Mathematics — CBSE Class 10 board question

Checking consistency:

Rewrite equations in standard form:

• $x + 3y - 6 = 0$ → $a_1=1,\ b_1=3,\ c_1=-6$
• $-2x + 3y + 12 = 0$ → $a_2=-2,\ b_2=3,\ c_2=12$

$$\frac{a_1}{a_2} = \frac{1}{-2}, \quad \frac{b_1}{b_2} = \frac{3}{3} = 1$$

Since $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$, the pair is consistent (lines intersect at a unique point).

Graphical Solution:

For $x + 3y = 6$:

| $x$ | 0 | 6 |
|-----|---|---|
| $y$ | 2 | 0 |

Plot points A(0, 2) and B(6, 0).

For $3y - 2x = -12$, i.e., $y = \dfrac{2x-12}{3}$:

| $x$ | 0 | 6 |
|-----|---|---|
| $y$ | −4 | 0 |

Plot points P(0, −4) and Q(6, 0).

Both lines pass through (6, 0).

∴ Solution: $x = 6,\ y = 0$

Source: Chapter 3, Section 3.2 (Graphical Method)

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• First check ratios $\frac{a_1}{a_2}$ and $\frac{b_1}{b_2}$; since they are unequal, lines intersect → consistent.
• For the graph, find two points per line (easiest: set $x=0$, then $y=0$).
• Both lines share the point (6, 0) — that is the unique solution.
• Always state the conclusion clearly. Examiners award marks for the ratio check, the table of values, and the correct solution point.