CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables Assignment Set 16

Question 1. On comparing the ratios \( \frac{a_1}{a_2},\ \frac{b_1}{b_2} \) and \( \frac{c_1}{c_2} \), find whether the pair \( x - 3 y = 4 \), \( 3 x + 2 y = 1 \) is consistent or inconsistent.
Answer:
\( x - 3 y - 4 = 0 \) and \( 3 x + 2 y - 1 = 0 \), so \( \frac{a_1}{a_2} = \frac{1}{3},\ \frac{b_1}{b_2} = - \frac{3}{2},\ \frac{c_1}{c_2} = 4 \).
Here \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \).
Hence the lines intersect at a point.
Therefore the given pair of linear equations is consistent.
In simple words: A pair of equations is consistent when it has at least one solution, i.e., when the lines are not parallel.

Teacher's Note:
1. Compare a1/a2 with b1/b2 first; only if they are equal do you need c1/c2.
2. Consistent covers both the unique-solution and the infinitely-many-solutions cases.

Question 2. On comparing the ratios \( \frac{a_1}{a_2},\ \frac{b_1}{b_2} \) and \( \frac{c_1}{c_2} \), find whether the pair \( \frac{4 x}{3} + 2 y = 8 \), \( 2 x + 3 y = 12 \) is consistent or inconsistent.
Answer:
\( 4 x + 6 y - 24 = 0 \) and \( 2 x + 3 y - 12 = 0 \), so \( \frac{a_1}{a_2} = 2,\ \frac{b_1}{b_2} = 2,\ \frac{c_1}{c_2} = 2 \).
Here \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \).
Hence the lines coincide.
Therefore the given pair of linear equations is consistent.
In simple words: A pair of equations is consistent when it has at least one solution, i.e., when the lines are not parallel.

Teacher's Note:
1. Compare a1/a2 with b1/b2 first; only if they are equal do you need c1/c2.
2. Consistent covers both the unique-solution and the infinitely-many-solutions cases.

Question 3. On comparing the ratios \( \frac{a_1}{a_2},\ \frac{b_1}{b_2} \) and \( \frac{c_1}{c_2} \), find whether the pair \( 4 x + 6 y = 7 \), \( 12 x + 18 y = 21 \) is consistent or inconsistent.
Answer:
\( 4 x + 6 y - 7 = 0 \) and \( 12 x + 18 y - 21 = 0 \), so \( \frac{a_1}{a_2} = \frac{1}{3},\ \frac{b_1}{b_2} = \frac{1}{3},\ \frac{c_1}{c_2} = \frac{1}{3} \).
Here \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \).
Hence the lines coincide.
Therefore the given pair of linear equations is consistent.
In simple words: A pair of equations is consistent when it has at least one solution, i.e., when the lines are not parallel.

Teacher's Note:
1. Compare a1/a2 with b1/b2 first; only if they are equal do you need c1/c2.
2. Consistent covers both the unique-solution and the infinitely-many-solutions cases.

Question 4. On comparing the ratios \( \frac{a_1}{a_2},\ \frac{b_1}{b_2} \) and \( \frac{c_1}{c_2} \), find whether the pair \( x - 2 y = 3 \), \( 3 x - 6 y = 1 \) is consistent or inconsistent.
Answer:
\( x - 2 y - 3 = 0 \) and \( 3 x - 6 y - 1 = 0 \), so \( \frac{a_1}{a_2} = \frac{1}{3},\ \frac{b_1}{b_2} = \frac{1}{3},\ \frac{c_1}{c_2} = 3 \).
Here \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \).
Hence the lines are parallel.
Therefore the given pair of linear equations is inconsistent.
In simple words: A pair of equations is consistent when it has at least one solution, i.e., when the lines are not parallel.

Teacher's Note:
1. Compare a1/a2 with b1/b2 first; only if they are equal do you need c1/c2.
2. Consistent covers both the unique-solution and the infinitely-many-solutions cases.

Question 5. On comparing the ratios, find out whether the lines representing \( 2 x - y = 3 \), \( 4 x - y = 5 \) intersect at a point, are parallel or coincide.
Answer:
\( 2 x - y - 3 = 0 \) and \( 4 x - y - 5 = 0 \), so \( \frac{a_1}{a_2} = \frac{1}{2},\ \frac{b_1}{b_2} = 1,\ \frac{c_1}{c_2} = \frac{3}{5} \).
Here \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \).
Hence the lines intersect at a point.
In simple words: Equal ratios of coefficients mean parallel or coincident lines; unequal ratios mean the lines cross.

