Algebraic methods: Substitution method and Elimination method.

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 direct answer is option C: \(x-y=8\). The second given equation already has the exact expression asked for. Start with \(\frac{x-y}{4}=2\). Multiply both sides by 4: \(x-y=2\times4=8\). The first equation, \(\frac{x+y}{3}=7\), gives \(x+y=21\), but that information is not needed for this particular question. Option A, 6, is wrong because it does not follow from multiplying 2 by 4. Option B, 7, is wrong for the same reason. Option C, 8, is correct because the denominator 4 is removed by multiplication by 4. Option D, 9, is also unsupported. A common mistake is to solve for both x and y even when the requested expression is given directly. Exam cue: if the question asks for the numerator of a fraction, first clear its denominator.

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.

The direct answer is option A: \(x=\frac{188}{11}\). We solve step by step. From \(x-y=5\), add y to both sides: \(x=y+5\), or equivalently \(y=x-5\). In the first equation, \(\frac{x}{4}+\frac{y}{7}=6\), multiply every term by 28, the least common denominator: \(7x+4y=168\). Substitute \(y=x-5\): \(7x+4(x-5)=168\). Thus \(7x+4x-20=168\), so \(11x=188\), and \(x=\frac{188}{11}\). Option A is correct. Option B, \(\frac{178}{11}\), would result from an arithmetic error after combining terms. Option C, \(\frac{168}{11}\), forgets the 20 produced by substitution. Option D, \(\frac{158}{11}\), has an even larger subtraction error. Always multiply the complete equation by 28, not just one term, and check the result by substituting back.

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.

Let (x+y=s) and (x-y=d). Solving gives (s=\frac{184}{13}) and (d=\frac{134}{13}), so (y=\frac{25}{13}).

Subtracting the first equation from the second directly gives (x-y=4). In such questions, the difference of equations gives the answer quickly.

Multiply the first equation by (2) and add it to the second. (x=\frac{126}{11}) and (y=\frac{73}{22}), so (x+y=\frac{325}{22}).

After clearing denominators, (2x+3y=60) and (2x-3y=6) are obtained. Adding gives (x=\frac{33}{2}), so none of the given options is correct.

Let the tens digit be \(x\) and the units digit be \(y\). The number is \(10x+y\), so \(10x+y=7(x+y)\) gives \(3x=6y\), or \(x=2y\). Also, \(x-y=2\). Thus, \(2y-y=2\), so \(y=2\) and \(x=4\). Therefore, the number is \(42\). Option \(64\) has digits differing by 2, but \(7(6+4)=70\), not 64. Exam tip: Always write a two-digit number as \(10\times\) tens digit \(+\) units digit.

Adding both equations gives (13x=85). Substituting (x=\frac{85}{13}) gives (y=\frac{93}{13}).

Substitute the given values \(x=6\) and \(y=\frac{7}{2}\) into the first equation \(3x+ky=40\). This gives \(3(6)+k\left(\frac{7}{2}\right)=40\), or \(18+\frac{7k}{2}=40\). Hence \(\frac{7k}{2}=22\), so \(k=\frac{44}{7}\). For example, \(\frac{48}{7}\) would not make the left-hand side equal to 40. Exam tip: when a solution is given, substitute it first into the equation containing the unknown parameter.

Multiplying the first equation by 4 gives \(3x-y=20\), and multiplying the second equation by 3 gives \(x+2y=21\). From \(3x-y=20\), we get \(y=3x-20\). Substituting this into \(x+2y=21\) gives \(x+2(3x-20)=21\), so \(7x=61\). Hence, \(x=\frac{61}{7}\). A close option such as \(\frac{62}{7}\) does not satisfy \(7x=61\). Exam tip: First clear the denominators in fractional equations to obtain simpler linear equations.

Subtracting the second equation from the first gives (7y=35), so (y=5). Then (x=\frac{15}{2}), hence (x-y=\frac{5}{2}).

Let the present ages of the mother and son be \(x\) years and \(y\) years respectively. Then \(x+y=64\). From the condition four years ago, \(x-4=3(y-4)\). Substituting \(x=64-y\) gives \(60-y=3y-12\), so \(72=4y\) and \(y=18\). Hence, the son is presently 18 years old. Although 16 may be considered from the total, it does not satisfy the condition that the mother was three times the son's age four years ago. Exam tip: In age problems, subtract the same number of years from each person's present age for a past-age condition.

Adding the two equations gives \((7x+6y)+(5x-6y)=5+31\), so \(12x=36\). Hence, \(x=3\). Substituting this into \(7x+6y=5\) gives \(21+6y=5\), so \(6y=-16\) and \(y=-\frac{8}{3}\). Therefore, the solution is \(x=3,\ y=-\frac{8}{3}\). In option A, \(x=2\) would give \(12x=24\), not \(36\), after adding the equations. Exam tip: when coefficients of a variable are opposites, add the equations to eliminate that variable directly.

To make coefficients proportional, (5:10=a:6) must hold. This gives (a=3), while (11:30) is not the same ratio.

Multiply both equations by 10 to remove decimals: \(4x+7y=53\) and \(8x-2y=38\). Dividing the second equation by 2 gives \(4x-y=19\), so \(y=4x-19\). Substituting this into the first equation, \(4x+7(4x-19)=53\), gives \(x=\frac{93}{16}\) and \(y=\frac{17}{4}\). Therefore, \(x+y=\frac{161}{16}\). Note that \(\frac{93}{16}\) is the value of \(x\) alone, not of the sum. Exam tip: remove decimals first when solving linear equations with decimal coefficients.

Adding both equations gives (12x=72), so (x=6). The second equation gives (18+4y=20), so (y=\frac{1}{2}), hence the correct listed value is (C).

After clearing denominators, (3x+5y=105) and (2x-y=16) are obtained. Elimination gives (x=\frac{185}{13}), so none of the given options is correct.