Find a and b for Infinitely Many Solutions | Class 10 Maths

Quick answer: A pair of linear equations has infinitely many solutions when $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$. For $2x + 3y = 7$ and $(a - b)x + (a + b)y = 3a + b - 2$, equating the ratios gives $a = 5b$ and $a - 9b + 4 = 0$, so a = 5 and b = 1.

The three conditions for a pair of linear equations

Write both equations in the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$. Then compare the ratios of the coefficients:

Key idea: For infinitely many solutions, the second equation must be just a multiple of the first. That is why all three ratios must be equal.

Problem 1: Find a and b for infinitely many solutions

This is a question from NCERT Exercise 3.2 (Class 10, Pair of Linear Equations in Two Variables).

For which values of $a$ and $b$ does the following pair of linear equations have an infinite number of solutions?

$2x + 3y = 7$

$(a - b)x + (a + b)y = 3a + b - 2$

Step 1: Write in standard form and read off the coefficients.

• First equation: $2x + 3y - 7 = 0$, so $a_1 = 2$, $b_1 = 3$, $c_1 = -7$.
• Second equation: $(a - b)x + (a + b)y - (3a + b - 2) = 0$, so $a_2 = a - b$, $b_2 = a + b$, $c_2 = -(3a + b - 2)$.

Step 2: Apply the condition for infinitely many solutions.

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

The two minus signs in $\frac{-7}{-(3a + b - 2)}$ cancel. This chain of equal ratios gives two equations in $a$ and $b$.

Step 3: Equate the first and second ratios.

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

Cross-multiply:

$2(a + b) = 3(a - b)$

$2a + 2b = 3a - 3b$

$a = 5b \qquad \text{(1)}$

Step 4: Equate the first and third ratios.

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

Cross-multiply:

$2(3a + b - 2) = 7(a - b)$

$6a + 2b - 4 = 7a - 7b$

$a - 9b + 4 = 0 \qquad \text{(2)}$

Step 5: Solve (1) and (2).

Put $a = 5b$ in (2):

$5b - 9b + 4 = 0$

$-4b = -4$

$b = 1$

Then from (1), $a = 5 \times 1 = 5$.

Step 6: Verify.

With $a = 5$ and $b = 1$:

• $\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$
• $\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}$
• $\frac{c_1}{c_2} = \frac{7}{15 + 1 - 2} = \frac{7}{14} = \frac{1}{2}$

All three ratios are equal. The second equation becomes $4x + 6y = 14$, which is exactly twice the first equation.

Answer: $a = 5$ and $b = 1$.

Problem 2: Find k for no solution

For which value of $k$ will the following pair of linear equations have no solution?

$3x + y = 1$

$(2k - 1)x + (k - 1)y = 2k + 1$

Step 1: Write the condition for no solution.

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

Here $a_1 = 3$, $b_1 = 1$, $c_1 = -1$ and $a_2 = 2k - 1$, $b_2 = k - 1$, $c_2 = -(2k + 1)$.

Step 2: Equate the first two ratios.

$\frac{3}{2k - 1} = \frac{1}{k - 1}$

$3(k - 1) = 2k - 1$

$3k - 3 = 2k - 1$

$k = 2$

Step 3: Check the third ratio.

For $k = 2$: $\frac{b_1}{b_2} = \frac{1}{1} = 1$, while $\frac{c_1}{c_2} = \frac{-1}{-5} = \frac{1}{5}$. Since $1 \ne \frac{1}{5}$, the lines are parallel and distinct.

Answer: $k = 2$. The second equation becomes $3x + y = 5$, which is parallel to $3x + y = 1$.

Summary table

Exam tips

• Always rewrite both equations in the form $ax + by + c = 0$ before reading the coefficients.
• Pairing the first two ratios usually gives the simplest equation (here $a = 5b$), so do that first and substitute.
• Write the verification step. It takes one line and catches arithmetic slips.
• For "no solution" questions, always check that the third ratio is not equal, and say so in words.

Common mistakes

• Forgetting the minus sign when moving the constant term: $c_2$ is $-(3a + b - 2)$, not $3a + b - 2$. Here the signs cancel, but in other questions they may not.
• Using only two of the three ratios for infinitely many solutions. You need two separate equations to find two unknowns.
• Errors while cross-multiplying, such as writing $2(a - b) = 3(a + b)$ instead of $2(a + b) = 3(a - b)$.
• Stopping at $k = 2$ without checking the third ratio in a "no solution" question.

Frequently Asked Questions

What is the condition for infinitely many solutions?

The ratios of the x-coefficients, the y-coefficients and the constant terms must all be equal: a₁/a₂ = b₁/b₂ = c₁/c₂. The two lines then coincide.

What are a and b if 2x + 3y = 7 and (a − b)x + (a + b)y = 3a + b − 2 have infinitely many solutions?

a = 5 and b = 1. Equating the ratios gives a = 5b and a − 9b + 4 = 0, which solve to these values.

What is the condition for no solution?

a₁/a₂ = b₁/b₂ but this is not equal to c₁/c₂. The lines are parallel and never meet.

Why do we need two equations to find a and b?

There are two unknowns, so we need two independent equations. The chain of three equal ratios gives exactly two such equations.

What do coincident lines mean?

Coincident lines lie exactly on top of each other. Every point on one line is on the other, so the pair has infinitely many solutions.