Teacher's Note:
1. Intersecting lines have a1/a2 different from b1/b2.
2. If all three ratios are equal the two equations describe the same line.

Question 6. On comparing the ratios, find out whether the lines representing \( x + 2 y = 8 \), \( 5 x - 10 y = 10 \) intersect at a point, are parallel or coincide.
Answer:
\( x + 2 y - 8 = 0 \) and \( 5 x - 10 y - 10 = 0 \), so \( \frac{a_1}{a_2} = \frac{1}{5},\ \frac{b_1}{b_2} = - \frac{1}{5},\ \frac{c_1}{c_2} = \frac{4}{5} \).
Here \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \).
Hence the lines intersect at a point.
In simple words: Equal ratios of coefficients mean parallel or coincident lines; unequal ratios mean the lines cross.

Teacher's Note:
1. Intersecting lines have a1/a2 different from b1/b2.
2. If all three ratios are equal the two equations describe the same line.

Question 7. On comparing the ratios, find out whether the lines representing \( 3 x + 4 y = -2 \), \( 12 x + 16 y = -8 \) intersect at a point, are parallel or coincide.
Answer:
\( 3 x + 4 y + 2 = 0 \) and \( 12 x + 16 y + 8 = 0 \), so \( \frac{a_1}{a_2} = \frac{1}{4},\ \frac{b_1}{b_2} = \frac{1}{4},\ \frac{c_1}{c_2} = \frac{1}{4} \).
Here \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \).
Hence the lines coincide.
In simple words: Equal ratios of coefficients mean parallel or coincident lines; unequal ratios mean the lines cross.

Teacher's Note:
1. Intersecting lines have a1/a2 different from b1/b2.
2. If all three ratios are equal the two equations describe the same line.

Question 8. On comparing the ratios, find out whether the lines representing \( 6 x + 3 y = 18 \), \( 2 x + y = 6 \) intersect at a point, are parallel or coincide.
Answer:
\( 6 x + 3 y - 18 = 0 \) and \( 2 x + y - 6 = 0 \), so \( \frac{a_1}{a_2} = 3,\ \frac{b_1}{b_2} = 3,\ \frac{c_1}{c_2} = 3 \).
Here \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \).
Hence the lines coincide.
In simple words: Equal ratios of coefficients mean parallel or coincident lines; unequal ratios mean the lines cross.

Teacher's Note:
1. Intersecting lines have a1/a2 different from b1/b2.
2. If all three ratios are equal the two equations describe the same line.

Question 9. On comparing the ratios, find out whether the lines representing \( x - 3 y = 3 \), \( 3 x - 9 y = 2 \) intersect at a point, are parallel or coincide.
Answer:
\( x - 3 y - 3 = 0 \) and \( 3 x - 9 y - 2 = 0 \), so \( \frac{a_1}{a_2} = \frac{1}{3},\ \frac{b_1}{b_2} = \frac{1}{3},\ \frac{c_1}{c_2} = \frac{3}{2} \).
Here \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \).
Hence the lines are parallel.
In simple words: Equal ratios of coefficients mean parallel or coincident lines; unequal ratios mean the lines cross.

Teacher's Note:
1. Intersecting lines have a1/a2 different from b1/b2.
2. If all three ratios are equal the two equations describe the same line.

Question 10. Find whether the lines representing \( ax - by = c_1,\ bx + ay = c_2 \), where \( a \neq 0,\ b \neq 0 \), intersect at a point, are parallel or coincide.
Answer:
\( ax - by = c_1 \Rightarrow ax - by - c_1 = 0 \) and \( bx + ay = c_2 \Rightarrow bx + ay - c_2 = 0 \).
\( \frac{a_1}{a_2} = \frac{a}{b} \) and \( \frac{b_1}{b_2} = \frac{-b}{a} \).
If \( \frac{a}{b} = \frac{-b}{a} \), then \( a^2 = -b^2 \), i.e., \( a^2 + b^2 = 0 \). This is impossible because \( a \neq 0 \) and \( b \neq 0 \).
So \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \) and the lines intersect at a point.
In simple words: The two ratios can never be equal because that would need a squared number to be negative.

Teacher's Note:
1. Squares of non-zero real numbers are positive.
2. So the ratios differ and the lines always cross.

