School Assignments for Class 10 Mathematics: Chapter 03 Pair of Linear Equations in Two Variables
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Practice Class 10 Mathematics Assignments: Chapter 03 Pair of Linear Equations in Two Variables
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Graphical Method and Pair of Linear Equations
Answer:
Let Aftab's present age be x years and his daughter's present age be y years.
Seven years ago: \( x - 7 = 7(y - 7) \Rightarrow x - 7y = -42 \) ...(i).
Three years from now: \( x + 3 = 3(y + 3) \Rightarrow x - 3y = 6 \) ...(ii).
Points for (i), \( y = \frac{x + 42}{7} \):
| x | 0 | 7 |
| y | 6 | 7 |
Points for (ii), \( y = \frac{x - 6}{3} \):
| x | 0 | 6 |
| y | -2 | 0 |
Plotting the two lines, they meet at \( (42, 12) \). The lines are intersecting, so the pair has a unique solution: Aftab is 42 years old and his daughter is 12 years old (check: 35 = 7 × 5 and 45 = 3 × 15).
In simple words: Turn each sentence into an equation and draw both lines; the meeting point gives the two ages.
Teacher's Note:
1. Convert each condition to an equation separately: write the past or future ages first.
2. The intersection point of the lines is the solution, so read it from the graph and then verify in both statements.
Answer:
Let the cost of a bat be Rs. x and the cost of a ball be Rs. y.
Algebraically: \( 3x + 6y = 3900 \), that is, \( x + 2y = 1300 \) ...(i), and \( x + 3y = 1300 \) ...(ii).
Points for (i):
| x | 300 | 100 | -100 |
| y | 500 | 600 | 700 |
Points for (ii):
| x | 1000 | 700 | 400 |
| y | 100 | 200 | 300 |
Geometrically the two lines are drawn below. They meet on the x-axis at \( (1300, 0) \), which would make the ball free. The data are only meant for practice in representing the situation, not for realistic prices.
In simple words: Write one equation for each purchase and draw the two lines on the same graph.
Teacher's Note:
1. Always define the variables first: x for a bat and y for a ball.
2. Simplify \( 3x + 6y = 3900 \) by dividing by 3 before choosing points.
Answer:
Let the cost of 1 kg of apples be Rs. x and of 1 kg of grapes be Rs. y.
\( 2x + y = 160 \) ...(i) and \( 4x + 2y = 300 \) ...(ii).
For (i), \( y = 160 - 2x \):
| x | 50 | 60 | 70 |
| y | 60 | 40 | 20 |
For (ii), \( y = \frac{300 - 4x}{2} \):
| x | 70 | 80 | 75 |
| y | 10 | -10 | 0 |
Here \( \frac{a_1}{a_2} = \frac{1}{2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} = \frac{160}{300} \), so the lines are parallel and the pair has no solution (the prices cannot satisfy both conditions).
In simple words: The two price conditions give parallel lines, so no single pair of prices fits both.
Teacher's Note:
1. Use the ratio test to predict the picture before plotting.
2. Choose points that are easy to plot, such as the intercepts, and use the same scale for both lines.
Answer:
Let the number of girls be x and the number of boys be y.
\( x + y = 10 \) ...(i) and \( x - y = 4 \) ...(ii).
For (i):
| x | 4 | 5 | 6 |
| y | 6 | 5 | 4 |
For (ii):
| x | 5 | 4 | 3 |
| y | 1 | 0 | -1 |
The lines intersect at \( (7, 3) \). Hence there are 7 girls and 3 boys.
In simple words: Plot both lines and read the meeting point: it gives the number of girls and boys.
Teacher's Note:
1. Check the answer: 7 + 3 = 10 and 7 − 3 = 4.
2. Label which variable stands for girls and which for boys before plotting.
Answer:
Let the cost of one pencil be Rs. x and of one pen be Rs. y.
\( 5x + 7y = 50 \) ...(i) and \( 7x + 5y = 46 \) ...(ii).
For (i):
| x | 3 | 10 | -4 |
| y | 5 | 0 | 10 |
For (ii):
| x | 3 | 8 | -2 |
| y | 5 | -2 | 12 |
The lines intersect at \( (3, 5) \). Hence one pencil costs Rs. 3 and one pen costs Rs. 5 (check: 15 + 35 = 50 and 21 + 25 = 46).
In simple words: Each purchase gives one equation; the intersection gives the prices.
