Graphical Method of Solution

The Big Idea — Lines and Intersections

Every linear equation in two variables is a straight line when drawn on graph paper. So a pair of equations is two lines. Solving the pair graphically means drawing both lines and finding where they meet.

The point of intersection $(x, y)$ satisfies both equations, so it is the solution.

Key Point: To draw a line, find at least two points on it (often the x- and y-intercepts), plot them, and join with a ruler. The coordinates of the intersection give the solution.

There are exactly three possibilities for how two lines can relate, and each tells us how many solutions the pair has. Let's look at all three.

[Board Important] When drawing graphs in the exam, make a neat table of at least 2 (preferably 3) points per line, and label both lines clearly.

Case 1: Intersecting Lines — Unique Solution

If the two lines cross at exactly one point, the pair has exactly one solution — the coordinates of that crossing point.

Example

$x + y = 4$ and $x - y = 2$. Plotting both, they cross at $(3, 1)$. Check: $3 + 1 = 4$ ✓ and $3 - 1 = 2$ ✓. So the unique solution is $x = 3$, $y = 1$.

This is the consistent and independent case.

Key Point: Intersecting lines ⇒ a unique solution ⇒ the system is consistent (with exactly one answer).

In terms of the coefficients, this happens when: $$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$$

[Board Important] The ratio test $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$ lets you predict a unique solution without drawing the graph.

Case 2 & 3: Coincident and Parallel Lines

Case 2: Coincident lines — infinitely many solutions

If the two equations represent the same line (one lies exactly on top of the other), every point on the line is common, so there are infinitely many solutions. This is the consistent and dependent case. It happens when: $$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$$

Case 3: Parallel lines — no solution

If the two lines are parallel (same slope, never meet), there is no common point, so there is no solution. This is the inconsistent case. It happens when: $$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$$

Key Point: Compare the three ratios $\dfrac{a_1}{a_2}$, $\dfrac{b_1}{b_2}$, $\dfrac{c_1}{c_2}$ to instantly tell the type of solution.

[Board Important] Memorise the three ratio conditions — they are asked directly and also save graph-drawing time.

Summary Table of the Three Cases

Here's everything in one place. For the pair $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:

Ratio comparison	Lines	Solutions	Consistency
$\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$	Intersecting	Unique	Consistent
$\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$	Coincident	Infinitely many	Consistent (dependent)
$\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$	Parallel	None	Inconsistent

Key Point: Write all equations in the form $ax + by + c = 0$ (move everything to one side) before computing the ratios — otherwise the $c$ comparison goes wrong.

[Board Important] This table is the single most useful summary of the chapter. If you remember only one thing, remember this.

Solved Examples

Example 1: Type of solution from ratios

Does $2x + 3y = 7$ and $4x + 6y = 14$ have a unique solution, no solution, or infinitely many?

Solution:

1. Write as $2x + 3y - 7 = 0$ and $4x + 6y - 14 = 0$.
2. Ratios: $\dfrac{2}{4} = \dfrac{3}{6} = \dfrac{-7}{-14} = \dfrac{1}{2}$. All equal.
3. So the lines are coincident.

Final Answer: Infinitely many solutions.

Takeaway: All three ratios equal ⇒ infinitely many solutions.

Example 2: Parallel lines

Classify $x + 2y = 4$ and $2x + 4y = 12$.

Solution:

1. As $x + 2y - 4 = 0$ and $2x + 4y - 12 = 0$.
2. $\dfrac{1}{2} = \dfrac{2}{4} = \dfrac{1}{2}$, but $\dfrac{-4}{-12} = \dfrac{1}{3}$.
3. So $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$.

Final Answer: No solution (parallel, inconsistent).

Takeaway: First two ratios equal but the third different ⇒ no solution.

Example 3: Unique solution

Classify $x - y = 1$ and $2x + y = 8$.