Question 11. For the linear equation \( 2x - 3y = 7 \), write another linear equation in two variables such that the geometrical representation of the pair so formed is (i) intersecting lines (ii) parallel lines (iii) coincident lines.
Answer:
(i) Take \( 3x + 5y - 7 = 0 \). Then \( \frac{a_1}{a_2} = \frac{2}{3} \) and \( \frac{b_1}{b_2} = \frac{-3}{5} \), which are unequal, so the lines intersect.
(ii) Take \( 4x - 6y - 8 = 0 \), i.e., \( 4x - 6y = 8 \). Then \( \frac{2}{4} = \frac{-3}{-6} = \frac{1}{2} \) but \( \frac{c_1}{c_2} = \frac{-7}{-8} = \frac{7}{8} \), so the lines are parallel.
(iii) Take \( 6x - 9y = 21 \). Then \( \frac{2}{6} = \frac{-3}{-9} = \frac{-7}{-21} = \frac{1}{3} \), so the lines coincide.
Many other equations are possible; any equation satisfying the stated ratio condition is correct.
In simple words: Change the coefficients in the same ratio and change or keep the constant term to control whether the lines meet, stay parallel or overlap.

Teacher's Note:
1. For coincident lines multiply the whole equation by a number.
2. For parallel lines multiply the coefficients of x and y but change the constant term differently.

Question 12. For the linear equation \( y = 2x + 3 \), write another linear equation in two variables such that the geometrical representation of the pair so formed is (i) intersecting lines (ii) parallel lines (iii) coincident lines.
Answer:
Write the given equation as \( 2x - y + 3 = 0 \).
(i) Take \( 2x + 3y - 4 = 0 \). Then \( \frac{2}{2} = 1 \neq \frac{-1}{3} \), so the lines intersect.
(ii) Take \( 4x - 2y + 8 = 0 \). Then \( \frac{2}{4} = \frac{-1}{-2} = \frac{1}{2} \neq \frac{3}{8} \), so the lines are parallel.
(iii) Take \( 8x - 4y + 12 = 0 \). Then \( \frac{2}{8} = \frac{-1}{-4} = \frac{3}{12} = \frac{1}{4} \), so the lines coincide.
Any other equation satisfying the same ratio condition is also correct.
In simple words: Choose coefficients that are not in the same ratio for crossing lines, in the same ratio with a different constant for parallel lines, and a multiple of the equation for coincident lines.

Teacher's Note:
1. Rewrite the given equation in the form ax + by + c = 0 first.
2. Check your new equation by computing the three ratios.

Question 13. Find the value of k for which the system of equations \( x \left(k - 3\right) + 3 y = k \), \( k x + k y = 12 \) has a unique solution.
Answer:
Write the equations as \( - k + x \left(k - 3\right) + 3 y = 0 \) and \( k x + k y - 12 = 0 \). Here \( a_1 = k - 3,\ b_1 = 3,\ c_1 = - k,\ a_2 = k,\ b_2 = k,\ c_2 = -12 \).
For a unique solution, \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), i.e., \( a_1b_2 - a_2b_1 \neq 0 \).
\( k^{2} - 6 k \neq 0 \).
Hence the system has a unique solution for \( k \neq 0 \) and \( k \neq 6 \).
In simple words: The lines cross when the ratio of x-coefficients differs from the ratio of y-coefficients.

Teacher's Note:
1. Cross-multiply the ratios to get a quadratic or linear condition in k.
2. Exclude every value of k that makes the ratios equal.

Question 14. Find the value of k for which the system of equations \( - k y + x = 2 \), \( 3 x + 2 y = -5 \) has a unique solution.
Answer:
Write the equations as \( - k y + x - 2 = 0 \) and \( 3 x + 2 y + 5 = 0 \). Here \( a_1 = 1,\ b_1 = - k,\ c_1 = -2,\ a_2 = 3,\ b_2 = 2,\ c_2 = 5 \).
For a unique solution, \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), i.e., \( a_1b_2 - a_2b_1 \neq 0 \).
\( 3 k + 2 \neq 0 \).
Hence the system has a unique solution for \( k \neq - \frac{2}{3} \).
In simple words: The lines cross when a1/a2 and b1/b2 are different.

Teacher's Note:
1. Cross-multiply a1/a2 = b1/b2 to find the one forbidden value of k.
2. Every other real value of k gives a unique solution.

Question 15. Find the value of k for which the system of equations \( k x + 2 y - 1 = 0 \), \( 5 x - 3 y + 2 = 0 \) has no solution.
Answer:
Write the equations as \( k x + 2 y - 1 = 0 \) and \( 5 x - 3 y + 2 = 0 \). Here \( a_1 = k,\ b_1 = 2,\ c_1 = -1,\ a_2 = 5,\ b_2 = -3,\ c_2 = 2 \).
The condition for no solution is \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \).
From \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \) we get \( - 3 k - 10 = 0 \), so \( k = - \frac{10}{3} \).
For \( k = - \frac{10}{3} \): \( \frac{b_1}{b_2} = - \frac{2}{3}\ ,\ \frac{c_1}{c_2} = - \frac{1}{2} \), which are not equal, so the condition holds.
Hence, \( k = - \frac{10}{3} \).
In simple words: No solution needs parallel lines.