Teacher's Note:
1. Pick points with integer coordinates so the lines can be drawn accurately.
2. Always substitute the answer back to verify both totals.
Question 6. On comparing the ratios \( \frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2} \), find out whether the lines representing the following pair of linear equations intersect at a point, are parallel or coincide: \( 5x - 4y + 8 = 0 and 7x + 6y - 9 = 0 \).
Answer:
Here \( \frac{a_1}{a_2} = \frac{5}{7},\ \frac{b_1}{b_2} = -\frac{2}{3},\ \frac{c_1}{c_2} = -\frac{8}{9} \).
Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), the lines intersect at exactly one point: the pair is consistent with a unique solution.
In simple words: Compare the three ratios; their equalities decide whether the lines meet, overlap or never meet.
Teacher's Note:
1. Write both equations in the form \( ax + by + c = 0 \) before reading a, b, c.
2. Compare \( \frac{a_1}{a_2} \) with \( \frac{b_1}{b_2} \) first; check \( \frac{c_1}{c_2} \) only if these two are equal.
Question 7. On comparing the ratios \( \frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2} \), find out whether the lines representing the following pair of linear equations intersect at a point, are parallel or coincide: \( 9x + 3y + 12 = 0 and 18x + 6y + 24 = 0 \).
Answer:
Here \( \frac{a_1}{a_2} = \frac{1}{2},\ \frac{b_1}{b_2} = \frac{1}{2},\ \frac{c_1}{c_2} = \frac{1}{2} \).
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), the lines are coincident: the pair is consistent with infinitely many solutions.
In simple words: Compare the three ratios; their equalities decide whether the lines meet, overlap or never meet.
Teacher's Note:
1. Write both equations in the form \( ax + by + c = 0 \) before reading a, b, c.
2. Compare \( \frac{a_1}{a_2} \) with \( \frac{b_1}{b_2} \) first; check \( \frac{c_1}{c_2} \) only if these two are equal.
Question 8. On comparing the ratios \( \frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2} \), find out whether the lines representing the following pair of linear equations intersect at a point, are parallel or coincide: \( 6x - 3y + 10 = 0 and 2x - y + 9 = 0 \).
Answer:
Here \( \frac{a_1}{a_2} = 3,\ \frac{b_1}{b_2} = 3,\ \frac{c_1}{c_2} = \frac{10}{9} \).
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \), the lines are parallel: the pair is inconsistent and has no solution.
In simple words: Compare the three ratios; their equalities decide whether the lines meet, overlap or never meet.
Teacher's Note:
1. Write both equations in the form \( ax + by + c = 0 \) before reading a, b, c.
2. Compare \( \frac{a_1}{a_2} \) with \( \frac{b_1}{b_2} \) first; check \( \frac{c_1}{c_2} \) only if these two are equal.
Question 9. On comparing the ratios \( \frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2} \), find out whether the following pair of linear equations is consistent or inconsistent: \( 3x + 2y = 5 and 2x - 3y = 7 \).
Answer:
Here \( \frac{a_1}{a_2} = \frac{3}{2},\ \frac{b_1}{b_2} = -\frac{2}{3},\ \frac{c_1}{c_2} = \frac{5}{7} \).
Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), the lines intersect at exactly one point: the pair is consistent with a unique solution.
Hence the pair is consistent (unique solution).
In simple words: A pair is consistent when it has at least one solution, that is, when the lines meet or coincide.
Teacher's Note:
1. Consistent covers both intersecting and coincident lines.
2. Inconsistent means parallel lines, so the system has no solution.
Question 10. On comparing the ratios \( \frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2} \), find out whether the following pair of linear equations is consistent or inconsistent: \( 2x - 3y = 8 and 4x - 6y = 9 \).
Answer:
Here \( \frac{a_1}{a_2} = \frac{1}{2},\ \frac{b_1}{b_2} = \frac{1}{2},\ \frac{c_1}{c_2} = \frac{8}{9} \).
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \), the lines are parallel: the pair is inconsistent and has no solution.
Hence the pair is inconsistent.
In simple words: A pair is consistent when it has at least one solution, that is, when the lines meet or coincide.
Teacher's Note:
1. Consistent covers both intersecting and coincident lines.
2. Inconsistent means parallel lines, so the system has no solution.
Question 11. On comparing the ratios \( \frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2} \), find out whether the following pair of linear equations is consistent or inconsistent: \( \frac{3}{2}x + \frac{5}{3}y = 7 and 9x - 10y = 14 \).
Answer:
Here \( \frac{a_1}{a_2} = \frac{1}{6},\ \frac{b_1}{b_2} = -\frac{1}{6},\ \frac{c_1}{c_2} = \frac{1}{2} \).
Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), the lines intersect at exactly one point: the pair is consistent with a unique solution.
Hence the pair is consistent (unique solution).
In simple words: A pair is consistent when it has at least one solution, that is, when the lines meet or coincide.
Teacher's Note:
1. Consistent covers both intersecting and coincident lines.
2. Inconsistent means parallel lines, so the system has no solution.
Question 12. On comparing the ratios \( \frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2} \), find out whether the following pair of linear equations is consistent or inconsistent: \( 5x - 3y = 11 and -10x + 6y = -22 \).
Answer:
Here \( \frac{a_1}{a_2} = -\frac{1}{2},\ \frac{b_1}{b_2} = -\frac{1}{2},\ \frac{c_1}{c_2} = -\frac{1}{2} \).
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), the lines are coincident: the pair is consistent with infinitely many solutions.
Hence the pair is consistent (dependent, infinitely many solutions).
In simple words: A pair is consistent when it has at least one solution, that is, when the lines meet or coincide.
Teacher's Note:
1. Consistent covers both intersecting and coincident lines.
2. Inconsistent means parallel lines, so the system has no solution.
Question 13. On comparing the ratios \( \frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2} \), find out whether the following pair of linear equations is consistent or inconsistent: \( \frac{4}{3}x + 2y = 8 and 2x + 3y = 12 \).
Answer:
Here \( \frac{a_1}{a_2} = \frac{2}{3},\ \frac{b_1}{b_2} = \frac{2}{3},\ \frac{c_1}{c_2} = \frac{2}{3} \).
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), the lines are coincident: the pair is consistent with infinitely many solutions.
Hence the pair is consistent (dependent, infinitely many solutions).
In simple words: A pair is consistent when it has at least one solution, that is, when the lines meet or coincide.
Teacher's Note:
1. Consistent covers both intersecting and coincident lines.
2. Inconsistent means parallel lines, so the system has no solution.
Answer:
\( \frac{a_1}{a_2} = \frac{1}{2},\ \frac{b_1}{b_2} = \frac{1}{2},\ \frac{c_1}{c_2} = \frac{5}{10} = \frac{1}{2} \): all three ratios are equal, so the lines are coincident and the pair is consistent with infinitely many solutions.
For \( x + y = 5 \):
| x | 2 | 3 | 4 |
| y | 3 | 2 | 1 |
For \( 2x + 2y = 10 \) the table is the same, so the lines coincide and every point of the line \( x + y = 5 \) is a solution.
In simple words: Both equations describe the same line, so every point on it is a solution.
Teacher's Note:
1. Divide the second equation by 2; it becomes the first one.
2. Coincident lines mean consistent and dependent equations.
Question 15. Which of the following pairs of linear equations are consistent or inconsistent? If consistent, obtain the solution graphically: (b) \( x - y = 8,\ 3x - 3y = 16 \)
Answer:
\( \frac{a_1}{a_2} = \frac{1}{3},\ \frac{b_1}{b_2} = \frac{-1}{-3} = \frac{1}{3},\ \frac{c_1}{c_2} = \frac{8}{16} = \frac{1}{2} \), so \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \).
The lines are parallel (see graph: \( y = x - 8 \) and \( y = x - \frac{16}{3} \)), so the pair is inconsistent: no solution.
In simple words: The two lines have the same slope but different intercepts, so they never meet.
Teacher's Note:
1. Equal slopes with different intercepts mean parallel lines.
2. Do not look for an intersection point when the pair is inconsistent.
Answer:
\( \frac{a_1}{a_2} = \frac{1}{2} \) and \( \frac{b_1}{b_2} = \frac{1}{-2} \), so \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \): the pair is consistent with a unique solution.
For \( 2x + y = 6 \):
| x | 1 | 2 | 3 |
| y | 4 | 2 | 0 |
For \( 4x - 2y = 4 \), \( y = 2x - 2 \):
| x | 1 | 2 | 3 |
| y | 0 | 2 | 4 |
The lines intersect at \( (2, 2) \); the solution is \( x = 2,\ y = 2 \).
In simple words: The lines are not parallel, so they cross at exactly one point.
Teacher's Note:
1. Verify: \( 2(2) + 2 = 6 \) and \( 4(2) - 2(2) = 4 \).
2. A unique solution means the ratios \( \frac{a_1}{a_2} \) and \( \frac{b_1}{b_2} \) differ.
Question 17. Which of the following pairs of linear equations are consistent or inconsistent? If consistent, obtain the solution graphically: (d) \( 2x - 2y - 2 = 0,\ 4x - 4y - 5 = 0 \)
Answer:
\( \frac{a_1}{a_2} = \frac{1}{2},\ \frac{b_1}{b_2} = \frac{1}{2},\ \frac{c_1}{c_2} = \frac{2}{5} \), so \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \).
