Algebraic methods: Substitution method and Elimination method.

From the second equation, put (y=19-4x), giving (x=4) and (y=3). In exams, isolate the easier variable first.

Adding the two equations eliminates y because the terms 4y and -4y cancel: 10x = 20, so x = 2. Substituting x = 2 into 7x + 4y = 2 gives 14 + 4y = 2, hence y = -3. Therefore, x - y = 2 - (-3) = 5. Exam tip: Add equations directly when one variable has equal and opposite coefficients.

This question needs careful substitution after elimination; careless cancellation gives a wrong value. Check each obtained value in both equations before marking.

Adding the two equations gives (10x=50), so (x=5). In exams, eliminate terms with opposite coefficients first.

Substitute (x=11-y) carefully in the first equation; incorrect simplification changes the answer. Always check the obtained values in both equations.

Adding gives (7x=35), so (x=5) and (y=3). Therefore (x+y=8); substitute back before choosing the option.

From the second equation, (y=2x-1). Substitute carefully and verify with both equations before using (xy).

Adding the two equations eliminates y: 12x = 36, so x = 3. Substituting x = 3 into 3x + 4y = 25 gives 9 + 4y = 25, hence y = 4. Therefore, x - y = 3 - 4 = -1. In the exam, check the sign carefully; 1 would be y - x, not x - y.

Subtracting the second equation from the first gives (12y=36), so (y=3). When coefficients are equal, subtraction is efficient.

Adding gives (8x=24), so (x=3) and (y=\frac{9}{2}). Thus (2x+y=\frac{21}{2}); solve fully before comparing options.

Substituting (x=2y+1) gives (3(2y+1)-y=17). Always simplify brackets carefully and verify the option.

Adding the two equations gives \((8x-3y)+(2x+3y)=13+17\), so \(10x=30\). Hence, \(x=3\). Substituting this into \(2x+3y=17\) gives \(6+3y=17\), and therefore \(y=\frac{11}{3}\). Option B does not satisfy the first equation when \(x=4\) and \(y=3\) are substituted. Exam tip: add equations first when a variable has opposite coefficients, so that variable is eliminated directly.

Subtracting gives (10y=50), so (y=5), and substitution gives (x=\frac{40}{11}). Fractional values are valid if both equations satisfy them.

Let the numbers be (x,y), so (x+y=23) and (x-y=7). Adding gives (2x=30), so the greater number is (15).

Let the denominator be (y), so the numerator is (y-3). From (\frac{y-1}{y+2}=\frac{4}{5}), (y=13), so the fraction is (\frac{10}{13}).

Subtracting the first equation from the second gives (3x=21), so (x=7). Then (y=\frac{13}{3}), so the ratio is (21:13).

Adding gives (18x=54), so (x=3) and (y=\frac{17}{5}). Hence (3x-y=\frac{28}{5}); do not guess from options.

Adding the two equations gives \((4x+7y)+(8x-7y)=1+35\), so \(12x=36\). Hence, \(x=3\). Substituting \(x=3\) into \(4x+7y=1\) gives \(12+7y=1\), so \(7y=-11\) and \(y=-\frac{11}{7}\). Therefore, the correct ordered pair is \((3,-\frac{11}{7})\). Option B does not satisfy both equations when \(x=2\). Exam tip: When coefficients of one variable have opposite signs, add the equations to eliminate that variable quickly.

From the second equation, (y=6x-27). Substitute carefully; expert questions may have fractional answers.

Adding gives (15x=45), so (x=3) and (y=2). Therefore (y-x=-1); sign reversal can change the answer.