Teacher's Note:
1. Parallel lines have equal ratios of x and y coefficients and a different ratio of constants.
2. Check the constant ratio for the value of k that you get.

Question 16. Find the value of k for which the system of equations \( x + 2 y = 3 \), \( k y + 5 x + 7 = 0 \) has no solution.
Answer:
Write the equations as \( x + 2 y - 3 = 0 \) and \( k y + 5 x + 7 = 0 \). Here \( a_1 = 1,\ b_1 = 2,\ c_1 = -3,\ a_2 = 5,\ b_2 = k,\ c_2 = 7 \).
The condition for no solution is \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \).
From \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \) we get \( k - 10 = 0 \), so \( k = 10 \).
For \( k = 10 \): \( \frac{b_1}{b_2} = \frac{1}{5}\ ,\ \frac{c_1}{c_2} = - \frac{3}{7} \), which are not equal, so the condition holds.
Hence, \( k = 10 \).
In simple words: No solution needs parallel lines.

Teacher's Note:
1. Parallel lines have equal ratios of x and y coefficients and a different ratio of constants.
2. Check the constant ratio for the value of k that you get.

Question 17. Find the value of k for which the system of equations \( k x + 3 y = k - 3 \), \( k y + 12 x = k \) has no solution.
Answer:
Write the equations as \( k x - k + 3 y + 3 = 0 \) and \( k y - k + 12 x = 0 \). Here \( a_1 = k,\ b_1 = 3,\ c_1 = 3 - k,\ a_2 = 12,\ b_2 = k,\ c_2 = - k \).
The condition for no solution is \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \).
From \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \) we get \( k^{2} - 36 = 0 \), so \( k = -6 \) or \( k = 6 \).
For \( k = -6 \): \( \frac{b_1}{b_2} = - \frac{1}{2}\ ,\ \frac{c_1}{c_2} = \frac{3}{2} \), which are not equal, so the condition holds.
For \( k = 6 \): \( \frac{b_1}{b_2} = \frac{1}{2}\ ,\ \frac{c_1}{c_2} = \frac{1}{2} \), which are equal, so the system is not inconsistent (rejected).
Hence, \( k = -6 \).
In simple words: No solution needs parallel lines.

Teacher's Note:
1. The equation k squared = 36 gives two values of k, so test both.
2. For each value verify that the constant ratio is different.

Question 18. Find the value of k for which the system of equations \( k x - y = 2 \), \( 6 x - 2 y = 3 \) has a unique solution.
Answer:
Write the equations as \( k x - y - 2 = 0 \) and \( 6 x - 2 y - 3 = 0 \). Here \( a_1 = k,\ b_1 = -1,\ c_1 = -2,\ a_2 = 6,\ b_2 = -2,\ c_2 = -3 \).
For a unique solution, \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), i.e., \( a_1b_2 - a_2b_1 \neq 0 \).
\( 6 - 2 k \neq 0 \).
Hence the system has a unique solution for \( k \neq 3 \).
In simple words: The lines cross when a1/a2 differs from b1/b2.

Teacher's Note:
1. Write the ratios before simplifying.
2. The excluded value comes from setting the two ratios equal.

Question 19. Find the value of k for which the system of equations \( k x - y = 2 \), \( 6 x - 2 y = 3 \) has no solution.
Answer:
Write the equations as \( k x - y - 2 = 0 \) and \( 6 x - 2 y - 3 = 0 \). Here \( a_1 = k,\ b_1 = -1,\ c_1 = -2,\ a_2 = 6,\ b_2 = -2,\ c_2 = -3 \).
The condition for no solution is \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \).
From \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \) we get \( 6 - 2 k = 0 \), so \( k = 3 \).
For \( k = 3 \): \( \frac{b_1}{b_2} = \frac{1}{2}\ ,\ \frac{c_1}{c_2} = \frac{2}{3} \), which are not equal, so the condition holds.
Hence, \( k = 3 \).
In simple words: No solution means parallel lines.

Teacher's Note:
1. For parallel lines the first two ratios are equal.
2. Then confirm that the constant ratio is different.