Solution:

1. As $x - y - 1 = 0$ and $2x + y - 8 = 0$.
2. $\dfrac{1}{2}$ vs $\dfrac{-1}{1} = -1$. Since $\dfrac{1}{2} \neq -1$, the ratios $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$.

Final Answer: Unique solution (intersecting, consistent).

Takeaway: $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$ ⇒ unique solution.

Example 4: Solve graphically (find the point)

Solve graphically: $x + y = 5$ and $x - y = 1$.

Solution:

1. For $x + y = 5$: points $(0,5), (5,0)$.
2. For $x - y = 1$: points $(0,-1), (1,0), (3,2)$.
3. The lines intersect at $(3, 2)$.

Final Answer: $x = 3$, $y = 2$.

Takeaway: Plot two points per line; read off the intersection.

Example 5: Find points for graphing

Find two points to plot the line $2x + y = 6$.

Solution:

1. Put $x = 0$: $y = 6$, giving $(0, 6)$.
2. Put $y = 0$: $2x = 6 \Rightarrow x = 3$, giving $(3, 0)$.

Final Answer: $(0, 6)$ and $(3, 0)$.

Takeaway: The intercepts (set $x=0$, then $y=0$) are the easiest two points.

Example 6: Find $k$ for no solution

Find $k$ so that $2x + 3y = 7$ and $6x + ky = 21$ has no solution.

Solution:

1. For no solution: $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$.
2. $\dfrac{2}{6} = \dfrac{3}{k} \Rightarrow 2k = 18 \Rightarrow k = 9$.
3. Check third ratio: $\dfrac{7}{21} = \dfrac{1}{3} = \dfrac{2}{6}$ — equal, so it would be infinitely many, not no solution.
4. Re-examine: with $k = 9$, all three ratios equal $\dfrac{1}{3}$, giving infinitely many solutions. So there is no value of $k$ giving 'no solution' here (the constants are proportional).

Final Answer: No value of $k$ gives 'no solution'; $k = 9$ gives infinitely many.

Takeaway: Always check the third ratio — proportional constants change the conclusion.

Example 7: Find $k$ for infinitely many solutions

Find $k$ so that $x + 2y = 3$ and $5x + ky = 15$ has infinitely many solutions.

Solution:

1. Need $\dfrac{1}{5} = \dfrac{2}{k} = \dfrac{3}{15}$.
2. $\dfrac{3}{15} = \dfrac{1}{5}$ ✓ (constants match).
3. $\dfrac{2}{k} = \dfrac{1}{5} \Rightarrow k = 10$.

Final Answer: $k = 10$.

Takeaway: For infinitely many, all three ratios must be equal.

Example 8: Find $k$ for a unique solution

For what value(s) of $k$ does $kx + 2y = 5$ and $3x + y = 1$ have a unique solution?

Solution:

1. Unique solution needs $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$, i.e. $\dfrac{k}{3} \neq \dfrac{2}{1}$.
2. $\dfrac{k}{3} \neq 2 \Rightarrow k \neq 6$.

Final Answer: All $k \neq 6$.

Takeaway: Unique solution fails only when the first two ratios become equal.

Example 9: Lines through the axes (graphing triangle)

The lines $x = 0$, $y = 0$ and $x + y = 4$ form a triangle. Find its vertices.

Solution:

1. $x = 0$ and $y = 0$ meet at the origin $(0, 0)$.
2. $x + y = 4$ meets $x = 0$ at $(0, 4)$ and $y = 0$ at $(4, 0)$.

Final Answer: Vertices $(0,0)$, $(0,4)$, $(4,0)$.

Takeaway: Intersect each pair of lines to get the triangle's corners.

Example 10: Read solution from a description

Two lines are drawn and found to be parallel. How many solutions does the corresponding pair have, and is it consistent?

Solution:

1. Parallel lines never meet, so there is no common point.
2. Hence no solution; the pair is inconsistent.

Final Answer: No solution; inconsistent.

Takeaway: Parallel ⇒ no solution ⇒ inconsistent.