CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables MCQs Set 25

Answer: (b) \(x=\dfrac{49}{29},\ y=\dfrac{19}{29}\)

Explanation:
1. From the second equation \(x=3-2y\)
2. \(7(3-2y)-15y=2\), so \(21-29y=2\) and \(y=\dfrac{19}{29}\)
3. \(x=3-\dfrac{38}{29}=\dfrac{49}{29}\)

Teacher's Note:
1. Substitute the simplest variable expression first.
2. Check: 49/29 + 38/29 = 3.

Answer: (c) \(x=2,\ y=3\)

Explanation:
1. Multiply both by 10: \(2x+3y=13\) and \(4x+5y=23\)
2. Double the first: \(4x+6y=26\); subtract the second: \(y=3\)
3. \(2x=13-9\), so \(x=2\)

Teacher's Note:
1. Clear decimals first.
2. Check: 0.8 + 1.5 = 2.3.

Answer: (d) \(x=-2,\ y=-1\)

Explanation:
1. From the second equation \(y=x+1\)
2. \(3x-5(x+1)+1=0\), so \(-2x-4=0\) and \(x=-2\)
3. \(y=-1\)

Teacher's Note:
1. Write one variable from the simpler equation.
2. Check: -6 + 5 + 1 = 0.

Answer: (a) \(x=1,\ y=-1\)

Explanation:
1. Write \(3x-4y-7=0\) and \(5x+2y-3=0\)
2. \(\dfrac{x}{(-4)(-3)-(2)(-7)}=\dfrac{y}{(-7)(5)-(-3)(3)}=\dfrac{1}{(3)(2)-(5)(-4)}\)
3. \(\dfrac x{26}=\dfrac y{-26}=\dfrac1{26}\), so \(x=1\), \(y=-1\)

Teacher's Note:
1. Bring both equations to the form ax + by + c = 0 before cross-multiplying.
2. Check: 3 + 4 = 7.

Answer: (c) infinitely many solutions

Explanation:
1. \(\dfrac{a_1}{a_2}=\dfrac23\), \(\dfrac{b_1}{b_2}=\dfrac{-6}{-9}=\dfrac23\), \(\dfrac{c_1}{c_2}=\dfrac{10}{15}=\dfrac23\)
2. All three ratios are equal, so the lines coincide

Teacher's Note:
1. Equal ratios of all three coefficients mean coincident lines.
2. Dividing each equation by its common factor gives the same equation \(x-3y+5=0\).

Answer: (d) \(x=\dfrac16,\ y=\dfrac14\)

Explanation:
1. Let \(p=\dfrac1x\) and \(q=\dfrac1y\): \(\dfrac p2-q=-1\) and \(p+\dfrac q2=8\)
2. From the first, \(q=\dfrac p2+1\); then \(p+\dfrac p4+\dfrac12=8\), so \(p=6\) and \(q=4\)
3. \(x=\dfrac16\), \(y=\dfrac14\)

Teacher's Note:
1. Reciprocal substitution turns the pair into linear equations.
2. Do not forget to invert p and q at the end.

Answer: (a) \(x=\dfrac1{14},\ y=\dfrac16\)

Explanation:
1. Let \(p=\dfrac1x\), \(q=\dfrac1y\): \(\dfrac p7+\dfrac q6=3\) gives \(6p+7q=126\); \(\dfrac p2-\dfrac q3=5\) gives \(3p-2q=30\)
2. Double the second: \(6p-4q=60\); subtract: \(11q=66\), so \(q=6\), \(p=14\)
3. \(x=\dfrac1{14}\), \(y=\dfrac16\)

Teacher's Note:
1. Clear the fractions by multiplying by the LCM.
2. Check: 2 + 1 = 3.

Answer: (b) \(x=2,\ y=1\)

Explanation:
1. Multiply both equations by xy: \(2x+3y=7\) and \(x+9y=11\)
2. Then \(x=11-9y\), so \(22-18y+3y=7\) and \(y=1\)
3. \(x=2\)

Teacher's Note:
1. Multiplying by xy removes all the fractions.
2. Check: 4 + 3 = 7 and 2 + 9 = 11.

Answer: (c) \(x=\dfrac12,\ y=\dfrac13\)

Explanation:
1. Divide by xy: \(\dfrac1y+\dfrac1x=5\) and \(\dfrac3y+\dfrac2x=13\)
2. With \(p=\dfrac1x\), \(q=\dfrac1y\): \(p+q=5\) and \(2p+3q=13\), so \(q=3\), \(p=2\)
3. \(x=\dfrac12\), \(y=\dfrac13\)

Teacher's Note:
1. Dividing by xy is allowed as xy is non-zero.
2. Check: 1/2 + 1/3 = 5/6 = 5 x 1/6.

