Multiple Choice Questions (MCQs) for Class 8 Mathematics: Linear Equations in Two Variables
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Practice Linear Equations in Two Variables MCQs for Class 8 Mathematics
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Multiple Choice Questions
(a) \( 2 \)
(b) \( -2 \)
(c) \( \frac{14}{11} \)
(d) \( - \frac{2}{11} \)
Show Answer & Explanation
Answer: (a) \( 2 \)
Explanation:
1. Multiply the first equation by 5: 25m - 5k = 20. Subtract the second: 22m = 44.
2. So m = 2 (and k = 6).
Teacher's Note:
1. Use elimination or substitution and check the answer in both equations.
2. A solution must satisfy both equations at the same time.
(a) \( -1 \)
(b) \( -3 \)
(c) \( 1 \)
(d) \( 14 \)
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Answer: (a) \( -1 \)
Explanation:
1. Subtract the second equation from the first: -6y = 18, so y = -3.
2. Then 2x = -5 + 3 = -2, so x = -1.
Teacher's Note:
1. Use elimination or substitution and check the answer in both equations.
2. A solution must satisfy both equations at the same time.
(a) \( \frac{4}{3} \)
(b) \( -3 \)
(c) \( 2 \)
(d) \( -2 \)
Show Answer & Explanation
Answer: (c) \( 2 \)
Explanation:
1. Subtract the second equation from the first: 3x = 9, so x = 3.
2. Then 3y = x + 3 = 6, so y = 2.
Teacher's Note:
1. Use elimination or substitution and check the answer in both equations.
2. A solution must satisfy both equations at the same time.
(a) \( \frac{13}{9} \)
(b) \( 19 \)
(c) \( 4 \)
(d) \( 22 \)
Show Answer & Explanation
Answer: (c) \( 4 \)
Explanation:
1. From 2y = -6, y = -3.
2. Then x = 13 + 3y = 13 - 9 = 4.
Teacher's Note:
1. Use elimination or substitution and check the answer in both equations.
2. A solution must satisfy both equations at the same time.
(a) x = \( 5 \), y = \( 6 \)
(b) x = \( 5 \), y = \( 3 \)
(c) x = \( 4 \), y = \( 1 \)
(d) x = \( 3 \), y = \( -2 \)
Show Answer & Explanation
Answer: (d) x = \( 3 \), y = \( -2 \)
Explanation:
1. From the first equation x = -1 - 2y. Substituting: -2 - 4y - 3y = 12, so y = -2.
2. x = -1 + 4 = 3.
Teacher's Note:
1. Use elimination or substitution and check the answer in both equations.
2. A solution must satisfy both equations at the same time.
(a) x = \( 4 \), y = \( 3 \)
(b) x = \( 1 \), y = \( -1 \)
(c) x = \( 11 \), y = \( -4 \)
(d) x = \( 15 \), y = \( 5 \)
Show Answer & Explanation
Answer: (c) x = \( 11 \), y = \( -4 \)
Explanation:
1. From the first, x = 7 - y. Substituting: 35 - 5y + 12y = 7, so y = -4.
2. x = 7 + 4 = 11.
Teacher's Note:
1. Use elimination or substitution and check the answer in both equations.
2. A solution must satisfy both equations at the same time.
(a) x = \( 11 \), y = \( -4 \)
(b) x = \( 4 \), y = \( 3 \)
(c) x = \( 15 \), y = \( 5 \)
(d) x = \( 1 \), y = \( -1 \)
Show Answer & Explanation
Answer: (b) x = \( 4 \), y = \( 3 \)
Explanation:
1. Multiply the first by 2 and the second by 3: 4x + 6y = 34 and 9x - 6y = 18.
2. Add: 13x = 52, so x = 4; then y = 3.
Teacher's Note:
1. Use elimination or substitution and check the answer in both equations.
2. A solution must satisfy both equations at the same time.
(a) x = \( 15 \), y = \( 5 \)
(b) x = \( 4 \), y = \( 3 \)
(c) x = \( 1 \), y = \( -1 \)
(d) x = \( 11 \), y = \( -4 \)
Show Answer & Explanation
Answer: (a) x = \( 15 \), y = \( 5 \)
Explanation:
1. Multiply the first by 4: 8x + 4y = 140. Subtract the second: 5x = 75, so x = 15.
2. Then y = 35 - 30 = 5.
Teacher's Note:
1. Use elimination or substitution and check the answer in both equations.
2. A solution must satisfy both equations at the same time.
(a) \( 8 \)
(b) \( 3 \)
(c) \( 4 \)
(d) \( 5 \)
Show Answer & Explanation
Answer: (b) \( 3 \)
Explanation:
1. Substitute y = 4: 3x + 16 = 25.
2. 3x = 9, so x = 3.
