Algebraic methods: Substitution method and Elimination method.

Form (l+b=29) and (l-b=7), then add or subtract. In exams, first derive (l+b) from the perimeter.

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 (x=2y-3) and (x+y=42), then solve. In exams, first convert the relation statement into an equation.

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.

Elimination gives (x=5) and (y=3). In exams, put both values in the final expression (x+y).