Algebraic methods: Substitution method and Elimination method.

Let (u=x+y) and (v=x-y). Solving (3u+2v=41), (2u-3v=-1) gives (u=7,v=10), so (x=\frac{17}{2}).

Let \(u=\frac{1}{x}\) and \(v=\frac{1}{y}\). The equations become \(2u+3v=13\) and \(3u-2v=4\). Multiply the first equation by 2 and the second by 3 to get \(4u+6v=26\) and \(9u-6v=12\). Adding them gives \(13u=38\), so \(u=\frac{38}{13}\). Hence, \(\frac{1}{x}=\frac{38}{13}\). The value \(\frac{31}{13}\) is for \(v=\frac{1}{y}\), not for \(\frac{1}{x}\). Exam tip: For reciprocal terms, substitute \(u=\frac{1}{x}\) and \(v=\frac{1}{y}\) first to form linear equations.

Let u=x-1 and v=y+2. The equations become 2u+3v=25 and 4u-3v=5. Adding them eliminates v and gives 6u=30, so u=5. Substituting into the first equation gives 10+3v=25, hence v=5. Therefore, x=6 and y=3, so x+y=9; option B is correct. Exam tip: when the coefficients of one variable have opposite signs, add the equations to eliminate that variable directly.

Let (u=2x-y) and (v=x+y). Solving (5u-3v=11), (2u+4v=50) gives (u=7,v=9), hence (y=\frac{11}{3}).

Adding the equations gives (13x=65), so (x=5). In exams, eliminate opposite coefficients first.

From the first equation, (y=4x-11). Substitution must be checked in both equations before selecting an option.

Adding gives (10x=40), so (x=4) and (y=\frac{11}{2}). Thus (x+y=\frac{19}{2}); evaluate the expression after solving.

Adding gives (11x=44), so (x=4) and (y=\frac{19}{5}). Therefore (x-y=\frac{1}{5}); check signs carefully.

Subtracting the second equation from the first gives (9y=45), so (y=5). Equal coefficients make subtraction faster.

The equations are \(3x+4y=26\) and \(5x-2y=22\). Multiplying the second equation by 2 gives \(10x-4y=44\). Adding it to the first equation eliminates \(y\): \(13x=70\), so \(x=\frac{70}{13}\). Substituting this into the second equation gives \(y=\frac{32}{13}\). Therefore, \(2x+y=2\left(\frac{70}{13}\right)+\frac{32}{13}=\frac{172}{13}\). Hence, option D is correct. Exam tip: Substitute the calculated values back into both original equations to verify the solution.

Adding the two equations gives \((11x+4y)+(7x-4y)=68+4\), so \(18x=72\). Hence, \(x=4\). Substituting \(x=4\) into \(7x-4y=4\) gives \(28-4y=4\), and therefore \(y=6\). Thus, the correct solution is \((4,6)\). In option A, substituting \(x=3\) does not satisfy the second equation. Exam tip: When the coefficients of one variable are opposites, add the equations to eliminate that variable directly.

Substitute (x=3y-2) in the second equation to get (7y-4=33). Verify the final value before choosing an option.

Subtracting the first equation from the second gives (15y=60), so (y=4). When (x)-coefficients are equal, subtract directly.

Adding the two equations eliminates y because 5y and -5y cancel: 12x=36, so x=3. Substituting x=3 into 4x+5y=7 gives 12+5y=7, hence y=-1. Therefore, 3x-y=3(3)-(-1)=10. In an exam, be careful when subtracting a negative value, as the sign changes; this is the most common error in this question.

Let the numbers be (x) and (y), so (x+y=41) and (x-y=9). Adding gives (2x=50), so the greater number is (25).

Let the numerator be (x) and denominator be (x+5). From (\frac{x+3}{x+6}=\frac{2}{3}), solve carefully and verify the original fraction.

Let the price of one pencil be \(p\) rupees and that of one eraser be \(e\) rupees. The equations are \(3p+2e=31\) and \(2p+5e=47\). Multiplying the first equation by 5 and the second by 2 gives \(15p+10e=155\) and \(4p+10e=94\). Subtracting, \(11p=61\), so \(p=\frac{61}{11}\) rupees. Therefore, option B is correct. The value 6 rupees in option C is close, but it does not satisfy the equations exactly. Exam tip: In the elimination method, make the coefficients of one variable equal before adding or subtracting the equations.

Subtracting the first equation from the second gives (3x=24), so (x=8). Compute (y) and reduce the ratio carefully.

Adding gives (15x=60), so (x=4) and (y=\frac{7}{4}). Hence (xy=7); do not depend only on options.

Let length be (l) and breadth be (b), so (l+b=37) and (l-b=11). Subtracting gives (2b=26), so (b=13).