Table of Contents
Exercise 4.1 Solutions – Class X Mathematics
These solutions are based on the Telangana State Class X Mathematics textbook, focusing on pairs of linear equations in two variables. Mathematical expressions are rendered using MathJax.
1. By comparing the ratios \( \frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2} \), state whether the lines represented by the following pairs of linear equations intersect at a point, are parallel, or are coincident.
(a) \( 5x – 4y + 8 = 0 \), \( 7x – 6y – 9 = 0 \)
Rewrite in standard form \( a_1 x + b_1 y + c_1 = 0 \), \( a_2 x + b_2 y + c_2 = 0 \):
Equation 1: \( 5x – 4y + 8 = 0 \), so \( a_1 = 5 \), \( b_1 = -4 \), \( c_1 = 8 \).
Equation 2: \( 7x – 6y – 9 = 0 \), so \( a_2 = 7 \), \( b_2 = -6 \), \( c_2 = -9 \).
Compute ratios: \( \frac{a_1}{a_2} = \frac{5}{7} \), \( \frac{b_1}{b_2} = \frac{-4}{-6} = \frac{2}{3} \), \( \frac{c_1}{c_2} = \frac{8}{-9} \).
Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), the lines intersect at a point (unique solution).
Conclusion: The lines intersect at a point.
(b) \( 9x + 3y + 12 = 0 \), \( 18x + 6y + 24 = 0 \)
Equation 1: \( 9x + 3y + 12 = 0 \), so \( a_1 = 9 \), \( b_1 = 3 \), \( c_1 = 12 \).
Equation 2: \( 18x + 6y + 24 = 0 \), so \( a_2 = 18 \), \( b_2 = 6 \), \( c_2 = 24 \).
Compute ratios: \( \frac{a_1}{a_2} = \frac{9}{18} = \frac{1}{2} \), \( \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2} \), \( \frac{c_1}{c_2} = \frac{12}{24} = \frac{1}{2} \).
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), the lines are coincident (infinite solutions).
Conclusion: The lines are coincident.
(c) \( 6x – 3y + 10 = 0 \), \( 2x – y + 9 = 0 \)
Equation 1: \( 6x – 3y + 10 = 0 \), so \( a_1 = 6 \), \( b_1 = -3 \), \( c_1 = 10 \).
Equation 2: \( 2x – y + 9 = 0 \), so \( a_2 = 2 \), \( b_2 = -1 \), \( c_2 = 9 \).
Compute ratios: \( \frac{a_1}{a_2} = \frac{6}{2} = 3 \), \( \frac{b_1}{b_2} = \frac{-3}{-1} = 3 \), \( \frac{c_1}{c_2} = \frac{10}{9} \).
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \), the lines are parallel (no solution).
Conclusion: The lines are parallel.
2. Check whether the following equations are consistent or inconsistent. Solve them graphically.
For graphical solution, plot each pair of equations as lines on a graph. Consistency is determined by whether they intersect (consistent, unique solution), are coincident (consistent, infinite solutions), or are parallel (inconsistent, no solution). Here, I’ll solve algebraically to determine consistency, then describe the graphical approach.
(a) \( 3x + 2y = 5 \), \( 2x – 3y = 7 \)
Rewrite: \( 3x + 2y – 5 = 0 \), \( 2x – 3y – 7 = 0 \).
Check ratios: \( a_1 = 3 \), \( b_1 = 2 \), \( c_1 = -5 \); \( a_2 = 2 \), \( b_2 = -3 \), \( c_2 = -7 \).
\( \frac{a_1}{a_2} = \frac{3}{2} \), \( \frac{b_1}{b_2} = \frac{2}{-3} \), \( \frac{c_1}{c_2} = \frac{-5}{-7} = \frac{5}{7} \).
Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), the equations are consistent (intersect at a point).
Solve algebraically: Multiply first by 3 and second by 2 to make \( y \)-coefficients opposites.
\( 9x + 6y = 15 \), \( 4x – 6y = 14 \).
Add: \( 13x = 29 \implies x = \frac{29}{13} \).
Substitute into \( 3x + 2y = 5 \): \( 3 \left(\frac{29}{13}\right) + 2y = 5 \implies \frac{87}{13} + 2y = 5 \implies 2y = 5 – \frac{87}{13} = \frac{65 – 87}{13} = -\frac{22}{13} \implies y = -\frac{11}{13} \).
Solution: \( (x, y) = \left(\frac{29}{13}, -\frac{11}{13}\right) \).
Graphically: For \( 3x + 2y = 5 \), points are \( (0, 2.5) \), \( (1, 1) \). For \( 2x – 3y = 7 \), points are \( (0, -\frac{7}{3}) \), \( (1, -1.67) \). The lines intersect at \( \left(\frac{29}{13}, -\frac{11}{13}\right) \), confirming consistency.
