Algebraic methods: Substitution method and Elimination method.

Subtract the second equation from the first: \((4x+7y)-(4x+3y)=41-25\). This gives \(4y=16\), so \(y=4\). If \(y=3\), then \(4y=12\), not 16. Exam tip: subtract equations directly when one variable has equal coefficients.

Adding the given equations gives \((9x-2y)+(3x+2y)=35+13\). The terms \(-2y\) and \(+2y\) cancel, so \(12x=48\), giving \(x=4\). Hence, option C is correct. If \(x=3\), the sum of the two equations would give \(12x=36\), not 48. Exam tip: add equations when a variable has equal coefficients with opposite signs.

Let the tens digit be (x) and units digit be (y). From (x+y=9) and (x-y=3), (x=6,\ y=3), so the number is (63).

Using (x=y+1) gives (3y+3+2y=20), so (y=\frac{17}{5}) and (x=\frac{22}{5}). Then (3x-y=\frac{49}{5}).

Adding the two equations cancels \(3y\) and \(-3y\): \(7x=35\), so \(x=5\). Substituting this into \(2x+3y=25\) gives \(10+3y=25\), hence \(y=5\). Therefore, the solution is \((x,y)=(5,5)\). Option B satisfies the first equation, but in the second equation it gives \(5x-3y=3\), not \(10\). Exam tip: add equations directly when one variable has opposite coefficients.

Use (y=22-4x) from the first equation. Substituting in the second gives (11x=45), then (y=\frac{62}{11}).

The first equation becomes (3x+2y=20), but the second is (3x+2y=22). Therefore there is no solution.

Let the number of boys be \(x\) and the number of girls be \(y\). Then \(x+y=45\) and \(x-y=9\). Adding the two equations gives \(2x=54\), so \(x=27\). Therefore, the number of boys is 27. Option 26 is close, but it would leave 19 girls, giving a difference of 7 rather than 9. Exam tip: for questions giving a total and a difference, add the equations to find the larger quantity directly.

On adding the two equations, \(-5y\) and \(+5y\) cancel: \(2x-5y+3x+5y=-1+31\). Thus, \(5x=30\), so \(x=6\). For example, \(x=5\) gives \(5x=25\), which does not satisfy the resulting equation \(5x=30\). Exam tip: Add equations directly when a variable has equal and opposite coefficients.

Adding the two equations eliminates \(2y\) and \(-2y\): \(8x=40\), so \(x=5\). Substituting \(x=5\) into \(5x+2y=29\) gives \(25+2y=29\), hence \(2y=4\) and \(y=2\). For example, \(y=3\) does not satisfy the first equation. Exam tip: add equations directly when a variable has opposite coefficients.

Substitute \(y=3x-7\) into \(2x+y=18\): \(2x+3x-7=18\). Thus, \(5x=25\), so \(x=5\). Then \(y=3(5)-7=8\). Hence, the correct pair is \(x=5,\ y=8\). Option C satisfies the first equation, but it gives \(2x+y=23\), not 18. Exam tip: Always verify the obtained pair in both equations.

Subtract the second equation from the first: \((3x+4y)-(3x-y)=38-13\). Thus, \(5y=25\), so \(y=5\). If \(y=4\), then \(5y=20\), which does not satisfy the resulting equation. Exam tip: subtract equations when a variable has the same coefficient to eliminate it quickly.

Use (y=4x-11) from the second equation. Substitution gives (14x-33=27), so (x=\frac{30}{7}).

Adding the two equations gives \((x+2y)+(3x-2y)=13+7\), so \(4x=20\). Hence, \(x=5\). Substituting this into \(x+2y=13\) gives \(5+2y=13\), and therefore \(y=4\). Thus, the solution is \(x=5,\ y=4\). In option C, \(x=6\) does not satisfy the first equation. Exam tip: adding equations is a quick elimination method when terms have opposite coefficients.

Use (y=2x-3) from the second equation. Substitution gives (7x=28), so (x=4,\ y=5) and (x+2y=14).

Substituting the given values in the equation gives \(2k+4(3)=22\). Thus, \(2k+12=22\), so \(2k=10\) and \(k=5\). If \(k=4\), the left-hand side becomes \(20\), not \(22\). Exam tip: In parameter questions, substitute the given values of \(x\) and \(y\) carefully before solving.

From the second equation, \(4x-y=3\), we get \(y=4x-3\). Substituting this into the first equation gives \(2x+3(4x-3)=19\), so \(14x=28\) and \(x=2\). Hence, \(y=4(2)-3=5\), and the solution is \(x=2,\ y=5\). Option A does not satisfy the second equation. Exam tip: always verify the obtained values in both original equations.

Adding the two equations eliminates the y-terms: 5x + 2y + 3x - 2y = 28 + 4, so 8x = 32. Hence, x = 4. Substituting x = 4 into 3x - 2y = 4 gives 12 - 2y = 4, so y = 4. Therefore, x + y = 4 + 4 = 8. Option 7 is not correct because the simultaneous solution is x = 4 and y = 4. Exam tip: when variable terms have opposite coefficients, add the equations to eliminate that variable quickly.

From the second equation, \(x=9-y\). Substituting this into \(3x+5y=31\) gives \(3(9-y)+5y=31\), or \(27+2y=31\). Hence \(y=2\) and \(x=9-2=7\). If \(x=6\), then the second equation gives \(y=3\), but the first equation becomes \(33\), not \(31\). Exam tip: verify the obtained values in both original equations.

From the first equation, \(y=6x-21\). Substituting this into \(2x+3y=17\) gives \(2x+3(6x-21)=17\), so \(20x=80\). Hence, \(x=4\) and \(y=3\), making option C correct. For example, \(x=5, y=1\) gives \(13=17\) in the second equation, so it is not a solution. Exam tip: Always verify the ordered pair in both original equations.