Teacher's Note:
1. Use elimination or substitution and check the answer in both equations.
2. A solution must satisfy both equations at the same time.
(a) x = \( 5 \), y = \( 6 \)
(b) x = \( 8 \), y = \( 1 \)
(c) x = \( 7 \), y = \( 3 \)
(d) x = \( 6 \), y = \( 5 \)
Show Answer & Explanation
Answer: (d) x = \( 6 \), y = \( 5 \)
Explanation:
1. Multiply the second by 5: 10x + 5y = 85. Add to the first: 16x = 96, so x = 6.
2. Then y = 17 - 12 = 5.
Teacher's Note:
1. Use elimination or substitution and check the answer in both equations.
2. A solution must satisfy both equations at the same time.
(a) \( 13x-95=0 \)
(b) \( 16x-96=0 \)
(c) \( 17x-85=0 \)
(d) \( 19x-91=0 \)
Show Answer & Explanation
Answer: (b) \( 16x-96=0 \)
Explanation:
1. Multiply the second equation by 5: 10x + 5y = 85.
2. Add it to 6x - 5y = 11: 16x = 96, i.e. 16x - 96 = 0.
Teacher's Note:
1. Use elimination or substitution and check the answer in both equations.
2. A solution must satisfy both equations at the same time.
(a) \( 8x-48=0 \)
(b) \( 7x-42=0 \)
(c) \( 6x-48=0 \)
(d) \( 5x-30=0 \)
Show Answer & Explanation
Answer: (b) \( 7x-42=0 \)
Explanation:
1. Multiply the first by 2: 4x - 2y = 20. Add the second: 7x = 42.
2. So the equation is 7x - 42 = 0 (x = 6).
Teacher's Note:
1. Use elimination or substitution and check the answer in both equations.
2. A solution must satisfy both equations at the same time.
(a) -1 and 4
(b) -1 and 2
(c) 4 and -2
(d) 2 and -1
Show Answer & Explanation
Answer: (b) -1 and 2
Explanation:
1. x/5 = 1/(-5) gives x = -1.
2. y/(-10) = 1/(-5) gives y = 2.
Teacher's Note:
1. Use elimination or substitution and check the answer in both equations.
2. A solution must satisfy both equations at the same time.
(a) \( x=a^2,\ y=b^2 \)
(b) \( x=a,\ y=2b \)
(c) \( x=a,\ y=b \)
(d) \( x=3a,\ y=b \)
Show Answer & Explanation
Answer: (a) \( x=a^2,\ y=b^2 \)
Explanation:
1. Put x = a² and y = b²: first equation gives a + b and the second gives 1 + 1 = 2, both satisfied.
2. The system has a unique solution, so x = a², y = b².
Teacher's Note:
1. Use elimination or substitution and check the answer in both equations.
2. A solution must satisfy both equations at the same time.
(a) \( x=\frac{1}{4},\ y=\frac{1}{2} \)
(b) \( x=\frac{1}{3},\ y=\frac{1}{2} \)
(c) \( x=\frac{1}{2},\ y=\frac{1}{4} \)
(d) \( x=\frac{1}{2},\ y=\frac{1}{3} \)
Show Answer & Explanation
Answer: (d) \( x=\frac{1}{2},\ y=\frac{1}{3} \)
Explanation:
1. Put u = 1/x, v = 1/y: 2u + 3v = 13 and 5u - 4v = -2.
2. Multiply by 4 and 3 and add: 23u = 46, so u = 2, v = 3.
3. Hence x = 1/2 and y = 1/3.
Teacher's Note:
1. Use elimination or substitution and check the answer in both equations.
2. A solution must satisfy both equations at the same time.
(a) \( x=3,\ y=-\frac{1}{3} \)
(b) \( x=\frac{5}{2},\ y=-\frac{1}{2} \)
(c) \( x=\frac{5}{2},\ y=\frac{1}{2} \)
(d) \( x=\frac{4}{3},\ y=-\frac{2}{3} \)
Show Answer & Explanation
Answer: (b) \( x=\frac{5}{2},\ y=-\frac{1}{2} \)
Explanation:
1. Put u = 1/(x + y), v = 1/(x - y): 6u - 3v = 2 and 4u + 6v = 4.
2. Doubling the first and adding gives 16u = 8, so u = 1/2 and v = 1/3.
3. x + y = 2 and x - y = 3, so x = 5/2 and y = -1/2.
Teacher's Note:
1. Use elimination or substitution and check the answer in both equations.
2. A solution must satisfy both equations at the same time.
(a) \( x=\frac{5}{3},\ y=-\frac{1}{3} \)
(b) \( x=\frac{5}{2},\ y=\frac{1}{2} \)
(c) \( x=\frac{5}{4},\ y=\frac{1}{4} \)
(d) \( x=2,\ y=\frac{1}{3} \)
Show Answer & Explanation
Answer: (b) \( x=\frac{5}{2},\ y=\frac{1}{2} \)
Explanation:
1. Put u = 1/(x + y), v = 1/(x - y): 12u + 8v = 8 and 27u - 12v = 3.
2. Multiply by 3 and 2 and add: 90u = 30, so u = 1/3 and v = 1/2.
3. x + y = 3 and x - y = 2, so x = 5/2 and y = 1/2.
