Linear Equations in Two Variables Ex 3.5 Class 10 NCERT Solutions

Understanding the Conditions for Consistency

Before applying any algebraic method to solve a pair of linear equations, we must analyze if a solution actually exists. For a pair of linear equations written in standard form:

• $a_1x + b_1y + c_1 = 0$
• $a_2x + b_2y + c_2 = 0$

We compare the ratios of their coefficients to determine the nature of their solutions:

1. Unique Solution (Consistent System): If $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the lines intersect at exactly one point. The system has a unique solution.
2. Infinitely Many Solutions (Consistent & Dependent System): If $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the lines coincide. The system has infinitely many solutions.
3. No Solution (Inconsistent System): If $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel. The system has no solution.

The Step-by-Step Cross-Multiplication Method

1. Step 1: Write in Standard Form — Arrange both equations in the standard general form: $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$. Move all constant terms to the left-hand side so that the right-hand side is zero.
2. Step 2: Arrange the Coefficients — Write down the variables $x$, $y$, and the constant $1$ in a row. Underneath, arrange the coefficients in a circular pattern starting from $b$: $x$ over $(b_1c_2 - b_2c_1)$, $y$ over $(c_1a_2 - c_2a_1)$, $1$ over $(a_1b_2 - a_2b_1)$.
3. Step 3: Set up the Proportion Formula — Write the proportion: $\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$. Ensure that the denominator of the constant term $(a_1b_2 - a_2b_1) \neq 0$ for a unique solution.
4. Step 4: Solve for x and y — Equate the first term to the third term to find $x$: $x = \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1}$. Equate the second term to the third term to find $y$: $y = \frac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1}$.

Common Board Exam Mistakes & Tips

1. The Standard Form Trap: The most common mistake students make is using equations in the form $ax + by = c$. If you keep the constant on the RHS, your sign for the constant column in cross-multiplication will be reversed, leading to wrong answers. Always convert to $ax + by + c = 0$ first!
2. Negative Sign Errors: Be extremely careful with brackets when multiplying negative values. For example, $b_1c_2 - b_2c_1$ should be written as $(b_1)(c_2) - (b_2)(c_1)$ with signs preserved inside the parentheses.
3. Check for Consistency First: Always calculate the ratios $\frac{a_1}{a_2}$ and $\frac{b_1}{b_2}$ before trying to solve. If they are equal, do not waste time applying the cross-multiplication formula; instead, determine if there are infinite solutions or no solutions.