Question 20. Find the value of k for which the system of equations \( k x + 2 y = 5 \), \( 3 x + y = 1 \) has a unique solution.
Answer:
Write the equations as \( k x + 2 y - 5 = 0 \) and \( 3 x + y - 1 = 0 \). Here \( a_1 = k,\ b_1 = 2,\ c_1 = -5,\ a_2 = 3,\ b_2 = 1,\ c_2 = -1 \).
For a unique solution, \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), i.e., \( a_1b_2 - a_2b_1 \neq 0 \).
\( k - 6 \neq 0 \).
Hence the system has a unique solution for \( k \neq 6 \).
In simple words: The lines cross when a1/a2 differs from b1/b2.

Teacher's Note:
1. Compare the ratios of the x and y coefficients first.
2. Exclude the value that makes them equal.

Question 21. Find the value of k for which the system of equations \( k x + 2 y = 5 \), \( 3 x + y = 1 \) has no solution.
Answer:
Write the equations as \( k x + 2 y - 5 = 0 \) and \( 3 x + y - 1 = 0 \). Here \( a_1 = k,\ b_1 = 2,\ c_1 = -5,\ a_2 = 3,\ b_2 = 1,\ c_2 = -1 \).
The condition for no solution is \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \).
From \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \) we get \( k - 6 = 0 \), so \( k = 6 \).
For \( k = 6 \): \( \frac{b_1}{b_2} = 2\ ,\ \frac{c_1}{c_2} = 5 \), which are not equal, so the condition holds.
Hence, \( k = 6 \).
In simple words: No solution means parallel lines.

Teacher's Note:
1. Equal x and y ratios with a different constant ratio give parallel lines.
2. Check 5/1 against the common ratio.

Question 22. Find the value of k for which the system of equations \( 5 x + 2 y = 2 k \), \( k y + 2 x \left(k + 1\right) = 3 k + 4 \) has infinitely many solutions.
Answer:
Write the equations as \( - 2 k + 5 x + 2 y = 0 \) and \( k y - 3 k + x \left(2 k + 2\right) - 4 = 0 \). Here \( a_1 = 5,\ b_1 = 2,\ c_1 = - 2 k,\ a_2 = 2 k + 2,\ b_2 = k,\ c_2 = - 3 k - 4 \).
The condition for infinitely many solutions is \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \).
From \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \) we get \( k - 4 = 0 \), so \( k = 4 \).
For \( k = 4 \): \( \frac{b_1}{b_2} = \frac{1}{2}\ ,\ \frac{c_1}{c_2} = \frac{1}{2} \), which are equal, so the condition holds.
Hence, \( k = 4 \).
In simple words: The lines must coincide, so all three ratios are equal.

Teacher's Note:
1. Take two ratios at a time and solve for k.
2. Make sure the same k satisfies all three ratios.

Question 23. Find the value of k for which the system of equations \( x + y \left(k + 1\right) = 5 \), \( x \left(k + 1\right) + 9 y = 8 k - 1 \) has infinitely many solutions.
Answer:
Write the equations as \( x + y \left(k + 1\right) - 5 = 0 \) and \( - 8 k + x \left(k + 1\right) + 9 y + 1 = 0 \). Here \( a_1 = 1,\ b_1 = k + 1,\ c_1 = -5,\ a_2 = k + 1,\ b_2 = 9,\ c_2 = 1 - 8 k \).
The condition for infinitely many solutions is \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \).
From \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \) we get \( - k^{2} - 2 k + 8 = 0 \), so \( k = -4 \) or \( k = 2 \).
For \( k = -4 \): \( \frac{b_1}{b_2} = - \frac{1}{3}\ ,\ \frac{c_1}{c_2} = - \frac{5}{33} \), which are not equal (rejected).
For \( k = 2 \): \( \frac{b_1}{b_2} = \frac{1}{3}\ ,\ \frac{c_1}{c_2} = \frac{1}{3} \), which are equal, so the condition holds.
Hence, \( k = 2 \).
In simple words: The lines must coincide, so all three ratios are equal.

Teacher's Note:
1. Equate the first two ratios to get a quadratic in k.
2. Reject any root for which the third ratio fails.

