Algebraic methods: Substitution method and Elimination method.

The equations become (x+2y=24) and (2x-y=15). The solution is (x=\frac{54}{5},\ y=\frac{33}{5}), so (x-y=\frac{21}{5}).

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

Twice the first equation is (4x+6y=22), but the second is (4x+6y=21). Therefore there is no solution.

Multiply the second equation by (3) and add the first. (x=5), then (4(5)+9y=71) gives (y=\frac{17}{3}).

Multiply the second equation, \(2a+5b=250\), by 2 to get \(4a+10b=500\). Subtracting \(4a+3b=276\) gives \(7b=224\), so \(b=32\). Substituting this into \(2a+5b=250\) gives \(2a+160=250\), hence \(a=45\). For example, \(a=40\) does not satisfy both equations together. Exam tip: in elimination, first make the coefficients of one variable equal.

Adding the two equations gives \(7x-2y+3x+2y=39+21\), so \(10x=60\) and hence \(x=6\). Substituting this in \(3x+2y=21\) gives \(18+2y=21\), so \(2y=3\). Therefore, \(x+2y=6+3=9\). Option 8 would require \(2y=2\), which does not satisfy the given equations. Exam tip: add equations directly when terms with opposite coefficients cancel.

Substitute \(x=3y-4\) into \(2x+5y=37\): \(2(3y-4)+5y=37\). Thus, \(6y-8+5y=37\), so \(11y=45\). Therefore, \(y=\frac{45}{11}\). For example, \(\frac{42}{11}\) would give \(11y=42\), not \(11y=45\). Exam tip: while substituting, distribute the coefficient \(2\) to both terms inside the bracket.

Let the numbers of boys and girls be x and y respectively. Then x+y=68 and 2x+y=112. Subtracting the first equation from the second gives x=44. Therefore, y=68-44=24. Hence, the correct answer is 24. If 26 were chosen, the total would become 70, so it cannot be correct. Exam tip: In such questions, subtracting the equations often eliminates one variable directly.

Adding both equations gives (8x=32), so (x=4). Then (y=\frac{7}{2}), hence (2x+y=\frac{23}{2}).

Dividing the first equation by 2 gives \(x+y=17\), and dividing the second equation by 3 gives \(x-y=5\). Adding these equations gives \(2x=22\), so \(x=11\). Substituting in \(x+y=17\) gives \(y=6\). Hence, the correct solution is \(x=11,\ y=6\). In option \(x=10,\ y=7\), the sum is 17, but the difference is not 5. Exam tip: first divide by any common factor to simplify the equations before eliminating a variable.

From (5x-y=11), put (y=5x-11) in the first equation and solve. In exams, combine terms carefully after substitution.

Multiply the second equation by (2) to eliminate (y), then find (x). In exams, eliminate one variable first and then calculate the required expression.

Multiply the first equation by (2) to eliminate (x) and get (y=5). In exams, watch the signs after multiplication.

From the first equation, y=14-x. Substituting this in 3x-2y=7 gives 3x-2(14-x)=7, so 5x=35 and x=7. Then y=14-7=7. Therefore, x-y=7-7=0. The values 1, 2, and 3 would be possible only if x and y were unequal. Exam tip: after finding both variables, calculate the exact expression asked in the question separately.

From (2x-y=1), put (y=2x-1) in the first equation. In exams, isolate one variable clearly first.

From the second equation, put (y=2x-3), giving (x=5) and (y=7). In exams, recheck both values before finding (xy).

Use (y=17-4x) from the second equation to get (x=3), (y=5). In exams, calculate the asked expression after finding the solution.

Multiply the first equation by (2) and the second by (5) to eliminate (y). In exams, making equal coefficients is an easy method.

Form (x+y=23) and (x-y=7), then add them. In exams, adding sum-difference equations quickly gives one variable.

Let the tens digit be (x) and the units digit be (y), giving (x+y=11) and (9x-9y=27). In exams, write a two-digit number as (10x+y).