5.4 Solving Linear Systems Graphically

5.4 Solving Linear Systems Graphically

Quick Summary: A linear system is two (or more) linear relations considered together. Solving it means finding the ordered pair $(x, y)$ that satisfies both relations at once — graphically, the point of intersection. EQAO lists solving linear systems graphically as one of the eleven fundamental knowledge and skills in the Relationships and Proportional Reasoning dimension, and it is the mathematical engine behind every "at what point do these two plans cost the same?" question on the test.

---

What a Solution Actually Is

For the system $y = 2x + 1 \qquad \text{and} \qquad y = -x + 7$ the solution is the single point that lies on both lines. Every point on the first line satisfies the first equation; every point on the second satisfies the second; only the intersection satisfies both.

graph LR
    L1["Line 1: y = 2x + 1<br/>every point satisfies equation 1"] --> INT["POINT OF INTERSECTION (2, 5)<br/>satisfies BOTH equations<br/>= the solution to the system"]
    L2["Line 2: y = -x + 7<br/>every point satisfies equation 2"] --> INT

Verification is the whole idea. Substitute $x = 2$, $y = 5$:

• Equation 1: $2(2) + 1 = 5$ ✓
• Equation 2: $-(2) + 7 = 5$ ✓

Because both check, $(2, 5)$ is the solution.

---

Method 1: Graph Both Lines and Read the Intersection

1. Write each relation in the form $y = mx + b$ if it is not already.
2. Plot the y-intercept $b$ for each line.
3. Use the slope $m = \frac{\text{rise}}{\text{run}}$ to step to a second point, and draw each line.
4. Read the coordinates of the crossing point.
5. Check the ordered pair in both original equations.

Worked example. Solve graphically: $y = 2x + 1$ and $y = -x + 7$.

The table shows the lines meeting at $x = 2$, where both produce $y = 5$. The solution is $(2, 5)$.

A table of values is a graph you can read without drawing. In a timed, multiple-choice environment, extending a small table until the two output columns agree is often the fastest reliable route to the intersection.

---

Method 2: Test the Options (the multiple-choice shortcut)

Every MPT question is multiple choice, so the answer is already on the screen. When the options are ordered pairs, substitute each one into both equations and keep the pair that satisfies both. This avoids graphing errors entirely and usually takes under a minute.

For $3x + 2y = 12$ and $y = x - 1$, testing $(2, 3)$:

• $3(2) + 2(3) = 6 + 6 = 12$ ✓
• $3 = 2 - 1 = 1$ ✗

$(2, 3)$ fails the second equation, so it is not the solution. Testing $(2.8, 1.8)$:

• $3(2.8) + 2(1.8) = 8.4 + 3.6 = 12$ ✓
• $1.8 = 2.8 - 1 = 1.8$ ✓

A pair must clear both equations. Checking only one is the most common way to select a distractor.

---

Classifying Systems: One, None, or Infinitely Many

The number of solutions is decided by the slopes and y-intercepts — a comparison you can make without plotting anything.

Examples.

• $y = 3x + 2$ and $y = 3x - 4$: same slope, different intercepts → parallel, no solution.
• $y = 3x + 2$ and $6x - 2y = -4$: rearranging the second gives $-2y = -6x - 4$, so $y = 3x + 2$ — the same line, therefore infinitely many solutions.
• $y = 3x + 2$ and $y = -x + 6$: different slopes → exactly one solution.

Rearrange before you judge. A system written as $2x + y = 5$ and $y = -2x + 9$ looks unrelated until you rewrite the first as $y = -2x + 5$. Both have slope $-2$ and different intercepts, so there is no solution.

---

Break-Even Problems Are Linear Systems

The most common contextual form on the MPT combines a partial variation with a direct variation — a fixed fee plus a rate, compared against a pure rate. The intersection is the break-even point: the value at which the two options cost exactly the same.

Worked example. A school is booking a bus for a field trip.

• Company A charges a $180 flat booking fee plus $1.20 per kilometre.
• Company B charges $2.70 per kilometre with no booking fee.

1. Model each option. Let $d$ be the distance in kilometres and $C$ the total cost: $C_A = 1.20d + 180 \qquad\qquad C_B = 2.70d$
2. Set the two expressions equal — algebraically, this is exactly what the graph does at the crossing point: $2.70d = 1.20d + 180$
3. Solve. $2.70d - 1.20d = 180 \;\Rightarrow\; 1.50d = 180 \;\Rightarrow\; d = 120 \text{ km}$
4. Find the shared cost (the y-coordinate of the intersection). $C_B = 2.70 \times 120 = \$324.00 \qquad C_A = 1.20(120) + 180 = 144 + 180 = \$324.00 \;\checkmark$
5. Interpret each side of the intersection — this is where the marks usually are.

Interpretation: below 120 km the pay-per-kilometre option wins because the flat fee is not yet worth paying; above 120 km the lower per-kilometre rate more than repays the $180 fee. Because Company A's line starts higher but rises more slowly, it must cross Company B's line exactly once — a direct consequence of the two lines having different slopes.

---

Common Errors to Avoid

1. Reading only the $x$-coordinate. "The solution is 120" is incomplete; the solution to the system is the ordered pair $(120, 324)$, and contextual questions often ask for the shared cost, not the distance.
2. Checking one equation only. An ordered pair on one line proves nothing about the other.
3. Assuming a crossing exists. Equal slopes mean the lines never meet.
4. Mis-scaling a hand-drawn graph. If the axes use different increments, the intersection lands in the wrong place. Verify algebraically or by substitution before committing.

---

Classroom Connection

Linear systems are where Ontario's Connecting and Representing processes meet: students move between a word problem, a table of values, a graph, and a pair of equations, and see that all four describe the same situation. Presenting a break-even scenario and asking "which company should the school choose?" is a low-floor, high-ceiling task — every student can extend a table of values, while stronger students generalize to the algebraic intersection.