Conclusion: Consistent, Solution: \( \left(\frac{29}{13}, -\frac{11}{13}\right) \)
(b) \( 2x – 3y = 8 \), \( 4x – 6y = 9 \)
Rewrite: \( 2x – 3y – 8 = 0 \), \( 4x – 6y – 9 = 0 \).
Check ratios: \( a_1 = 2 \), \( b_1 = -3 \), \( c_1 = -8 \); \( a_2 = 4 \), \( b_2 = -6 \), \( c_2 = -9 \).
\( \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2} \), \( \frac{b_1}{b_2} = \frac{-3}{-6} = \frac{1}{2} \), \( \frac{c_1}{c_2} = \frac{-8}{-9} = \frac{8}{9} \).
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \), the equations are inconsistent (parallel lines).
Graphically: For \( 2x – 3y = 8 \), points are \( (0, -\frac{8}{3}) \), \( (1, -2) \). For \( 4x – 6y = 9 \), points are \( (0, -\frac{3}{2}) \), \( (1, -\frac{5}{6}) \). The lines are parallel, confirming inconsistency.
Conclusion: Inconsistent
(c) \( \frac{3}{2}x – \frac{5}{3}y = 7 \), \( 9x – 10y = 12 \)
Clear fractions in the first equation: \( 3x – \frac{5}{3}y = 7 \implies 9x – 5y = 21 \).
Second equation: \( 9x – 10y = 12 \).
Check ratios: \( a_1 = 9 \), \( b_1 = -5 \), \( c_1 = -21 \); \( a_2 = 9 \), \( b_2 = -10 \), \( c_2 = -12 \).
\( \frac{a_1}{a_2} = \frac{9}{9} = 1 \), \( \frac{b_1}{b_2} = \frac{-5}{-10} = \frac{1}{2} \), \( \frac{c_1}{c_2} = \frac{-21}{-12} = \frac{7}{4} \).
Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), the equations are consistent.
Solve: Subtract the second from the first: \( (9x – 5y) – (9x – 10y) = 21 – 12 \implies 5y = 9 \implies y = \frac{9}{5} \).
Substitute into \( 9x – 10y = 12 \): \( 9x – 10 \left(\frac{9}{5}\right) = 12 \implies 9x – 18 = 12 \implies 9x = 30 \implies x = \frac{10}{3} \).
Graphically: For \( 9x – 5y = 21 \), points are \( (0, -\frac{21}{5}) \), \( (1, -\frac{12}{5}) \). For \( 9x – 10y = 12 \), points are \( (0, -\frac{6}{5}) \), \( (1, -\frac{3}{10}) \). Intersect at \( \left(\frac{10}{3}, \frac{9}{5}\right) \).
Conclusion: Consistent, Solution: \( \left(\frac{10}{3}, \frac{9}{5}\right) \)
(d) \( 5x – 3y = 11 \), \( -10x + 6y = -22 \)
Check ratios: \( a_1 = 5 \), \( b_1 = -3 \), \( c_1 = -11 \); \( a_2 = -10 \), \( b_2 = 6 \), \( c_2 = 22 \).
\( \frac{a_1}{a_2} = \frac{5}{-10} = -\frac{1}{2} \), \( \frac{b_1}{b_2} = \frac{-3}{6} = -\frac{1}{2} \), \( \frac{c_1}{c_2} = \frac{-11}{22} = -\frac{1}{2} \).
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), the equations are coincident (consistent, infinite solutions).
Graphically: For \( 5x – 3y = 11 \), points are \( (0, -\frac{11}{3}) \), \( (1, -2) \). The second equation is a multiple (\( -2 \times \text{first} \)), so the lines coincide.
Conclusion: Consistent (coincident)
(e) \( \frac{4}{3}x + 2y = 8 \), \( 2x + 3y = 12 \)
Clear fraction: \( \frac{4}{3}x + 2y = 8 \implies 4x + 6y = 24 \).
Second equation: \( 2x + 3y = 12 \).
Check ratios: \( a_1 = 4 \), \( b_1 = 6 \), \( c_1 = -24 \); \( a_2 = 2 \), \( b_2 = 3 \), \( c_2 = -12 \).
\( \frac{a_1}{a_2} = \frac{4}{2} = 2 \), \( \frac{b_1}{b_2} = \frac{6}{3} = 2 \), \( \frac{c_1}{c_2} = \frac{-24}{-12} = 2 \). Coincident, but solve to confirm.
Solve: Multiply second by 2: \( 4x + 6y = 24 \), which is the first equation. Thus, coincident.
Graphically: For \( 2x + 3y = 12 \), points are \( (0, 4) \), \( (6, 0) \). The first equation plots the same line, confirming infinite solutions.