Question 24. Find the value of k for which the system of equations \( 10 x + 5 y - \left(k - 5\right) = 0 \), \( - k + 20 x + 10 y = 0 \) has infinitely many solutions.
Answer:
Write the equations as \( - k + 10 x + 5 y + 5 = 0 \) and \( - k + 20 x + 10 y = 0 \). Here \( a_1 = 10,\ b_1 = 5,\ c_1 = 5 - k,\ a_2 = 20,\ b_2 = 10,\ c_2 = - k \).
The condition for infinitely many solutions is \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \).
Here \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \) holds for every k, so we use the third ratio.
From \( \frac{b_1}{b_2} = \frac{c_1}{c_2} \) we get \( 5 k - 50 = 0 \), so \( k = 10 \).
For \( k = 10 \): \( \frac{b_1}{b_2} = \frac{1}{2}\ ,\ \frac{c_1}{c_2} = \frac{1}{2} \), which are equal, so the condition holds.
Hence, \( k = 10 \).
In simple words: The lines must coincide, so all three ratios are equal.

Teacher's Note:
1. Both x and y ratios are 1/2 already.
2. Set the constant ratio equal to 1/2.

Question 25. Find the value of k for which the system of equations \( k x + 3 y = k - 3 \), \( k y + 12 x = k \) has infinitely many solutions.
Answer:
Write the equations as \( k x - k + 3 y + 3 = 0 \) and \( k y - k + 12 x = 0 \). Here \( a_1 = k,\ b_1 = 3,\ c_1 = 3 - k,\ a_2 = 12,\ b_2 = k,\ c_2 = - k \).
The condition for infinitely many solutions is \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \).
From \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \) we get \( k^{2} - 36 = 0 \), so \( k = -6 \) or \( k = 6 \).
For \( k = -6 \): \( \frac{b_1}{b_2} = - \frac{1}{2}\ ,\ \frac{c_1}{c_2} = \frac{3}{2} \), which are not equal (rejected).
For \( k = 6 \): \( \frac{b_1}{b_2} = \frac{1}{2}\ ,\ \frac{c_1}{c_2} = \frac{1}{2} \), which are equal, so the condition holds.
Hence, \( k = 6 \).
In simple words: The lines must coincide, so all three ratios are equal.

Teacher's Note:
1. Solve k squared = 36 and test each root in the third ratio.
2. Only the root that satisfies all three ratios is acceptable.

Question 26. Find the values of a and b for which the system of linear equations \( 2 x + 3 y = 7 \), \( x \left(a - b\right) + y \left(a + b\right) = 3 a + b - 2 \) has an infinite number of solutions.
Answer:
Write the equations as \( 2 x + 3 y - 7 = 0 \) and \( - 3 a - b + x \left(a - b\right) + y \left(a + b\right) + 2 = 0 \). Here \( a_1 = 2,\ b_1 = 3,\ c_1 = -7,\ a_2 = a - b,\ b_2 = a + b,\ c_2 = - 3 a - b + 2 \).
The condition for infinitely many solutions is \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \).
This gives \( - a + 5 b = 0 \) and \( - 2 a + 4 b + 6 = 0 \).
Solving, \( a = 5 \), \( b = 1 \).
In simple words: All three ratios of coefficients must be equal, which gives two equations in a and b.

Teacher's Note:
1. Equate the first ratio with the second and the second with the third.
2. Solve the two resulting linear equations for a and b.

Question 27. Find the values of a and b for which the system of linear equations \( - 2 b y + x \left(a + b\right) = 5 a + 2 b + 1 \), \( 3 x - y = 14 \) has an infinite number of solutions.
Answer:
Write the equations as \( - 5 a - 2 b y - 2 b + x \left(a + b\right) - 1 = 0 \) and \( 3 x - y - 14 = 0 \). Here \( a_1 = a + b,\ b_1 = - 2 b,\ c_1 = - 5 a - 2 b - 1,\ a_2 = 3,\ b_2 = -1,\ c_2 = -14 \).
The condition for infinitely many solutions is \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \).
This gives \( - a + 5 b = 0 \) and \( - 5 a + 26 b - 1 = 0 \).
Solving, \( a = 5 \), \( b = 1 \).
In simple words: All three ratios of coefficients must be equal, which gives two equations in a and b.

Teacher's Note:
1. Equate the first ratio with the second and the second with the third.
2. Solve the two resulting linear equations for a and b.

Question 28. Find the values of a and b for which the system of linear equations \( x \left(2 a - 1\right) + 3 y - 5 = 0 \), \( 3 x + y \left(b - 1\right) - 2 = 0 \) has an infinite number of solutions.
Answer:
Write the equations as \( x \left(2 a - 1\right) + 3 y - 5 = 0 \) and \( 3 x + y \left(b - 1\right) - 2 = 0 \). Here \( a_1 = 2 a - 1,\ b_1 = 3,\ c_1 = -5,\ a_2 = 3,\ b_2 = b - 1,\ c_2 = -2 \).
The condition for infinitely many solutions is \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \).
This gives \( 2 a b - 2 a - b - 8 = 0 \) and \( 5 b - 11 = 0 \).
Solving, \( a = \frac{17}{4} \), \( b = \frac{11}{5} \).
In simple words: All three ratios of coefficients must be equal, which gives two equations in a and b.

