Algebraic methods: Substitution method and Elimination method.

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).

The direct answer is option C: the breadth is 11 cm. Let length be l and breadth be b. The perimeter formula is 2(l+b)=58, so l+b=29. The statement that length is 7 cm more gives l-b=7. Add these equations: 2l=36, so l=18. Then b=29-18=11 cm. Option C is correct, and checking gives perimeter 2(18+11)=58 and length difference 18-11=7. Option A, 9 cm, would give length 16 cm and perimeter 50 cm. Option B, 10 cm, would give length 17 cm and perimeter 54 cm. Option D, 12 cm, would give length 19 cm and perimeter 62 cm. The common mistake is to use l+b=58 instead of half the perimeter; remember that a rectangle has two lengths and two breadths.

Using elimination, multiply the first equation by 5 and the second by 7: \(20x+35y=290\) and \(42x-35y=42\). Adding them gives \(62x=332\), so \(x=166/31\). Substituting this into \(4x+7y=58\) gives \(y=162/31\). Therefore, \(x:y=166:162=83:81\). The option \(3:4\) is incorrect because it does not satisfy the given pair of equations. Exam tip: always reduce a ratio by dividing both terms by their greatest common factor.

Multiply the first equation by (2) and the second by (3) to eliminate (y). In exams, eliminating variables with opposite signs is faster.

Make the coefficients equal to eliminate (y) and get (x=5), (y=7). In exams, check the answer in both original equations.

Let the numerator be (x) and denominator be (y), so (y=x+5) and (\frac{x+1}{y+1}=\frac{2}{3}). In exams, cross multiply when converting a fraction into an equation.

Substituting \(x=5\) in the second equation gives \(5+y=7\), so \(y=2\). Now substitute \(x=5\) and \(y=2\) in the first equation: \(2(5)+2k=19\). Thus, \(10+2k=19\), so \(2k=9\) and \(k=\frac{9}{2}\). Option \(5\) may result from solving \(2k=9\) incorrectly. In exams, verify the given solution in both equations.

Multiply the second equation by (2), add it to the first, and get (x=4), (y=5). In exams, first find the solution and then evaluate the expression.

Multiply the second equation by (4) to eliminate (x) and get (y=5). In exams, create equal coefficients instead of worrying about large numbers.

Downstream speed is (12) and upstream speed is (8), so (b+s=12), (b-s=8). In exams, first find speeds from distance and time.

Form (x+y=32) and (10x+5y=245), then solve. In exams, money problems usually need separate equations for count and value.

Given \(y=3\), substitute it into \(2x-y=5\): \(2x-3=5\), so \(2x=8\) and \(x=4\). Now put \(x=4\) and \(y=3\) in \(ax+3y=25\): \(4a+9=25\). Thus \(4a=16\), giving \(a=4\). Exam tip: First substitute the given variable into the equation that lets you find the other variable most easily.

Multiply the second equation by (2) and add it to the first. In exams, add equations when opposite signs appear.

Multiply the first equation by (2) and subtract from the second to get (y=4). In exams, subtract in the right direction to avoid sign errors.

Form (f=4s+2) and (f+8=3(s+8)), then solve. In exams, add the same number of years to both ages for future age.

The second equation is (3) times the first, so both lines are the same. In exams, compare ratios to identify dependent equations.

The ratio of variable coefficients is equal, but the ratio of constants is different. In exams, this means parallel lines and no solution.

Putting (x=4) in the second equation gives (y=7). Then the first equation gives (4k+14=16), so (k=\frac{1}{2}); check options carefully in exams.

From (2x+y=11), put (y=11-2x) in the first equation. In exams, combine all terms correctly after substitution.