Practice Questions with Solutions

• Q: Determine if the pair of equations $2x + y = 5$ and $3x + 2y = 8$ has a unique solution, no solution, or infinitely many solutions. If unique, find the solution using the cross-multiplication method. A: Step 1: Write both equations in standard form: $2x + y - 5 = 0$ (Equation 1) $3x + 2y - 8 = 0$ (Equation 2) Step 2: Identify the coefficients: $a_1 = 2$, $b_1 = 1$, $c_1 = -5$ $a_2 = 3$, $b_2 = 2$, $c_2 = -8$ Step 3: Compare ratios: $\frac{a_1}{a_2} = \frac{2}{3}$ $\frac{b_1}{b_2} = \frac{1}{2}$ Since $\frac{2}{3} \neq \frac{1}{2}$, the system has a unique solution. Step 4: Apply the cross-multiplication formula: $\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$ Substitute the values: $\frac{x}{(1)(-8) - (2)(-5)} = \frac{y}{(-5)(3) - (-8)(2)} = \frac{1}{(2)(2) - (3)(1)}$ $\frac{x}{-8 + 10} = \frac{y}{-15 + 16} = \frac{1}{4 - 3}$ $\frac{x}{2} = \frac{y}{1} = \frac{1}{1}$ Step 5: Solve for $x$ and $y$: $\frac{x}{2} = 1 \Rightarrow x = 2$ $\frac{y}{1} = 1 \Rightarrow y = 1$ Final answer: The system has a unique solution where $x = 2$ and $y = 1$.
• Q: For which values of $a$ and $b$ does the following pair of linear equations have an infinite number of solutions? $2x + 3y = 7$ $(a-b)x + (a+b)y = 3a+b-2$ A: Step 1: Write equations in standard form: $2x + 3y - 7 = 0$ $(a-b)x + (a+b)y - (3a+b-2) = 0$ Step 2: Identify coefficients: $a_1 = 2$, $b_1 = 3$, $c_1 = -7$ $a_2 = (a-b)$, $b_2 = (a+b)$, $c_2 = -(3a+b-2)$ Step 3: For infinitely many solutions, the ratio condition is: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ $\frac{2}{a-b} = \frac{3}{a+b} = \frac{-7}{-(3a+b-2)} = \frac{7}{3a+b-2}$ Step 4: Set up two independent equations: From first and second ratios: $2(a+b) = 3(a-b) \Rightarrow 2a + 2b = 3a - 3b \Rightarrow a = 5b$ (Equation A) From second and third ratios: $3(3a+b-2) = 7(a+b) \Rightarrow 9a + 3b - 6 = 7a + 7b \Rightarrow 2a - 4b = 6$ (Equation B) Step 5: Substitute $a = 5b$ into Equation B: $2(5b) - 4b = 6 \Rightarrow 10b - 4b = 6 \Rightarrow 6b = 6 \Rightarrow b = 1$ Step 6: Solve for $a$: $a = 5(1) = 5$ Final answer: $a = 5$ and $b = 1$.
• Q: Find the value of $k$ for which the system of equations has no solution: $3x + y = 1$ $(2k-1)x + (k-1)y = 2k+1$ A: Step 1: Write in standard form: $3x + y - 1 = 0$ $(2k-1)x + (k-1)y - (2k+1) = 0$ Step 2: Identify coefficients: $a_1 = 3$, $b_1 = 1$, $c_1 = -1$ $a_2 = (2k-1)$, $b_2 = (k-1)$, $c_2 = -(2k+1)$ Step 3: For no solution, the condition is: $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ $\frac{3}{2k-1} = \frac{1}{k-1} \neq \frac{-1}{-(2k+1)}$ Step 4: Solve the equality part: $\frac{3}{2k-1} = \frac{1}{k-1}$ $3(k-1) = 1(2k-1)$ $3k - 3 = 2k - 1$ $k = 2$ Step 5: Check inequality part with $k=2$: $\frac{b_1}{b_2} = \frac{1}{2-1} = 1$ $\frac{c_1}{c_2} = \frac{1}{2(2)+1} = \frac{1}{5}$ Since $1 \neq \frac{1}{5}$, the condition holds. Final answer: The system of equations has no solution when $k = 2$.
• Q: A fraction becomes $\frac{1}{3}$ when $1$ is subtracted from the numerator and it becomes $\frac{1}{4}$ when $8$ is added to its denominator. Find the fraction. A: Step 1: Let the fraction be $\frac{x}{y}$, where $x$ is the numerator and $y$ is the denominator. Step 2: Formulate the equations from the given conditions: Condition 1: $\frac{x-1}{y} = \frac{1}{3} \Rightarrow 3(x-1) = y \Rightarrow 3x - y - 3 = 0$ (Equation 1) Condition 2: $\frac{x}{y+8} = \frac{1}{4} \Rightarrow 4x = y + 8 \Rightarrow 4x - y - 8 = 0$ (Equation 2) Step 3: Solve the system. Let's use subtraction since the $y$ terms have identical coefficients: Subtract Equation 1 from Equation 2: $(4x - y - 8) - (3x - y - 3) = 0$ $x - 5 = 0 \Rightarrow x = 5$ Step 4: Substitute $x = 5$ into Equation 1: $3(5) - y - 3 = 0$ $15 - 3 = y \Rightarrow y = 12$ Final answer: The fraction is $\frac{5}{12}$.

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