Mathematics — CBSE Class 10 board question

For infinitely many solutions, the condition is:

$$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$$

Here: $a_1 = 2,\ b_1 = 3,\ c_1 = -7$ and $a_2 = 2a,\ b_2 = (a+b),\ c_2 = -28$

$$\frac{2}{2a} = \frac{3}{a+b} = \frac{-7}{-28}$$

From $\dfrac{2}{2a} = \dfrac{7}{28} = \dfrac{1}{4}$:

$$2a = 8 \implies a = 4$$

From $\dfrac{3}{a+b} = \dfrac{1}{4}$:

$$a + b = 12 \implies 4 + b = 12 \implies b = 8$$

Therefore, $a = 4$ and $b = 8$.

Source: Chapter 3, Section 3.2 (Graphical Method / Condition for infinitely many solutions)

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• The key condition for infinitely many solutions (coincident lines) is $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$.
• Rewrite both equations in standard form first to correctly identify $c_1$ and $c_2$ (i.e., $-7$ and $-28$).
• Solve the ratios one pair at a time — first find $a$, then substitute to find $b$.
• Show all steps clearly; examiners award marks for setting up the ratio condition, finding $a$, and finding $b$ (roughly 1 mark each).