Algebraic methods: Substitution method and Elimination method.

Given \(y=2\), substitute it in \(x-y=5\): \(x-2=5\). Adding 2 to both sides gives \(x=7\), so option C is correct. For example, if \(x=6\), then \(6-2=4\), not 5. Exam tip: verify the obtained value by substituting it back into the original equation.

The second equation gives \(y=6\). Substituting it into \(x+y=15\) gives \(x+6=15\). Subtracting 6 from both sides gives \(x=9\), so option C is correct. If \(x=8\), the sum would be 14, not 15. Exam tip: verify the obtained value in the original equation.

Adding both equations gives (4x=20), so (x=5). When (y) terms are opposite, addition is the fastest method.

From the first equation, \(x=10-2y\). Substituting it into \(x+y=7\) gives \((10-2y)+y=7\), or \(10-y=7\). Hence, \(y=3\). If \(y=2\), then \(x=6\) and \(x+y=8\), so it does not satisfy the second equation. Exam tip: combine like terms carefully after substitution, especially their signs.

Subtracting the second equation from the first gives (3x=12), so (x=4). Equal (-y) terms cancel by subtraction.

Substituting (y=2x) gives (2x+2x=6), so (x=\frac{3}{2}); hence none of the listed pairs would be correct.

Adding both equations gives (2x=8), so (x=4) and (y=3). In exams, first eliminate the easy variable.

Substitute \(x=2\) into \(2x+y=8\): \(2(2)+y=8\). Thus, \(4+y=8\), so \(y=4\). If \(y=2\), the left-hand side becomes \(6\), not \(8\). Exam tip: verify the obtained value in the original equation.

Given \(x+y=9\) and \(y=4\), substitute 4 for \(y\) to get \(x+4=9\). Subtracting 4 from both sides gives \(x=5\). Therefore, option C is correct. If \(x=4\), the sum would be 8, not 9. Exam tip: after substitution, perform the same operation on both sides to isolate the unknown.

Adding the equations gives (2x=14), so (x=7) and (y=4). Terms with opposite signs cancel quickly.

Given \(3x+y=10\) and \(x=2\), substitute 2 for \(x\): \(3(2)+y=10\), so \(6+y=10\). Subtracting 6 from both sides gives \(y=4\). If \(y=3\), the left-hand side becomes 9, so it is not correct. Exam tip: after substitution, simplify the equation and isolate the unknown term.

Substitute \(x=4\) into \(x+2y=12\): \(4+2y=12\). Subtracting 4 from both sides gives \(2y=8\), so \(y=4\). If \(y=2\), the left-hand side becomes 8, so it is not correct. Exam tip: verify the obtained value in the original equation.

On adding, (y) and (-y) cancel, so (4x=12). In elimination, check signs of like terms carefully.

Given \(x=3\). Substituting it into \(x+y=8\) gives \(3+y=8\), so \(y=5\). Therefore, the solution pair is \((3,5)\). In \((5,3)\), the numbers are present but their order is reversed; an ordered pair must be written as \((x,y)\). Exam tip: always check the order of the coordinates before choosing an answer.

Substituting (y=x+1) gives (x+x+1=7), so (x=3) and (y=4). Put the given expression into the other equation first.

Using substitution, the first equation gives \(y=2x\). Substituting it into \(x+y=12\) gives \(x+2x=12\), so \(3x=12\). Hence \(x=4\) and \(y=8\), making \((4,8)\) the solution. Although \((6,6)\) satisfies the second equation, it does not satisfy \(y=2x\). Exam tip: verify the ordered pair in both equations.

On subtraction, (x) cancels and (3y-(-y)=4y), so (4y=12). Be careful with negative signs while subtracting.

Given \(x=4\), substitute it in \(2x-y=5\): \(2(4)-y=5\), so \(8-y=5\). Hence, \(-y=-3\) and therefore \(y=3\). If \(y=5\), the left-hand side becomes \(8-5=3\), not 5. Exam tip: After substitution, change signs carefully when removing a negative term.

Given \(y=6\), substitute it into \(x+y=10\) to get \(x+6=10\). Subtracting 6 from both sides gives \(x=4\). Therefore, option C is correct. For example, if \(x=5\), the sum would be 11, not 10. Exam tip: after substitution, perform the same operation on both sides to isolate the unknown.

Adding gives (6x=18), so (x=3) and (y=2). In elimination, remove the variable with equal coefficients first.