Welcome to StudyMats. This comprehensive educational guide for Class 10 Mathematics Chapter 3 Linear Equations Solutions is prepared to assist students in preparing effectively for their school, board, and competitive examinations. All questions, exercise solutions, and concept notes strictly adhere to the latest NCERT curriculum and CBSE guidelines.
Exercise 3.1
Question: (i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
Solution:
Let the number of boys = \(x\)
Let the number of girls = \(y\)
We have:
\[
x + y = 10
\]
\[
y - x = 4
\]
Question: (ii) 5 pencils and 7 pens together cost ₹50, whereas 7 pencils and 5 pens together cost ₹46. Find the cost of one pencil and one pen.
Solution:
Let the cost of a pencil = \(x\)
Let the cost of a pen = \(y\)
We have:
\[
5x + 7y = 50
\]
\[
7x + 5y = 46
\]
Question: 2. On comparing the ratios \(\tfrac{a_1}{a_2}, \tfrac{b_1}{b_2}, \tfrac{c_1}{c_2}\), find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
(i) \(5x - 4y + 8 = 0\) and \(7x + 6y - 9 = 0\)
Solution:
Compare:
\[
\frac{a_1}{a_2} = \frac{5}{7},\quad
\frac{b_1}{b_2} = \frac{-4}{6} = -\tfrac{2}{3},\quad
\frac{c_1}{c_2} = \frac{8}{-9}
\]
Since these ratios are not equal, the lines intersect at a point.
Question: (ii) \(9x + 3y + 12 = 0\) and \(18x + 6y + 24 = 0\)
Solution:
\[
\frac{a_1}{a_2} = \frac{9}{18} = \tfrac{1}{2},\quad
\frac{b_1}{b_2} = \frac{3}{6} = \tfrac{1}{2},\quad
\frac{c_1}{c_2} = \frac{12}{24} = \tfrac{1}{2}
\]
All ratios are equal → the lines are coincident.
Question: (iii) \(6x - 3y + 10 = 0\) and \(2x - y + 9 = 0\)
Solution:
\[
\frac{a_1}{a_2} = \frac{6}{2} = 3,\quad
\frac{b_1}{b_2} = \frac{-3}{-1} = 3,\quad
\frac{c_1}{c_2} = \frac{10}{9}
\]
Here, \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}\) → lines are parallel.
Question: 3. On comparing the ratios, find out whether the following pair of linear equations are consistent or inconsistent:
(i) \(3x + 2y = 5\); \(2x - 3y = 7\)
Solution:
\[
\frac{a_1}{a_2} = \frac{3}{2},\quad
\frac{b_1}{b_2} = \frac{2}{-3},\quad
\frac{c_1}{c_2} = \frac{5}{7}
\]
Ratios are unequal → lines are intersecting → 'consistent with a unique solution'.
Question: (ii) \(2x - 3y = 8\); \(4x - 6y = 9\)
Solution:
\[
\frac{a_1}{a_2} = \frac{2}{4} = \tfrac{1}{2},\quad
\frac{b_1}{b_2} = \frac{-3}{-6} = \tfrac{1}{2},\quad
\frac{c_1}{c_2} = \frac{8}{9}
\]
Here, \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}\) → 'parallel' → 'inconsistent'.
Question: (iii) \(\frac{3}{2}x + \frac{5}{3}y = 7\); \(9x - 10y = 14\)
Solution:
\[
\frac{a_1}{a_2} = \frac{3/2}{9} = \frac{1}{6},\quad
\frac{b_1}{b_2} = \frac{5/3}{-10} = -\frac{1}{6},\quad
\frac{c_1}{c_2} = \frac{7}{14} = \tfrac{1}{2}
\]
Ratios not consistent → lines intersect → 'consistent'.
Question: (iv) \(5x - 3y = 11\); \(-10x + 6y = -22\)
Solution:
\[
\frac{a_1}{a_2} = \frac{5}{-10} = -\tfrac{1}{2},\quad
\frac{b_1}{b_2} = \frac{-3}{6} = -\tfrac{1}{2},\quad
\frac{c_1}{c_2} = \frac{11}{-22} = -\tfrac{1}{2}
\]
All ratios equal → lines are coincident → 'infinite solutions' → 'consistent'.
