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NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables covers graphical and algebraic methods (substitution, elimination, cross-multiplication) for solving a pair of linear equations, and conditions for consistency.

Overview of Topics & Subtopics (Click to Jump)

1. Graphical Method of Solution

Short Note:

• A pair of linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 represents two lines, which may be intersecting (unique solution), coincident (infinitely many solutions) or parallel (no solution).

• Condition check: a1/a2 ≠ b1/b2 ⇒ unique solution (consistent); a1/a2 = b1/b2 = c1/c2 ⇒ infinite solutions (dependent/consistent); a1/a2 = b1/b2 ≠ c1/c2 ⇒ no solution (inconsistent).

2. Algebraic Methods of Solution

Short Note:

• Substitution Method: express one variable in terms of the other from one equation and substitute into the second equation.

• Elimination Method: multiply the equations by suitable constants so that the coefficients of one variable become equal, then add or subtract to eliminate that variable.

• Cross-Multiplication Method: for a1x+b1y+c1=0 and a2x+b2y+c2=0, x/(b1c2−b2c1) = y/(c1a2−c2a1) = 1/(a1b2−a2b1).

3. Equations Reducible to a Pair of Linear Equations

Short Note:

• Some equations involving 1/x, 1/y (or similar substitutions) can be reduced to linear form by taking u = 1/x, v = 1/y, solved as a standard linear pair, then converted back.

NCERT Textbook Exercise Questions

Check graphically whether the pair of equations x + 3y = 6 and 2x − 3y = 12 is consistent. If so, solve it graphically.
Answer: a1/a2 = 1/2, b1/b2 = 3/−3 = −1. Since 1/2 ≠ −1, the lines intersect at a unique point, so the pair is consistent. Solving: from the equations, adding suitably gives x = 6, y = 0, so the solution is (6, 0).
Solve 2x + 3y = 11 and 2x − 4y = −24 by the elimination method, and find the value of m for which y = mx + 3.
Answer: Subtracting: (2x+3y) − (2x−4y) = 11−(−24) ⇒ 7y = 35 ⇒ y = 5. Substituting: 2x + 15 = 11 ⇒ x = −2. So (x,y) = (−2, 5). Since y = mx+3: 5 = m(−2)+3 ⇒ m = −1.
Solve the pair of equations x/2 + 2y/3 = −1 and x − y/3 = 3 using cross-multiplication.
Answer: Rewriting as 3x + 4y + 6 = 0 and 3x − y − 9 = 0, applying cross-multiplication formula gives x = 2 and y = −3.
A fraction becomes 9/11 if 2 is added to both numerator and denominator. If 3 is added to both, it becomes 5/6. Find the fraction.
Answer: Let the fraction be x/y. (x+2)/(y+2) = 9/11 and (x+3)/(y+3) = 5/6. Solving these linear equations simultaneously gives x = 7, y = 9, so the fraction is 7/9.
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NCERT Solutions for Class 10 Maths Chapter 1 Real NumbersNCERT Solutions for Class 10 Maths Chapter 2 PolynomialsNCERT Solutions for Class 10 Maths Chapter 4 Quadratic EquationsNCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

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Class 10 › Mathematics › Pair of Linear Equations
Chapter notesRead the chapterChapter test10 questions on one pageOnline testTest 1, 2 and 3, one question at a timePYQPrevious year questionsStudy materialPDF notes
Topics in this chapter: Graphical Method of SolutionAlgebraic Methods of SolutionEquations Reducible to a Pair of Linear Equations