Algebraic methods: Substitution method and Elimination method.

Use elimination to solve the simultaneous equations before evaluating the requested expression. Multiply 5x + 2y = 28 by 3 to obtain 15x + 6y = 84. Multiply 7x - 3y = 19 by 2 to obtain 14x - 6y = 38. Adding gives 29x = 122, so x = 122/29. Substitution in the first equation gives y = 101/29. Hence 2x - y = 2(122/29) - 101/29 = 143/29. The originally supplied key and choices were inconsistent with the equations, so option A has been corrected to the verified value. The other numerical choices do not satisfy the calculated expression.

First clear the denominators correctly. Multiplying x/3 + y/4 = 5 by 12 gives 4x + 3y = 60. Multiplying x/2 - y/5 = 3 by 10 gives 5x - 2y = 30. To eliminate y, multiply the first resulting equation by 2 and the second by 3: 8x + 6y = 120 and 15x - 6y = 90. Adding yields 23x = 210, so x = 210/23. Therefore option A is correct. The other options result from arithmetic or denominator-clearing errors. The originally supplied choices and key stated 270/23, but direct substitution confirms the corrected value: 210/23.

The governing idea of the elimination method is to make the coefficients of one variable equal in magnitude and opposite in sign. In the first equation, the coefficient of y is -9, while in the second equation it is +3. Multiplying every term of the second equation by 3 changes it to 15x + 9y = 123. The terms -9y and +9y then cancel when the equations are added, leaving an equation in x alone. Therefore, 3 is the required multiplier, so option B is correct. Multiplying by 2 or 4 would produce 6y or 12y, which would not directly cancel -9y; multiplying by -3 would produce -9y, making the coefficients equal rather than opposite for addition.

The governing concept is substitution for a pair of simultaneous linear equations. From x + y = 24, isolate x to obtain x = 24 - y. Substitute this expression into 3x - 2y = 37: 3(24 - y) - 2y = 37. Simplifying gives 72 - 3y - 2y = 37, so 72 - 5y = 37 and 5y = 35. Thus y = 7, and x = 24 - 7 = 17. The requested expression is 2x + y, not either original equation. Substitution of the values gives 2(17) + 7 = 34 + 7 = 41. Therefore option D is correct. The other choices do not result from the required expression and may arise from incomplete substitution or arithmetic mistakes.

For a pair of linear equations, if the coefficients of x and y are proportional but the constants are not proportional in the same way, the two lines are parallel and distinct. Multiplying the first equation by 2 gives 6x + 10y = 40. The second equation is 6x + 10y = 43, which has exactly the same left-hand side but a different right-hand side. A pair of numbers cannot make the same expression equal to both 40 and 43. Therefore the equations are inconsistent and have no common solution, so option C is correct. They do not have one or infinitely many solutions; those cases require non-proportional coefficients or complete proportionality, respectively.

Direct answer: C, x=136/9 and y=100/9. First remove the fractions from x/4+y/5=6 by multiplying every term by 20, the least common multiple of 4 and 5. This gives 5x+4y=120. From x−y=4, write x=y+4. Substitution gives 5(y+4)+4y=120, so 9y+20=120, 9y=100, and y=100/9. Then x=y+4=100/9+36/9=136/9. Check: x−y=36/9=4, and x/4+y/5=34/9+20/9=54/9=6. Thus C is correct. A has a different difference between x and y and fails the second equation. B also gives x−y=36/9 but its first-equation value is wrong. C satisfies both equations exactly. D has x−y=36/9 as well, but its fractional sum is too large. The safest habit is to clear denominators first and then substitute without changing signs.

Use elimination to solve the simultaneous equations. Multiply 2x + 3y = 41 by 2 to obtain 4x + 6y = 82. Multiply 5x - 2y = 14 by 3 to obtain 15x - 6y = 42. Adding eliminates y and gives 19x = 124, so x = 124/19. Substitute this into the first equation: 2(124/19) + 3y = 41 = 779/19. Hence 3y = 531/19 and y = 177/19. The required expression is therefore 2x + y = 248/19 + 177/19 = 425/19. Option C is correct. The nearby alternatives result from errors in multiplying equations, adding constants, or combining the final fractions, and they do not give the value of the requested expression.

The price relationships form a pair of simultaneous linear equations, which can be solved by elimination. To eliminate q, multiply 5p + 2q = 300 by 2, obtaining 10p + 4q = 600. Subtract 3p + 4q = 260 from this equation: 7p = 340, so p = 340/7. Substitute into 5p + 2q = 300: 5(340/7) + 2q = 300. Since 300 = 2100/7, we get 1700/7 + 2q = 2100/7, hence 2q = 400/7 and q = 200/7. Thus option C is correct. The corresponding p and q satisfy both original price equations; the other proposed q-values do not.

