Algebraic methods: Substitution method and Elimination method.

Elimination gives (x=4) and (y=3), so (x-y=1). In exams, first make coefficients equal to eliminate one variable.

Solving gives (x=4) and (y=4), so (2x+y=12). In exams, compute the required expression after finding the variables.

Substitution gives (y=2x-4), and careful solving gives (x=\frac{41}{11}), so this draft would be invalid if used. Always verify both equations and options.

Clear denominators to get (5x+2y=60) and use (x=y+4). This gives (y=\frac{40}{7}) and (x=\frac{68}{7}).

Multiplying both equations by 10 gives \(3x+2y=27\) and \(5x-y=14\). From the second equation, \(y=5x-14\). Substituting this into the first equation gives \(3x+2(5x-14)=27\), so \(13x=55\). Hence, \(y=5\times\frac{55}{13}-14=\frac{93}{13}\). The value \(\frac{105}{13}\) does not satisfy both equations simultaneously. Exam tip: For linear equations containing decimals, first multiply by 10 or a suitable power of 10 to remove decimals.

Adding both equations gives (10x=100), so (x=10). Then (y=\frac{31}{7}), hence (x+2y=\frac{132}{7}), so no integer option is correct.

After clearing denominators, (4x+3y=60) and (5x-2y=30) are obtained. Elimination gives (x=\frac{270}{23}).

The given solution \(x=8, y=5\) must satisfy \(kx+5y=42\). Substituting gives \(8k+5(5)=42\), or \(8k+25=42\). Hence \(8k=17\), so \(k=\frac{17}{8}\). For example, \(\frac{19}{8}\) would not make the left-hand side equal to 42. Exam tip: when a solution is given, substitute it into the equation containing the unknown parameter.

For no solution, coefficients must be proportional and constants not proportional. Since (6:2=3), (a=9), and (18:11) is not the same ratio.

Adding the two equations gives \(7x-3y+2x+3y=22+23\), so \(9x=45\). Hence, \(x=5\). Substituting this into \(2x+3y=23\) gives \(10+3y=23\), and therefore \(y=\frac{13}{3}\). Thus, the correct ordered pair is \(\left(5,\frac{13}{3}\right)\). In option A, substituting \(x=4\) does not satisfy the first equation. Exam tip: When coefficients have opposite signs, add the equations first to eliminate that variable quickly.

Let the tens digit be (x) and units digit be (y). From (x+y=13) and (9(x-y)=45), (x=9,\ y=4), so the number is (94).

Let the larger number be \(x\) and the smaller number be \(y\). Then \(x+y=52\) and \(3x+2y=136\). Multiplying the first equation by 2 gives \(2x+2y=104\). Subtracting this from the second equation gives \(x=32\). Therefore, the larger number is 32. For example, 30 does not satisfy the second condition. Exam tip: multiply one equation suitably to eliminate one variable quickly.

The coefficients of \(y\) are \(-4\) and \(+4\), so adding the equations eliminates \(y\): \(12x=60\). Hence, \(x=5\), making option B correct. For example, \(x=4\) does not satisfy the resulting equation \(12x=60\). Exam tip: Add equations directly when a variable has opposite coefficients.

Multiply the second equation by (2) and add it to the first. This gives (x=\frac{111}{11}) and then (y=\frac{23}{11}).

The second equation directly gives (x-y=8). In exams, the asked expression is sometimes obtained directly.

Multiply the second equation by (2) and subtract the first. (q=23) and then (p=27), so none of the options is correct.

Substitute \(x=6\) and \(y=2\) into \(2x+my=26\): \(2(6)+m(2)=26\). Thus, \(12+2m=26\), so \(2m=14\) and \(m=7\). If \(m=6\), the left-hand side becomes \(24\), not \(26\). Exam tip: When a value pair is given, substitute each value into its corresponding variable carefully.

The first equation becomes (2x+3y=10). At (c=11), the left side is the same but the right side is different, so there is no solution.

Multiply the first equation by (28) to get (7x+4y=168). Using (x=y+5) gives (x=\frac{188}{11}).

Multiply both equations by 10 to remove decimals: \(6x-3y=27\) and \(2x+5y=31\). Dividing the first equation by 3 gives \(2x-y=9\), so \(y=2x-9\). Substituting this into the second equation gives \(2x+5(2x-9)=31\), hence \(12x=76\) and \(x=\frac{19}{3}\). \(\frac{38}{7}\) is incorrect because correct substitution or elimination gives a denominator of 3. Exam tip: remove decimals first when solving linear equations to reduce calculation errors.