Conclusion: Consistent (coincident)
(f) \( x + y = 5 \), \( 2x + 2y = 10 \)
Check ratios: \( a_1 = 1 \), \( b_1 = 1 \), \( c_1 = -5 \); \( a_2 = 2 \), \( b_2 = 2 \), \( c_2 = -10 \).
\( \frac{a_1}{a_2} = \frac{1}{2} \), \( \frac{b_1}{b_2} = \frac{1}{2} \), \( \frac{c_1}{c_2} = \frac{-5}{-10} = \frac{1}{2} \). Coincident.
Graphically: For \( x + y = 5 \), points are \( (0, 5) \), \( (5, 0) \). The second equation is the same line, confirming infinite solutions.
Conclusion: Consistent (coincident)
(g) \( x – y = 8 \), \( 3x – 3y = 16 \)
Check ratios: \( a_1 = 1 \), \( b_1 = -1 \), \( c_1 = -8 \); \( a_2 = 3 \), \( b_2 = -3 \), \( c_2 = -16 \).
\( \frac{a_1}{a_2} = \frac{1}{3} \), \( \frac{b_1}{b_2} = \frac{-1}{-3} = \frac{1}{3} \), \( \frac{c_1}{c_2} = \frac{-8}{-16} = \frac{1}{2} \). Parallel, inconsistent.
Graphically: For \( x – y = 8 \), points are \( (0, -8) \), \( (8, 0) \). For \( 3x – 3y = 16 \), points are \( (0, -\frac{16}{3}) \), \( (\frac{16}{3}, 0) \). Parallel lines.
Conclusion: Inconsistent
(h) \( 2x + y – 6 = 0 \), \( 4x + 2y – 4 = 0 \)
Check ratios: \( a_1 = 2 \), \( b_1 = 1 \), \( c_1 = -6 \); \( a_2 = 4 \), \( b_2 = 2 \), \( c_2 = -4 \).
\( \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2} \), \( \frac{b_1}{b_2} = \frac{1}{2} \), \( \frac{c_1}{c_2} = \frac{-6}{-4} = \frac{3}{2} \). Parallel, inconsistent.
Graphically: For \( 2x + y = 6 \), points are \( (0, 6) \), \( (3, 0) \). For \( 4x + 2y = 4 \), points are \( (0, 2) \), \( (1, 0) \). Parallel lines.
Conclusion: Inconsistent
(i) \( 2x – 2y – 2 = 0 \), \( 4x – 4y – 5 = 0 \)
Check ratios: \( a_1 = 2 \), \( b_1 = -2 \), \( c_1 = -2 \); \( a_2 = 4 \), \( b_2 = -4 \), \( c_2 = -5 \).
\( \frac{a_1}{a_2} = \frac{1}{2} \), \( \frac{b_1}{b_2} = \frac{1}{2} \), \( \frac{c_1}{c_2} = \frac{-2}{-5} = \frac{2}{5} \). Parallel, inconsistent.
Graphically: For \( x – y = 1 \), points are \( (0, -1) \), \( (1, 0) \). For \( 4x – 4y = 5 \), points are \( (0, -\frac{5}{4}) \), \( (1, -\frac{1}{4}) \). Parallel lines.
Conclusion: Inconsistent
3. Neha went to a ‘sale’ to purchase some pants and skirts. When her friend asked her how many of each she had bought, she answered, “the number of skirts are two less than twice the number of pants purchased and the number of skirts is four less than four times the number of pants purchased.” Help her friend to find how many pants and skirts Neha bought.
Let \( x \) be the number of pants, \( y \) be the number of skirts.
First statement: \( y = 2x – 2 \).
Second statement: \( y = 4x – 4 \).
Equate the two expressions for \( y \): \( 2x – 2 = 4x – 4 \).
Solve: \( 2x – 4x = -4 + 2 \implies -2x = -2 \implies x = 1 \).
Substitute \( x = 1 \) into \( y = 2x – 2 \): \( y = 2(1) – 2 = 0 \).
So, Neha bought 1 pant and 0 skirts. Check the second equation: \( y = 4(1) – 4 = 0 \), which matches.
Number of pants: 1, Number of skirts: 0
4. 10 students of Class-X took part in a mathematics quiz. If the number of girls is 4 more than the number of boys then, find the number of boys and the number of girls who took part in the quiz.
Let \( x \) be the number of boys, \( y \) be the number of girls.
Equation 1: \( x + y = 10 \) (total students).
Equation 2: \( y = x + 4 \) (girls are 4 more than boys).
Substitute \( y = x + 4 \) into the first equation: \( x + (x + 4) = 10 \).
Solve: \( 2x + 4 = 10 \implies 2x = 6 \implies x = 3 \).
Then, \( y = x + 4 = 3 + 4 = 7 \).
