Mathematics — Standard (041) — CBSE Class 10 board question

For Line 1: $x - 2y = 3$

| $x$ | 3 | 1 | -1 |
|-----|---|---|----|
| $y$ | 0 | -1 | -2 |

For Line 2: $3x - 8y = 7$

| $x$ | 1 | 3 | -1 |
|-----|---|---|----|
| $y$ | -\frac{1}{2} | \frac{1}{4} | ... |

Since the coordinates are non-integral, we also try:

| $x$ | 9 | 1 |
|-----|---|---|
| $y$ | \frac{5}{2} | -\frac{1}{2} |

Plotting: Plot both lines on the same graph using the above points.

Finding the solution algebraically to verify the intersection point:

From equation 1: $x = 3 + 2y$

Substituting in equation 2:
$$3(3 + 2y) - 8y = 7$$
$$9 + 6y - 8y = 7$$
$$-2y = -2 \Rightarrow y = 1$$

$$x = 3 + 2(1) = 5$$

The two lines intersect at the point $(5, 1)$.

∴ Solution: $x = 5$, $y = 1$

Verification: $5 - 2(1) = 3$ ✓ and $3(5) - 8(1) = 15 - 8 = 7$ ✓

Source: Chapter 3, Section 3.2 and 3.3.1

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• In a graphical question, examiners expect a table of values for each line (minimum 2–3 points), a description of plotting, and identification of the intersection point.
• Since both lines here give a non-integral intersection when read off a graph roughly, CBSE expects you to state the intersection point clearly and verify it — marks are awarded for the table, graph description, intersection point, and verification.
• The substitution step shown here both solves and verifies, which is accepted. Always show the verification step — it carries 1 mark typically.
• The key answer is $x = 5,\ y = 1$.