Teacher's Note:
1. Equate the first ratio with the second and the second with the third.
2. Solve the two resulting linear equations for a and b.

Question 29. Find the value of k so that the system of linear equations \( x + y \left(k + 2\right) = 4 \), \( x \left(2 k - 1\right) + 25 y = 6 k + 2 \) has infinitely many solutions.
Answer:
Write the equations as \( x + y \left(k + 2\right) - 4 = 0 \) and \( - 6 k + x \left(2 k - 1\right) + 25 y - 2 = 0 \). Here \( a_1 = 1,\ b_1 = k + 2,\ c_1 = -4,\ a_2 = 2 k - 1,\ b_2 = 25,\ c_2 = - 6 k - 2 \).
The condition for infinitely many solutions is \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \).
From \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \) we get \( - 2 k^{2} - 3 k + 27 = 0 \), so \( k = - \frac{9}{2} \) or \( k = 3 \).
For \( k = - \frac{9}{2} \): \( \frac{b_1}{b_2} = - \frac{1}{10}\ ,\ \frac{c_1}{c_2} = - \frac{4}{25} \), which are not equal (rejected).
For \( k = 3 \): \( \frac{b_1}{b_2} = \frac{1}{5}\ ,\ \frac{c_1}{c_2} = \frac{1}{5} \), which are equal, so the condition holds.
Hence, \( k = 3 \).
In simple words: The lines must coincide.

Teacher's Note:
1. Write both equations as ax + by + c = 0 and compare the ratios.
2. Always check the constant-term ratio.

Question 30. Find the value of k so that the system of linear equations \( 3 x - y = 5 \), \( - k + 6 x - 2 y = 0 \) has no solution.
Answer:
Write the equations as \( 3 x - y - 5 = 0 \) and \( - k + 6 x - 2 y = 0 \). Here \( a_1 = 3,\ b_1 = -1,\ c_1 = -5,\ a_2 = 6,\ b_2 = -2,\ c_2 = - k \).
The condition for no solution is \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \).
Here \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \) holds for every k, so we use the third ratio.
The system has no solution when \( \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \). Equality gives \( k - 10 = 0 \), i.e., \( k = 10 \).
Hence the system has no solution for \( k \neq 10 \).
In simple words: The lines must be parallel and different.

Teacher's Note:
1. Write both equations as ax + by + c = 0 and compare the ratios.
2. Always check the constant-term ratio.

Question 31. Find the value of k so that the system of linear equations \( 3 x - y - 5 = 0 \), \( k + 6 x - 2 y = 0 \) has infinitely many solutions.
Answer:
Write the equations as \( 3 x - y - 5 = 0 \) and \( k + 6 x - 2 y = 0 \). Here \( a_1 = 3,\ b_1 = -1,\ c_1 = -5,\ a_2 = 6,\ b_2 = -2,\ c_2 = k \).
The condition for infinitely many solutions is \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \).
Here \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \) holds for every k, so we use the third ratio.
From \( \frac{b_1}{b_2} = \frac{c_1}{c_2} \) we get \( - k - 10 = 0 \), so \( k = -10 \).
For \( k = -10 \): \( \frac{b_1}{b_2} = \frac{1}{2}\ ,\ \frac{c_1}{c_2} = \frac{1}{2} \), which are equal, so the condition holds.
Hence, \( k = -10 \).
In simple words: The lines must coincide.

Teacher's Note:
1. Write both equations as ax + by + c = 0 and compare the ratios.
2. Always check the constant-term ratio.

Question 32. Find the value of k so that the system of linear equations \( k x + 3 y = k - 2 \), \( k y + 12 x = k \) has no solution.
Answer:
Write the equations as \( k x - k + 3 y + 2 = 0 \) and \( k y - k + 12 x = 0 \). Here \( a_1 = k,\ b_1 = 3,\ c_1 = 2 - k,\ a_2 = 12,\ b_2 = k,\ c_2 = - k \).
The condition for no solution is \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \).
From \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \) we get \( k^{2} - 36 = 0 \), so \( k = -6 \) or \( k = 6 \).
For \( k = -6 \): \( \frac{b_1}{b_2} = - \frac{1}{2}\ ,\ \frac{c_1}{c_2} = \frac{4}{3} \), which are not equal, so the condition holds.
For \( k = 6 \): \( \frac{b_1}{b_2} = \frac{1}{2}\ ,\ \frac{c_1}{c_2} = \frac{2}{3} \), which are not equal, so the condition holds.
Hence, \( k = -6 \) or \( k = 6 \).
In simple words: The lines must be parallel and different.