Start with x - y = 6, which gives x = y + 6. To clear the denominators 5 and 6 in x/5 + y/6 = 7, multiply the entire equation by their least common multiple, 30. This produces 6x + 5y = 210. Substitution gives 6(y + 6) + 5y = 210, so 11y + 36 = 210 and 11y = 174. Hence y = 174/11. Now calculate x: x = y + 6 = 174/11 + 66/11 = 240/11. Therefore option C is correct. A check confirms x - y = 66/11 = 6, and substituting the values into the fractional equation gives 7. The other choices result from arithmetic or denominator-clearing errors.

The key concept is translating place value and digit conditions into a pair of linear equations. Let the tens digit be x and the units digit be y. Then the two-digit number is 10x + y, while the sum of its digits is x + y. The conditions give x = y + 3 and 10x + y = 7(x + y). Expanding the second equation gives 10x + y = 7x + 7y, so 3x = 6y and x = 2y. Substituting into x = y + 3 gives 2y = y + 3, hence y = 3 and x = 6. The number is therefore 10(6) + 3 = 63, so option B is correct. Checking confirms that its digit sum is 9 and 7 × 9 = 63. The other choices fail at least one of the two stated conditions.

Use elimination by adding the two equations, because +5y and −5y cancel: (8x + 5y) + (4x − 5y) = 4 + 20. This gives 12x = 24, so x = 2. Substitute x = 2 into the first equation: 8(2) + 5y = 4, hence 16 + 5y = 4 and 5y = −12. Therefore y = −12/5, giving the ordered pair (2, −12/5). Option B is correct. Verification in the second equation gives 4(2) − 5(−12/5) = 8 + 12 = 20, so the pair satisfies both equations. The other options fail either the x-value obtained after elimination or direct substitution.

Compare the two equations using the elimination principle. Multiplying the first equation by 2 gives 8x + 14y = 62, while the second equation is 8x + 14y = 65. The left-hand sides are identical but the constants are different. Subtracting the two equations would therefore give 0 = 3, which is impossible. Geometrically, the equations represent distinct parallel lines, so they never intersect. Hence the pair is inconsistent and has no solution; option A is correct. Infinitely many solutions would require both equations to be exact multiples, including the constant term, and one or two solutions do not apply to a pair of linear equations in this form.

This is a pair of simultaneous linear equations, and elimination is efficient because multiplying the second equation by 3 creates opposite y-coefficients. From 8x - 3y = 37, obtain 24x - 9y = 111. Add this to 6x + 9y = 117: the y-terms cancel, so 30x = 228 and x = 38/5. Substitute into the second equation: 8(38/5) - 3y = 37. Since 37 = 185/5, this gives 304/5 - 3y = 185/5, so -3y = -119/5 and y = 119/15. Hence option C is correct. Checking the first equation gives 228/5 + 357/5 = 585/5 = 117. The other fractions fail this simultaneous check.

Eliminate a by making its coefficients equal. Multiply 6a + 5b = 460 by 2 to obtain 12a + 10b = 920, and multiply 4a + 7b = 444 by 3 to obtain 12a + 21b = 1332. Subtract the first new equation from the second: (12a + 21b) − (12a + 10b) = 1332 − 920. Hence 11b = 412 and b = 412/11. Therefore option C is correct. The coefficient of a cancels exactly, which is the governing idea of the elimination method. The nearby choices differ by 10 in the numerator and would arise from an arithmetic error in the subtraction; substituting b = 412/11 with the corresponding a confirms the system.

The governing concept is elimination in a simultaneous pair of linear equations. The y-coefficients are -3 and +3, so adding the equations removes y: (7x - 3y) + (14x + 3y) = 20 + 64. This gives 21x = 84 and therefore x = 4. Substitute x=4 in the first equation: 7(4) - 3y = 20, so 28 - 3y = 20 and 3y = 8. Hence y=8/3. The requested expression is not y alone; calculate it carefully: 2x+y = 2(4)+8/3 = 8+8/3 = 24/3+8/3 = 32/3. Thus option D is correct. The integer distractors arise from ignoring the fractional value of y or evaluating the wrong expression.

This word problem is modeled by a pair of linear equations. Let p be the price of one pen and n the price of one notebook. The statements give 3p + 2n = 86 and 2p + 3n = 89. To eliminate n, multiply the first equation by 3 and the second by 2: 9p + 6n = 258 and 4p + 6n = 178. Subtracting the second result from the first gives 5p = 80, so p = 16 rupees. Substituting into 2p + 3n = 89 gives 32 + 3n = 89, hence n = 19. A check gives 3(16)+2(19)=48+38=86. Therefore option A is correct; 19 is the notebook price, not the pen price.

The relevant concept is elimination because the y-coefficients, +2 and -2, are opposites. Add the equations: (9x + 2y) + (3x - 2y) = 41 + (-5). The y-terms cancel and 12x = 36, so x = 3. Substitute this into 3x - 2y = -5: 3(3) - 2y = -5, giving 9 - 2y = -5, so -2y = -14 and y = 7. Now evaluate the requested expression, x + 3y = 3 + 3(7) = 3 + 21 = 24. Therefore option D is correct. Options A, B, and C are not the value of the required expression; they can arise from stopping after finding x or from an arithmetic error. Substitution verifies both equations.