Check: \( 3 + 7 = 10 \), and \( 7 = 3 + 4 \), both true.
Number of boys: 3, Number of girls: 7
5. 5 pencils and 7 pens together cost ₹50 whereas 7 pencils and 5 pens together cost ₹46. Find the cost of one pencil and that of one pen.
Let \( x \) be the cost of one pencil, \( y \) be the cost of one pen (in ₹).
Equation 1: \( 5x + 7y = 50 \).
Equation 2: \( 7x + 5y = 46 \).
Add the equations: \( (5x + 7y) + (7x + 5y) = 50 + 46 \implies 12x + 12y = 96 \implies x + y = 8 \).
Subtract the second from the first: \( (5x + 7y) – (7x + 5y) = 50 – 46 \implies -2x + 2y = 4 \implies -x + y = 2 \).
Solve the system: \( x + y = 8 \), \( -x + y = 2 \).
Add: \( 2y = 10 \implies y = 5 \).
Substitute \( y = 5 \) into \( x + y = 8 \): \( x + 5 = 8 \implies x = 3 \).
Check: \( 5(3) + 7(5) = 15 + 35 = 50 \), \( 7(3) + 5(5) = 21 + 25 = 46 \), both true.
Cost of one pencil: ₹3, Cost of one pen: ₹5
6. Half the perimeter of a rectangular garden is 36 m. If the length is 4 m more than its width, find the dimensions of the garden.
Let the width be \( x \) m, length be \( y \) m.
Half the perimeter: \( x + y = 36 \).
Length is 4 m more than width: \( y = x + 4 \).
Substitute \( y = x + 4 \) into \( x + y = 36 \): \( x + (x + 4) = 36 \implies 2x + 4 = 36 \implies 2x = 32 \implies x = 16 \).
Then, \( y = x + 4 = 16 + 4 = 20 \).
Check: Perimeter = \( 2(16 + 20) = 72 \), half = 36, and \( 20 = 16 + 4 \), both true.
Dimensions: Length = 20 m, Width = 16 m
7. We have a linear equation \( 2x + 3y – 8 = 0 \). Write another linear equation in two variables \( x \) and \( y \) such that the geometrical representation of the pair so formed is intersecting lines. Now, write two more linear equations so that one forms a pair of parallel lines and the second forms coincident line with the given equation.
Given equation: \( 2x + 3y – 8 = 0 \).
For intersecting lines, \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \). Choose \( a_2 = 3 \), \( b_2 = 2 \), \( c_2 = -7 \): \( 3x + 2y – 7 = 0 \).
Check: \( \frac{2}{3} \neq \frac{3}{2} \), so they intersect.
For parallel lines, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \). Multiply by 2 and change the constant: \( 4x + 6y – 10 = 0 \). Check: \( \frac{2}{4} = \frac{3}{6} \neq \frac{-8}{-10} \).
For coincident lines, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \). Multiply by 3: \( 6x + 9y – 24 = 0 \). Check: \( \frac{2}{6} = \frac{3}{9} = \frac{-8}{-24} = \frac{1}{3} \).
Intersecting: \( 3x + 2y – 7 = 0 \), Parallel: \( 4x + 6y – 10 = 0 \), Coincident: \( 6x + 9y – 24 = 0 \)
8. The area of a rectangle gets reduced by 80 sq units if its length is reduced by 5 units and breadth is increased by 2 units. If we increase the length by 10 units and decrease the breadth by 5 units, the area increases by 50 sq units. Find the length and breadth of the rectangle.
Let length be \( x \) units, breadth be \( y \) units. Area = \( xy \).
First condition: \( (x – 5)(y + 2) = xy – 80 \).
Expand: \( xy + 2x – 5y – 10 = xy – 80 \implies 2x – 5y = -70 \).
Second condition: \( (x + 10)(y – 5) = xy + 50 \).
Expand: \( xy – 5x + 10y – 50 = xy + 50 \implies -5x + 10y = 100 \implies -x + 2y = 20 \).
Solve the system: \( 2x – 5y = -70 \), \( -x + 2y = 20 \).
Multiply the second by 2: \( -2x + 4y = 40 \).
Add to the first: \( (2x – 5y) + (-2x + 4y) = -70 + 40 \implies -y = -30 \implies y = 30 \).
Substitute \( y = 30 \) into \( -x + 2y = 20 \): \( -x + 2(30) = 20 \implies -x + 60 = 20 \implies -x = -40 \implies x = 40 \).
Check: First: \( (40 – 5)(30 + 2) = 35 \cdot 32 = 1120 \), \( 40 \cdot 30 – 80 = 1200 – 80 = 1120 \), matches. Second: \( (40 + 10)(30 – 5) = 50 \cdot 25 = 1250 \), \( 1200 + 50 = 1250 \), matches.
Length: 40 units, Breadth: 30 units