Teacher's Note:
1. Use elimination or substitution and check the answer in both equations.
2. A solution must satisfy both equations at the same time.
(a) \( x=\frac{1}{6},\ y=1 \)
(b) \( x=\frac{1}{4},\ y=\frac{1}{3} \)
(c) \( x=\frac{1}{3},\ y=\frac{1}{4} \)
(d) \( x=2,\ y=\frac{1}{5} \)
Show Answer & Explanation
Answer: (b) \( x=\frac{1}{4},\ y=\frac{1}{3} \)
Explanation:
1. Dividing x + y = 7xy by xy: 1/x + 1/y = 7.
2. With 2/x + 3/y = 17: subtract twice the first: 1/y = 3, so y = 1/3.
3. 1/x = 4, so x = 1/4.
Teacher's Note:
1. Use elimination or substitution and check the answer in both equations.
2. A solution must satisfy both equations at the same time.
(a) \( \frac{1}{2},\ \frac{1}{3},\ \frac{1}{4} \)
(b) \( \frac{1}{2},\ \frac{1}{3},\ \frac{1}{5} \)
(c) \( \frac{1}{2},\ \frac{1}{4},\ \frac{1}{3} \)
(d) \( \frac{1}{3},\ \frac{1}{4},\ \frac{1}{2} \)
Show Answer & Explanation
Answer: (a) \( \frac{1}{2},\ \frac{1}{3},\ \frac{1}{4} \)
Explanation:
1. Divide each equation by the product: 1/x + 1/y = 5, 1/y + 1/z = 7, 1/z + 1/x = 6.
2. Adding: 1/x + 1/y + 1/z = 9, so 1/z = 4, 1/x = 2 and 1/y = 3.
3. x = 1/2, y = 1/3, z = 1/4.
Teacher's Note:
1. Use elimination or substitution and check the answer in both equations.
2. A solution must satisfy both equations at the same time.
(a) \( k\neq 3 \)
(b) \( k=-3 \)
(c) \( k=3 \)
(d) \( k=6 \)
Show Answer & Explanation
Answer: (a) \( k\neq 3 \)
Explanation:
1. Unique solution requires a₁/a₂ ≠ b₁/b₂, i.e. 6/k ≠ 2/1.
2. So k ≠ 3.
Teacher's Note:
1. Compare the ratios of the coefficients.
2. For k = 3 the lines are parallel.
(a) 6
(b) -6
(c) 12
(d) 3
Show Answer & Explanation
Answer: (a) 6
Explanation:
1. Infinitely many solutions need c/12 = 3/c = (3 - c)/(-c).
2. c/12 = 3/c gives c² = 36; 3/c = (3 - c)/(-c) gives -3 = 3 - c, so c = 6.
3. c = 6 satisfies all the ratios (c = -6 does not).
Teacher's Note:
1. All three ratios must be equal for coincident lines.
2. Check c = 6: 6/12 = 3/6 = -3/-6 = 1/2.
(a) 6
(b) 9
(c) 12
(d) 3
Show Answer & Explanation
Answer: (a) 6
Explanation:
1. Coincident lines: (k - 3)/k = 3/k = k/12.
2. 3/k = k/12 gives k² = 36 and (k - 3)/k = 3/k gives k = 6.
3. So k = 6.
Teacher's Note:
1. The three ratios must be equal.
2. Check: 3/6 = 3/6 = 6/12.
(a) no solution
(b) exactly two solutions
(c) infinitely many solutions
(d) a unique solution
Show Answer & Explanation
Answer: (a) no solution
Explanation:
1. Compare a₁/a₂, b₁/b₂ and c₁/c₂.
2. 1/3 = -3/-9 = 1/3 but this differs from 3/2, so the lines are parallel and distinct.
Teacher's Note:
1. Use elimination or substitution and check the answer in both equations.
2. A solution must satisfy both equations at the same time.
(a) no solution
(b) exactly two solutions
(c) infinitely many solutions
(d) a unique solution
Show Answer & Explanation
Answer: (c) infinitely many solutions
Explanation:
1. Compare a₁/a₂, b₁/b₂ and c₁/c₂.
2. 2/4 = 1/2 = 5/10, all ratios equal, so the lines coincide.
Teacher's Note:
1. Use elimination or substitution and check the answer in both equations.
2. A solution must satisfy both equations at the same time.
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Multiple Choice Questions (MCQs) for Class 8 Mathematics Linear Equations in Two Variables
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