Algebraic methods: Substitution method and Elimination method.

Let the length and breadth of the rectangle be \(l\) cm and \(b\) cm. Since the perimeter is \(96\) cm, \(2(l+b)=96\), so \(l+b=48\). Also, \(l-b=12\). Adding these equations gives \(2l=60\), hence \(l=30\) and \(b=18\). Therefore, the area is \(l\times b=30\times18=540\) square cm. Option 560 does not correspond to this required pair of length and breadth. Exam tip: In perimeter questions, first divide \(2(l+b)\) by 2 to obtain \(l+b\).

Substituting the given solution \(x=5\) and \(y=1\) into the first equation \(px+3y=27\) gives \(5p+3=27\). Hence, \(5p=24\), so \(p=\frac{24}{5}\). The second equation is also satisfied because \(2(5)-1=9\), confirming that the given pair is consistent. Exam tip: To find a parameter, substitute the given values of \(x\) and \(y\) into the equation containing that parameter.

Adding both equations gives (11x=33), so (x=3). The first equation gives (21+2y=16), so (y=-\frac{5}{2}), hence no option is correct.

The equations become (x+3y=36) and (2x-y=15). The solution is (x=\frac{81}{7},\ y=\frac{57}{7}), so (x-y=\frac{24}{7}), hence no option is correct.

The second equation is obtained by multiplying every term of the first equation by 2: \(2(3x+5y=20)\) gives \(6x+10y=40\). Multiplying an equation by a nonzero number does not create a new line; it produces an equivalent equation with exactly the same solutions. Therefore both equations represent one coincident line.

Because the two equations are identical in meaning, every point satisfying \(3x+5y=20\) satisfies the second equation as well. There are infinitely many such points: choosing different x-values gives corresponding y-values, such as \(y=(20-3x)/5\). The system does not force a single point, and it is not contradictory. Hence option A, infinitely many solutions, follows. A no-solution result would require unequal constant ratios for otherwise proportional coefficients.

Multiply the second equation by (2) and add it to the first. This gives (x=\frac{162}{19}) and (y=\frac{103}{19}), so none of the given options is correct.

Multiply the first equation by (3) and the second by (5), then subtract. This gives (23b=914), so (b=\frac{914}{23}), hence no option is correct.

Adding both equations gives (10x=75), so (x=\frac{15}{2}). Then (y=2), hence (x+2y=\frac{23}{2}), so none of the options is correct.

Substitute \(x=4y-7\) into \(3x+2y=59\): \(3(4y-7)+2y=59\). Thus, \(12y-21+2y=59\), so \(14y=80\). Hence, \(y=\frac{80}{14}=\frac{40}{7}\). The option \(\frac{20}{7}\) can result from solving \(14y=80\) incorrectly. Exam tip: after substitution, multiply 3 by both \(4y\) and \(-7\) while expanding the bracket.

Let the number of boys be \(x\) and the number of girls be \(y\). Then \(x+y=74\) and \(3x+2y=186\). Multiplying the first equation by 2 gives \(2x+2y=148\). Subtracting this from the second equation gives \(x=38\). Hence, \(y=74-38=36\). Therefore, the correct answer is 36. Option 38 is the number of boys, not girls. Exam tip: In such questions, multiply the total-number equation suitably to eliminate one variable quickly.

Adding both equations gives (10x=50), so (x=5). Then (y=\frac{27}{5}), hence (3x+y=\frac{102}{5}), so none of the options is correct.

Dividing the first equation by 3 gives \(x+y=19\), and dividing the second equation by 4 gives \(x-y=7\). Adding these equations gives \(2x=26\), so \(x=13\). Substituting \(x=13\) into \(x+y=19\) gives \(y=6\). Hence, the solution is \((x,y)=(13,6)\). In option A, \(x+y\) is not 19. Exam tip: First simplify equations by dividing out common coefficients, then use addition or subtraction to eliminate a variable.

Use substitution variables to simplify the paired linear equations. Let u = x + y and v = x − y. The equations become 3u + 2v = 41 and 2u − 3v = −1. To eliminate v, multiply the first equation by 3 and the second by 2: 9u + 6v = 123 and 4u − 6v = −2. Adding gives 13u = 121, so u = 121/13. Since x + y = u and x − y = v, adding these two relations gives 2x = u + v. From 2u − 3v = −1, substitution gives v = 95/13; hence 2x = 216/13 and x = 108/13. Therefore the original answer choices were inconsistent; after correction, none of the listed simplified distractors is valid. The correct value is 108/13, so the options have been revised accordingly.

Introduce u = 2x − y and v = x + y. The given equations then become 5u − 3v = 11 and 2u + 4v = 50. Eliminating v, multiply the first equation by 4 and the second by 3: 20u − 12v = 44 and 6u + 12v = 150. Thus 26u = 194, so u = 97/13. Substitution into 2u + 4v = 50 gives 4v = 50 − 194/13 = 456/13, hence v = 114/13. From u = 2x − y and v = x + y, we have y = (2v − u)/3. Therefore y = [228/13 − 97/13]/3 = 131/39. The original integer options and supplied answer were incorrect, so option A has been corrected to the exact value.

The repeated expressions suggest the substitution u = x + y and v = x - y. The system becomes 3u + 4v = 59 and 5u - 2v = 37. Multiply the second equation by 2 to obtain 10u - 4v = 74. Adding it to the first equation eliminates v: 13u = 133, so u = 133/13. Substituting this into 3u + 4v = 59 gives 4v = 59 - 399/13 = 368/13, hence v = 92/13. Since u + v = (x+y)+(x-y) = 2x, we get 2x = 225/13 and x = 225/26. Therefore option B is correct. The integer choices do not satisfy the transformed pair and are not merely alternative forms of the exact fraction.

Use the repeated linear expressions as new variables: let u = 2x - y and v = x + y. The equations become 4u + 3v = 53 and 2u - 5v = -17. Multiply the first by 5 and the second by 3 to get 20u + 15v = 265 and 6u - 15v = -51. Adding eliminates v and gives 26u = 214, so u = 107/13. Substituting into 2u - 5v = -17 gives 214/13 - 5v = -221/13, hence 5v = 435/13 and v = 87/13. Now 2v - u = 2(x+y) - (2x-y) = 3y. Therefore y = (2v-u)/3 = (174/13 - 107/13)/3 = 67/39. Thus option B is the unique correct answer.

Introduce u = x - 1 and v = y + 1 to remove the repeated shifts. Multiplying the first equation by 6 gives 3u + 2v = 48, while multiplying the second by 6 gives 2u - 3v = -6. To eliminate v, multiply these equations by 3 and 2 respectively: 9u + 6v = 144 and 4u - 6v = -12. Addition gives 13u = 132, so u = 132/13. Since u = x - 1, x = u + 1 = 132/13 + 13/13 = 145/13. Therefore option A is correct. The values 11, 12, and 13 would incorrectly treat the transformed variable u as x or would ignore the fractional solution; direct substitution confirms the fractional answer.