Algebraic methods: Substitution method and Elimination method.

Putting (x=2y) gives (3y=15), so (y=5) and (x=10). Substitution is easiest when a ratio relation is given.

Subtracting the second equation from the first gives \((2x+5y)-(2x+y)=29-13\). Thus, \(4y=16\), so \(y=4\). The close distractor \(y=3\) would require \(4y=12\), which is not obtained here. Exam tip: while subtracting equations, carefully keep track of the signs of all terms.

Given \(y=2\), substitute it into \(x-3y=1\): \(x-3(2)=1\), so \(x-6=1\). Adding 6 to both sides gives \(x=7\). If \(x=8\), the left-hand side becomes \(8-6=2\), not 1. Exam tip: after substitution, simplify brackets and negative signs carefully.

Adding the two equations gives \((3x+2y)+(3x-2y)=20+4\). The \(+2y\) and \(-2y\) terms cancel, so \(6x=24\). Hence, \(x=4\). \(x=6\) may seem close, but it results from solving \(6x=24\) incorrectly. Exam tip: in elimination, first check whether the coefficients of one variable are opposites.

Given y=12-2x and x=5, substitute 5 for x: y=12-2(5)=12-10=2. Hence, the correct answer is 2. A value such as 3 can result from an error in multiplication or subtraction. Exam tip: after substitution, perform multiplication before addition or subtraction.

Putting (y=x+3) gives (2x+3=17), so (x=7) and (y=10). Solve the one-variable equation formed after substitution.

Subtracting the second equation, \(x+2y=10\), from the first equation, \(x+5y=22\), gives \(3y=12\). Hence, \(y=4\). Substituting this into \(x+2y=10\) gives \(x+8=10\), so \(x=2\). Therefore, the solution is \((2,4)\). The close distractor \((4,3)\) does not satisfy either equation. Exam tip: Always substitute the ordered pair back into an original equation to verify the answer.

Adding both equations gives (3x=12), so (x=4) and (y=3). In exams, eliminate the variable with opposite signs first.

Substitute the given value \(x=6\) into \(x+y=14\). This gives \(6+y=14\). Subtracting 6 from both sides gives \(y=8\). Therefore, \(y=8\) is correct; if \(y=7\), the sum would be 13, not 14. Exam tip: after substitution, use the same operation on both sides to isolate the unknown.

Substituting (y=x+5) gives (2x+5=17), so (x=6) and (y=11). Using the isolated variable is easier.

This pair is best solved by elimination because the coefficient of \(y\) is the same in both equations. Subtract the second equation from the first: \((3x+y)-(x+y)=18-10\). The \(y\)-terms cancel, giving \(2x=8\). Dividing by 2 gives \(x=4\), so option D is correct. Substitution confirms the result: putting \(x=4\) in \(x+y=10\) gives \(y=6\), and then \(3(4)+6=18\). The other options do not satisfy both equations simultaneously; for example, each would lead to a different value of \(y\) in the two equations. This illustrates the elimination method for a pair of linear equations.

Adding both equations gives (2x=22), so (x=11) and (y=4). After finding one variable, substitute it in a simple equation.

Adding both equations gives (5x=15), so (x=3) and (y=1). In elimination, opposite (y) terms cancel quickly.

Substituting (x=3y) gives (4y=16), so (y=4) and (x=12). Substitution is fast when one variable is a multiple of another.

Substituting (y=2x+3) gives (3x+3=15), so (x=4) and (y=11). Combine like terms after substitution.

Subtracting the second equation from the first gives (3x=15), so (x=5). If (y) terms are equal, subtraction is the correct method.

From the second equation (x=y+2); substituting gives (3y+2=20), so (y=6) and (x=8). Isolate a variable from the simpler equation.

Using (x=8-y) gives (16-2y+3y=21), so (y=5). It is useful to isolate (x) from the smaller equation.

Add the equations \(3x-y=10\) and \(x+y=10\). The \(y\)-terms cancel, giving \(4x=20\), so \(x=5\). If \(x=4\), the sum of the left sides would give \(4x=16\), not 20. Exam tip: add equations directly when a variable has opposite coefficients.

Adding both equations gives (3x=18), so (x=6) and (y=4). Check the solution in both equations.