Answer: (d) \(x=1,\ y=1\)

Explanation:
1. Let \(w=\dfrac1y\): \(5x+4w=9\) and \(7x-2w=5\)
2. Double the second: \(14x-4w=10\); add to the first: \(19x=19\), so \(x=1\)
3. \(w=1\), so \(y=1\)

Teacher's Note:
1. Treat 1/y as a single unknown.
2. Check: 5 + 4 = 9 and 7 - 2 = 5.

Answer: (b) \(a=1,\ b=\dfrac32\)

Explanation:
1. Divide by ab: \(\dfrac3b+\dfrac9a=11\) and \(\dfrac6b+\dfrac3a=7\)
2. With \(p=\dfrac1a\), \(q=\dfrac1b\): \(9p+3q=11\) and \(3p+6q=7\)
3. Solving gives \(p=1\), \(q=\dfrac23\), so \(a=1\), \(b=\dfrac32\)

Teacher's Note:
1. Reciprocal substitution again reduces it to a linear pair.
2. Check: 3(1 + 4.5) = 16.5 = 11 x 1.5.

Answer: (c) \(k\ne-6\)

Explanation:
1. Unique solution requires \(\dfrac{a_1}{a_2}\ne\dfrac{b_1}{b_2}\)
2. \(\dfrac13\ne\dfrac{-2}{k}\), so \(k\ne-6\)

Teacher's Note:
1. Unequal ratios of x and y coefficients mean intersecting lines.
2. At k = -6 the lines are parallel.

Answer: (a) \(k=15\)

Explanation:
1. \(\dfrac7{21}=\dfrac{-1}{-3}=\dfrac5k\)
2. \(\dfrac13=\dfrac5k\), so \(k=15\)

Teacher's Note:
1. All three ratios must be equal.
2. The first two ratios are already equal to 1/3.

Answer: (b) \(k=8\)

Explanation:
1. \(\dfrac k{16}=\dfrac4k=\dfrac{k-4}{k}\)
2. From the last two: \(k-4=4\), so \(k=8\)
3. Check the first ratio: \(\dfrac8{16}=\dfrac48\), true
4. (\(k=-8\) fails the third ratio)

Teacher's Note:
1. Check all three equalities, not just two.
2. The condition \(k^2=64\) alone gives \(k=\pm8\), and only 8 works.

Answer: (d) \(k=3\)

Explanation:
1. \(\dfrac42=\dfrac6k\), so \(k=3\)
2. Check \(\dfrac42=2\ne\dfrac{11}7\), so the lines are parallel

Teacher's Note:
1. Parallel but not coincident lines give no solution.
2. Confirm with the constants.

Answer: (a) \(k=-\dfrac{14}{5}\)

Explanation:
1. \(\dfrac25=\dfrac{k}{-7}\), so \(k=-\dfrac{14}5\)
2. \(\dfrac25\ne\dfrac{11}5\), so the lines are parallel

Teacher's Note:
1. Watch the sign of -7 in the ratio.
2. The constant ratio is different, so no solution.

Answer: (b) \(k=-12\)

Explanation:
1. \(\dfrac13=\dfrac{-4}{k}\), so \(k=-12\)
2. \(\dfrac13\ne\dfrac65\), so the lines are parallel

Teacher's Note:
1. Equate the ratios of the x and y coefficients.
2. Verify that the constant ratio differs.

Answer: (c) \(k=-15\)

Explanation:
1. \(\dfrac k6=\dfrac{-5}{2}\), so \(k=-15\)
2. \(\dfrac{-5}{2}\ne\dfrac27\), confirming parallel lines

Teacher's Note:
1. Cross-multiply the equal ratios.
2. The constants confirm there is no coincidence.

Answer: (d) 11

Explanation:
1. Substitute \(x=-1,\ y=-1\): \(-2+k=9\)
2. \(k=11\)

Teacher's Note:
1. A point lies on a line when its coordinates satisfy the equation.
2. Check: 2(-1) - 11(-1) = 9.

Answer: (a) \(x=a,\ y=-b\)

Explanation:
1. From the first, \(x=y+a+b\)
2. \(a(y+a+b)+by=a^2-b^2\) gives \((a+b)y=-b(a+b)\), so \(y=-b\)
3. \(x=a\)

Teacher's Note:
1. Substitute and factor out the common factor \(a+b\).
2. Check: a - (-b) = a + b.

Answer: (b) \(\left(-\dfrac27,\ 0\right)\)

Explanation:
1. On the x-axis \(y=0\)
2. \(7x=-2\), so \(x=-\dfrac27\)

Teacher's Note:
1. Set y = 0 for the x-axis intercept and x = 0 for the y-axis intercept.
2. The y-intercept here is (0, -2).