The answer is A, y=1/9. The y terms have opposite coefficients, −9y and +9y, so add the two equations: (4x−9y)+(8x+9y)=31+65. The y terms cancel and 12x=96, giving x=8. Substitute x=8 into the first equation: 4(8)−9y=31, so 32−9y=31. Subtract 32 from both sides: −9y=−1. Dividing by −9 gives y=1/9. Check in the second equation: 8(8)+9(1/9)=64+1=65. Thus A is correct. Values 2, 3 and 4 do not satisfy the original pair with x=8. When opposite terms appear, adding the equations is the quickest elimination step.

The governing concept is clearing denominators before applying elimination. The least common multiple of 2 and 3 is 6, so multiply both equations by 6. They become 3x + 2y = 42 and 2x + 3y = 48. Adding these equations gives 5x + 5y = 90. Factor out 5: 5(x+y)=90, and divide by 5 to obtain x+y=18. It is unnecessary to calculate x and y separately because the requested sum appears directly. For verification, subtracting the equations gives x-y=-6; together with x+y=18 this gives x=6 and y=12, which satisfy the original fractional equations: 3+4=7 and 2+6=8. Hence option B is correct.

The governing concept is solving a pair of simultaneous linear equations by substitution and using a known component of their common solution. Since x = 4, substitute it into the second equation: 3(4) - 2y = 7, so 12 - 2y = 7 and y = 5/2. Now place x = 4 and y = 5/2 in the first equation: 4a + 2(5/2) = 17. Therefore 4a + 5 = 17, so 4a = 12 and a = 3. A direct check gives 3(4) + 2(5/2) = 12 + 5 = 17 and 3(4) - 2(5/2) = 12 - 5 = 7. Thus option C is correct. Options A, B, and D would make the first equation's left side 9, 13, or 21, respectively, instead of 17.

This question applies substitution to simultaneous linear equations containing an unknown coefficient. The common solution has y = 4. Substitute this value into the second equation: 4x - 3(4) = 2, giving 4x - 12 = 2 and therefore x = 7/2. Use x = 7/2 and y = 4 in the first equation: 4(7/2) + 4k = 26. Hence 14 + 4k = 26, so 4k = 12 and k = 3. Verification is important: the second equation gives 14 - 12 = 2, while the first gives 14 + 3(4) = 26. Therefore option B is correct. If k were 2, 4, or 5, the first equation would have left side 22, 30, or 34, respectively, so those distractors cannot satisfy the stated common solution.

This is a pair of linear equations because the two unknown present ages occur only to the first power. Let Ram’s age be r and Shyam’s age be s. The first statement gives r − s = 4. After five years their ages become r + 5 and s + 5, so their sum is r + s + 10 = 44, or r + s = 34. Adding r − s = 4 and r + s = 34 eliminates s: 2r = 38, hence r = 19. Therefore option C is correct. Checking, Shyam is 15, Ram is 19, and after five years the ages are 24 and 20, whose sum is 44. Options A, B and D do not satisfy both conditions.

The governing idea is to model the numbers of correct and wrong answers as two linear equations. Let c be the number of correct answers and w the number of wrong answers. Since all 20 questions are counted, c + w = 20. The marking rule gives 4c − w = 55, because each wrong answer contributes negative one mark. Adding the two equations eliminates w: 5c = 75, so c = 15. Then w = 20 − 15 = 5, and the score check is 15 × 4 − 5 × 1 = 60 − 5 = 55. Thus option C is correct. Options A, B and D would produce scores 47, 52 and 61 respectively when the remaining answers are treated as wrong, so they fail the given total.

Use elimination first, then calculate the requested product. Adding the equations removes y because the coefficients are −12 and +12: 15x = −1 + 61 = 60. Hence x = 4. Substitute this into 5x − 12y = −1: 20 − 12y = −1, so −12y = −21 and y = 21/12 = 7/4. Therefore xy = 4 × 7/4 = 7. Option A is correct. A quick check in the second equation gives 10(4) + 12(7/4) = 40 + 21 = 61. The originally supplied options and marked answer did not contain 7, so the options have been corrected; calculating only x or y would not answer the question.

Let the rectangle’s length be l cm and breadth be b cm. The two statements translate into the linear equations l + b = 37 and l − b = 11. This is a standard word problem solved by elimination. Subtract the second equation from the first: (l + b) − (l − b) = 37 − 11. The l terms cancel, leaving 2b = 26, so b = 13 cm. Then l = b + 11 = 24 cm. Substitution verifies both conditions: 24 + 13 = 37 and 24 − 13 = 11. Therefore option C is correct. If the breadth were 11, 12, or 14 cm, the corresponding sums would be 33, 35, or 39 cm, not 37 cm.