Algebraic methods: Substitution method and Elimination method.

The equations become (x+3y=36) and (2x-y=15). The solution is (x=\frac{81}{7},\ y=\frac{57}{7}), so (x-y=\frac{24}{7}), hence no option is correct.

The second equation is (2) times the first. Therefore both represent the same line and have infinitely many solutions.

Twice the first equation is (6x+10y=40), but the second is (6x+10y=43). Therefore there is no solution.

Multiply the second equation by (2) and add it to the first. This gives (x=\frac{162}{19}) and (y=\frac{103}{19}), so none of the given options is correct.

Multiply the first equation by (3) and the second by (5), then subtract. This gives (23b=914), so (b=\frac{914}{23}), hence no option is correct.

Adding both equations gives (10x=75), so (x=\frac{15}{2}). Then (y=2), hence (x+2y=\frac{23}{2}), so none of the options is correct.

Substitute \(x=4y-7\) into \(3x+2y=59\): \(3(4y-7)+2y=59\). Thus, \(12y-21+2y=59\), so \(14y=80\). Hence, \(y=\frac{80}{14}=\frac{40}{7}\). The option \(\frac{20}{7}\) can result from solving \(14y=80\) incorrectly. Exam tip: after substitution, multiply 3 by both \(4y\) and \(-7\) while expanding the bracket.

Let the number of boys be \(x\) and the number of girls be \(y\). Then \(x+y=74\) and \(3x+2y=186\). Multiplying the first equation by 2 gives \(2x+2y=148\). Subtracting this from the second equation gives \(x=38\). Hence, \(y=74-38=36\). Therefore, the correct answer is 36. Option 38 is the number of boys, not girls. Exam tip: In such questions, multiply the total-number equation suitably to eliminate one variable quickly.

Adding both equations gives (10x=50), so (x=5). Then (y=\frac{27}{5}), hence (3x+y=\frac{102}{5}), so none of the options is correct.

Dividing the first equation by 3 gives \(x+y=19\), and dividing the second equation by 4 gives \(x-y=7\). Adding these equations gives \(2x=26\), so \(x=13\). Substituting \(x=13\) into \(x+y=19\) gives \(y=6\). Hence, the solution is \((x,y)=(13,6)\). In option A, \(x+y\) is not 19. Exam tip: First simplify equations by dividing out common coefficients, then use addition or subtraction to eliminate a variable.

In option A, terms such as \(ay\) and \(-ay\) give \(ay+(-ay)=0\) on addition, so \(y\) is eliminated. Equal coefficients with the same sign require subtraction instead. Exam tip: check signs before choosing addition.

Multiply both equations by 10 to remove the decimals, obtaining \(4x+7y=62\) and \(3x-2y=11\). Multiply the first equation by 3 and the second by 4: \(12x+21y=186\) and \(12x-8y=44\). Subtracting gives \(29y=142\), so \(y=\frac{142}{29}\). Therefore, option B is correct. Exam tip: For decimal coefficients, clear the decimals first and then use elimination to reduce calculation errors.

Adding the two equations gives \(5x-3y+2x+3y=19+26\), so \(7x=45\). Hence, \(x=\frac{45}{7}\). Substituting this into \(2x+3y=26\) gives \(3y=26-\frac{90}{7}=\frac{92}{7}\), and thus \(y=\frac{92}{21}\). Therefore, \(x-y=\frac{135}{21}-\frac{92}{21}=\frac{43}{21}\). A value such as \(\frac{47}{21}\) can result from an error while subtracting fractions. Exam tip: add equations first when one variable has opposite coefficients.

The first equation becomes (5x+4y=120). Using (x=y+4) gives (y=\frac{100}{9}) and (x=\frac{136}{9}).

Elimination gives (x=\frac{124}{19}) and (y=\frac{177}{19}). Therefore (2x+y=\frac{425}{19}).

Substitute the given solution \(x=7\) and \(y=4\) into \(kx+4y=38\): \(7k+4(4)=38\). Thus, \(7k+16=38\), so \(7k=22\) and \(k=\frac{22}{7}\). Hence, option B is correct. The equation \(x-y=3\) only verifies the given solution, since \(7-4=3\). Exam tip: To find a parameter, substitute the given solution into the equation containing that parameter.

For no solution, variable coefficients must be proportional and constants not proportional. Since (6:2=3), (a=9).

Adding the equations gives \(8x-5y+3x+5y=7+48\), so \(11x=55\). Hence, \(x=5\). Substituting this into \(3x+5y=48\) gives \(15+5y=48\), and therefore \(y=\frac{33}{5}\). Thus, the required ordered pair is \(\left(5,\frac{33}{5}\right)\). For example, option C cannot be correct because its \(x\)-value is 6, whereas elimination requires \(x=5\). Exam tip: add equations directly when one variable has equal and opposite coefficients.

Let the tens and units digits be \(x\) and \(y\), respectively. Then \(x+y=14\). The difference between the number and its reverse is \((10x+y)-(10y+x)=9(x-y)=36\), so \(x-y=4\). Adding the two equations gives \(2x=18\), hence \(x=9\) and \(y=5\). Therefore, the original number is \(95\). Exam tip: When a two-digit number is reversed, the difference between the numbers is always \(9\times\) the difference between their digits.

Let the present ages of the father and son be \(x\) and \(y\) years, respectively. Then \(x+y=56\). Six years ago, \(x-6=3(y-6)\), which gives \(x=3y-12\). Substituting this in the first equation, \(3y-12+y=56\), so \(4y=68\), \(y=17\), and \(x=39\). Therefore, the father’s present age is (39) years.