Question: (v) \(\frac{4}{3}x + 2y = 8\); \(2x + 3y = 12\)
Solution:
\[
\frac{a_1}{a_2} = \frac{4/3}{2} = \frac{2}{3},\quad
\frac{b_1}{b_2} = \frac{2}{3},\quad
\frac{c_1}{c_2} = \frac{8}{12} = \frac{2}{3}
\]
RatiOS equal → lines are coincident → 'infinite solutions' → 'consistent'.
Question: 4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:
(i) \(x + y = 5\); \(2x + 2y = 10\)
Solution:
Second equation is just 2× first → equations are coincident → 'infinite solutions' (consistent).
Question: (ii) \(x - y = 8\); \(3x - 3y = 16\)
Solution:
Second is 3× first except constant (16 ≠ 24) → lines are parallel → 'no solution' (inconsistent).
Question: (iii) \(2x + y - 6 = 0\); \(4x - 2y - 4 = 0\)
Solution:
Compute ratios:
\[
\frac{a_1}{a_2} = \frac{2}{4} = \tfrac{1}{2},\quad
\frac{b_1}{b_2} = \frac{1}{-2} = -\tfrac{1}{2},\quad
\frac{c_1}{c_2} = \frac{-6}{-4} = \tfrac{3}{2}
\]
Lines intersect → 'unique solution' (consistent).
Question: (iv) \(2x - 2y - 2 = 0\); \(4x - 4y - 5 = 0\)
Solution:
\[
\frac{a_1}{a_2} = \frac{2}{4} = \tfrac{1}{2},\quad
\frac{b_1}{b_2} = \frac{-2}{-4} = \tfrac{1}{2},\quad
\frac{c_1}{c_2} = \frac{-2}{-5} = \tfrac{2}{5}
\]
Lines are parallel → 'no solution' (inconsistent).
Question: 5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.
Solution:
Let width = \(x\), length = \(x + 4\).
Perimeter = \(2(x + x + 4)\). Half perimeter = \(x + x + 4 = 36\).
\[
2x + 4 = 36
\]
\[
2x = 32
\]
\[
x = 16
\]
Width = 16 m, Length = 20 m.
Question: 6. Given the linear equation \(2x + 3y - 8 = 0\), write another linear equation in two variables such that the geometrical representation of the pair so formed is: (i) intersecting (ii) parallel (iii) coincident lines.
Solution:
(i) For intersecting: pick non-proportional coefficients, e.g., \(x - y = 0\).
(ii) For parallel: multiply LHS but change constant, e.g., \(4x + 6y + 5 = 0\).
(iii) For coincident: multiply entire equation by a nonzero constant, e.g., \(4x + 6y - 16 = 0\).
Question: 7. Draw the graphs of the equations \(x - y + 1 = 0\) and \(3x + 2y - 12 = 0\). Determine the coordinates of the vertices of the triangle formed by these lines and the x‑axis, and shade the triangular region.
Solution:
Convert to slope‑intercept form:
(i) \(y = x + 1\) → intersects x-axis at \(y=0 \Rightarrow x = -1\).
(ii) \(2y = 12 - 3x\) → \(y = 6 - \frac{3}{2}x\); x‑axis at \(y=0 \Rightarrow x = 4\).
Intersection point: solve \(x + 1 = 6 - 1.5x\):
\[
x + 1.5x = 6 - 1 \Rightarrow 2.5x = 5 \Rightarrow x = 2
\]
Then \(y = 3\).
Triangle vertices: \((-1,0), (4,0), (2,3)\).
💡 Important Exam & Study Tips
- Master NCERT Fundamentals: Ensure thorough familiarity with standard textbook definitions, formulas, and in-text examples before solving advanced problems.
- Step-by-Step Presentation: In board examinations, neat presentation, correct formula application, and labeled steps earn maximum step-marking.
- Practice Under Timed Conditions: Regular practice with previous years' question papers (PYQs) builds exam speed and accuracy.
📚 Related Study Materials on StudyMats
Access more chapter-wise notes and practice papers:
❓ Frequently Asked Questions (FAQs)
Q1: Are these solutions aligned with the latest rationalized syllabus?
Yes, all study notes and practice questions on StudyMats strictly follow the updated NCERT textbook pattern and CBSE assessment guidelines.
Q2: How can I prepare this chapter for highest marks?
Review the theory thoroughly, solve all textbook exercise questions, and repeatedly practice the high-frequency questions provided in this guide.