The lines are parallel (\( y = x - 1 \) and \( y = x - 1.25 \)), so the pair is inconsistent.
In simple words: Same slope, different intercepts: parallel lines, no solution.
Teacher's Note:
1. Notice that 4x − 4y is twice 2x − 2y but the constants 5 and 2 do not match in the same ratio.
2. Drawing the lines confirms that they never meet.
Answer:
Let the width be x m and the length be y m.
Length is 4 m more than the width: \( y = x + 4 \), i.e., \( y - x = 4 \) ...(i). Half the perimeter is 36 m: \( x + y = 36 \) ...(ii).
For (i):
| x | 0 | 12 | 8 |
| y | 4 | 16 | 12 |
For (ii):
| x | 0 | 16 | 36 |
| y | 36 | 20 | 0 |
The lines intersect at \( (16, 20) \). Hence the width is 16 m and the length is 20 m.
In simple words: Half the perimeter equals length plus width, which gives the second equation.
Teacher's Note:
1. Half the perimeter is \( \text{length} + \text{width} \).
2. Check: 20 − 16 = 4 and 20 + 16 = 36.
Question 19. 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: (a) intersecting lines
Answer:
For intersecting lines we need \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \). Take \( 2x - 7y + 9 = 0 \): \( \frac{a_1}{a_2} = 1 \) and \( \frac{b_1}{b_2} = -\frac{3}{7} \), which are unequal.
Hence \( 2x - 7y + 9 = 0 \) (many other answers are possible).
In simple words: Pick any equation whose x and y coefficients are not in the same ratio as 2 and 3.
Teacher's Note:
1. Any equation with different coefficient ratios works.
2. Verify by computing the two ratios.
Question 20. 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: (b) parallel lines
Answer:
For parallel lines we need \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \). Take \( 6x + 9y + 9 = 0 \): \( \frac{a_1}{a_2} = \frac{2}{6} = \frac{1}{3},\ \frac{b_1}{b_2} = \frac{3}{9} = \frac{1}{3},\ \frac{c_1}{c_2} = \frac{-8}{9} \).
Hence \( 6x + 9y + 9 = 0 \) (many other answers are possible).
In simple words: Multiply the x and y parts by the same number but change the constant in a different ratio.
Teacher's Note:
1. Keep the same ratio for x and y coefficients but break the ratio for the constant.
2. Check that \( \frac{1}{3} \neq \frac{-8}{9} \).
Question 21. 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: (c) coincident lines
Answer:
For coincident lines we need \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \). Multiply the given equation by 2: \( 4x + 6y - 16 = 0 \), for which all three ratios equal \( \frac{1}{2} \).
In simple words: Multiply the whole equation by any non-zero number to get the same line again.
Teacher's Note:
1. A multiple of the same equation always gives coincident lines.
2. All three ratios must be equal, including the constant term.
Answer:
For \( x - y + 1 = 0 \), \( y = x + 1 \):
| x | 0 | 1 | 2 |
| y | 1 | 2 | 3 |
For \( 3x + 2y - 12 = 0 \), \( y = \frac{12 - 3x}{2} \):
| x | 0 | 2 | 4 |
| y | 6 | 3 | 0 |
The lines meet at \( (2, 3) \). The first line cuts the x-axis at \( (-1, 0) \) and the second at \( (4, 0) \). The vertices of the triangle are \( (2, 3),\ (-1, 0),\ (4, 0) \). The triangular region is shaded in the graph; its area is \( \frac{1}{2} \times 5 \times 3 = 7.5 \) square units.
In simple words: The triangle is bounded by the two lines and the x-axis; its corners are where the lines meet each other and the axis.
Teacher's Note:
1. Find the x-intercept of each line by putting y = 0.
2. The height of the triangle is the y-coordinate of the intersection point.
Free study material for Mathematics
CBSE Class 10 Mathematics Assignments for Chapter 03 Pair of Linear Equations in Two Variables
Chapter Practice Questions for Class 10 Mathematics
Access structured practice assignments for Chapter 03 Pair of Linear Equations in Two Variables designed in alignment with the latest CBSE curriculum for Class 10 Mathematics. These printable sets cover objective and descriptive problem types to support thorough revision.
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