Teacher's Note:
1. Write both equations as ax + by + c = 0 and compare the ratios.
2. Always check the constant-term ratio.

Question 33. Find the number of solutions of the following pair of linear equations: \( x + 2 y - 8 = 0 \), \( 2 x + 4 y = 16 \)
Answer:
\( x + 2 y - 8 = 0 \) and \( 2 x + 4 y - 16 = 0 \), so \( \frac{a_1}{a_2} = \frac{1}{2},\ \frac{b_1}{b_2} = \frac{1}{2},\ \frac{c_1}{c_2} = \frac{1}{2} \).
Here \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \).
Hence the lines coincide.
So the system has infinitely many solutions.
In simple words: Compare the ratios of coefficients to decide how the two lines are placed.

Teacher's Note:
1. Equal ratios of all three coefficients mean infinitely many solutions.
2. Unequal x and y ratios mean exactly one solution.

Question 34. Write whether the following pair of linear equations is consistent or not: \( x + y = 14 \), \( x - y = 4 \)
Answer:
\( x + y - 14 = 0 \) and \( x - y - 4 = 0 \), so \( \frac{a_1}{a_2} = 1,\ \frac{b_1}{b_2} = -1,\ \frac{c_1}{c_2} = \frac{7}{2} \).
Here \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \).
Hence the lines intersect at a point.
A unique solution exists, so the pair is consistent.
In simple words: Compare the ratios of coefficients to decide how the two lines are placed.

Teacher's Note:
1. Equal ratios of all three coefficients mean infinitely many solutions.
2. Unequal x and y ratios mean exactly one solution.

Question 35. Without drawing the graph, find out whether the lines representing the following pair of linear equations intersect at a point, are parallel or coincident: \( 9 x - 10 y = 21 \), \( \frac{3 x}{2} - \frac{5 y}{3} = \frac{7}{2} \)
Answer:
\( 9 x - 10 y - 21 = 0 \) and \( 9 x - 10 y - 21 = 0 \), so \( \frac{a_1}{a_2} = 1,\ \frac{b_1}{b_2} = 1,\ \frac{c_1}{c_2} = 1 \).
Here \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \).
Hence the lines coincide.
So the lines are coincident.
In simple words: Compare the ratios of coefficients to decide how the two lines are placed.

Teacher's Note:
1. Equal ratios of all three coefficients mean infinitely many solutions.
2. Unequal x and y ratios mean exactly one solution.

Question 36. Without drawing the graph, find out whether the lines representing the following pair of linear equations intersect at a point, are parallel or coincident: \( 18 x - 7 y = 24 \), \( \frac{9 x}{5} - \frac{7 y}{10} = \frac{9}{10} \)
Answer:
\( 18 x - 7 y - 24 = 0 \) and \( 18 x - 7 y - 9 = 0 \), so \( \frac{a_1}{a_2} = 1,\ \frac{b_1}{b_2} = 1,\ \frac{c_1}{c_2} = \frac{8}{3} \).
Here \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \).
Hence the lines are parallel.
So the lines are parallel.
In simple words: Compare the ratios of coefficients to decide how the two lines are placed.

Teacher's Note:
1. Equal ratios of all three coefficients mean infinitely many solutions.
2. Unequal x and y ratios mean exactly one solution.

Question 37. Without drawing the graph, find out whether the lines representing the following pair of linear equations intersect at a point, are parallel or coincident: \( 5 x + 3 y - 6 = 0 \), \( \frac{9 x}{5} + 3 y = 6 \)
Answer:
\( 5 x + 3 y - 6 = 0 \) and \( 9 x + 15 y - 30 = 0 \), so \( \frac{a_1}{a_2} = \frac{5}{9},\ \frac{b_1}{b_2} = \frac{1}{5},\ \frac{c_1}{c_2} = \frac{1}{5} \).
Here \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \).
Hence the lines intersect at a point.
So the lines intersect at a point and the system has a unique solution.
In simple words: Compare the ratios of coefficients to decide how the two lines are placed.

Teacher's Note:
1. Equal ratios of all three coefficients mean infinitely many solutions.
2. Unequal x and y ratios mean exactly one solution.