Algebraic methods: Substitution method and Elimination method.

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.

Multiplying the second equation by (3) gives (9y). It cancels with (-9y) in the first equation.

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.

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.

Using (x=24-y) gives (72-5y=37), so (y=7) and (x=17). Hence (2x+y=41), so the correct option is (D).