/5)." - Why it loses credit: The student didn’t check the solution in the second equation. Plugging (6/5, 8/5) into 2x + y gives 2(6/5) + 8/5 = 20/5 = 4, which is correct—but the student might have made a sign error (e.g., forgetting to distribute the negative) and not caught it.
- Correct approach: 1. Substitute y = 3x – 2 into 2x + y = 4.
2. Solve for x: 2x + 3x – 2 = 4 → 5x = 6 → x = 6/5.
3. Find y: y = 3(6/5) – 2 = 8/5.
4. Always check: 2(6/5) + 8/5 = 4. Correct!

Mistake 3: Assuming parallel lines intersect (inconsistent systems)
- Prompt: How many solutions does this system have? y = 2x + 3 and y = 2x – 1.
- Common wrong response: "One solution, because all lines intersect." - Why it loses credit: The student assumes all lines intersect, but parallel lines (same slope, different y-intercepts) never meet. This system has no solution.
- Correct approach: 1. Compare slopes: Both equations have slope = 2.
2. Compare y-intercepts: 3 ≠ -1.
3. Since slopes are equal and y-intercepts are different, the lines are parallel and never intersect. No solution.

5. Connection Layer

1. Within math: Systems of linear equations → linear programming (optimization).
   Why it matters: Linear programming uses systems to find the best possible outcome (e.g., maximizing profit or minimizing cost) under constraints, like the lemonade stand problem but scaled up to businesses.

2. Across subjects: Systems of linear equations → chemistry (stoichiometry).
   Why it matters: Balancing chemical equations is like solving a system where the number of atoms of each element must be equal on both sides of the reaction. For example, finding how many grams of hydrogen and oxygen combine to form water is a system of equations.

3. Outside school: Systems of linear equations → personal finance (budgeting).
   Why it matters: If you’re saving for two goals (e.g., a phone and a concert ticket) with different savings rates, a system of equations can tell you when you’ll reach both goals at the same time—or if it’s impossible.

6. The Stretch Question

If two lines intersect at (3, -2), and one line has a slope of 4, what’s the equation of the other line if it passes through (0, 5)?

Pointer toward the answer: - You know the point of intersection (3, -2) lies on both lines. For the first line, use point-slope form: y – (-2) = 4(x – 3) → y + 2 = 4x – 12 → y = 4x – 14.
- The second line must also pass through (3, -2) and (0, 5). Find its slope: (5 – (-2))/(0 – 3) = 7/-3 = -7/3.
- Now write its equation using point-slope form: y – 5 = (-7/3)(x – 0) → y = (-7/3)x + 5.
- But here’s the twist: What if the question didn’t give you the slope of the first line? Could you still find the second line’s equation? (Hint: You’d need another point or the y-intercept of the first line.)