Algebraic methods: Substitution method and Elimination method.

Given y=x+2 and x=5, substitute 5 for x: y=5+2=7. Option 6 would result only if 1 were added instead of 2, so it is incorrect. Exam tip: after substitution, carefully perform the indicated arithmetic operation.

From the second equation, \(x=6\). Substituting it into \(x-y=2\) gives \(6-y=2\). Thus, \(-y=-4\), so \(y=4\). The value \(y=8\) can result from handling the negative sign incorrectly. Exam tip: after substitution, perform the same operation on both sides while isolating the variable.

Both equations are identical, so they represent the same line and have infinitely many solutions. Identifying identical equations gives easy marks.

The left sides are the same but the right sides are different, so there is no solution. In exams, spot contradictory constants quickly.

Given \(2x+3y=18\), substituting \(y=4\) gives \(2x+3(4)=18\), or \(2x+12=18\). Thus, \(2x=6\) and \(x=3\). \(x=6\) would result from forgetting to divide \(2x=6\) by 2. Exam tip: After substitution, simplify multiplication and addition first, then isolate the variable.

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

Subtracting the second equation from the first gives \((x+4y)-(x+y)=17-8\). Thus, \(3y=9\), so \(y=3\). For example, \(y=4\) cannot satisfy both equations simultaneously. Exam tip: In elimination, subtract every term on both sides carefully to avoid sign errors.

Given y=3, substitute it into x=2y: x=2\times 3=6. Therefore, the ordered pair is (x,y)=(6,3). Option (3,6) is a close distractor because it reverses the order of x and y. Exam tip: always write the x-value first and the y-value second in an ordered pair.

Given \(4x-y=10\). Substituting \(x=3\) gives \(4(3)-y=10\), or \(12-y=10\). Subtracting 12 from both sides gives \(-y=-2\), hence \(y=2\). If \(y=3\), the left-hand side becomes 9, so it is not correct. Exam tip: when \(-y=-2\), multiply both sides by \(-1\) to change the signs and get \(y=2\).

Dividing every term by (2) gives (x+y=7). Simplifying equations before elimination is useful.

Adding the two equations eliminates the \(y\)-terms: \((5x+y)+(5x-y)=16+4\), so \(10x=20\). Hence, \(x=2\). Substituting \(x=2\) into \(5x+y=16\) gives \(10+y=16\), so \(y=6\). Therefore, the correct solution is \((2,6)\). The pair \((2,4)\) gives \(14\) in the first equation, so it is not correct. Exam tip: in elimination, add or subtract equations when the coefficients of one variable are opposites.

Given \(x+y=13\) and \(x=9\), substitute 9 for \(x\) to get \(9+y=13\). Subtracting 9 from both sides gives \(y=4\). If \(y=5\), the sum would be 14, so it is not correct. Exam tip: after substitution, isolate the unknown term using the inverse operation.

Given \(x=10-y\) and \(y=7\), substitute 7 for \(y\) in the first equation: \(x=10-7=3\). Therefore, the correct option is \(x=3\). \(x=4\) would result only if \(y=6\), so it is a close but incorrect distractor. Exam tip: after substitution, check the subtraction sign and arithmetic carefully.

Substitute \(y=4\) into \(x+2y=11\): \(x+2(4)=11\). Thus, \(x+8=11\), so \(x=3\). If \(x=5\), the left-hand side becomes \(5+8=13\), not 11. Exam tip: after substitution, simplify the resulting one-variable equation carefully.

Subtracting the second equation from the first gives \((2x+3y)-(2x+y)=19-11\). Thus, \(2y=8\), so \(y=4\). If \(y=3\), then \(2y=6\), which does not match the obtained equation \(2y=8\). Exam tip: while eliminating, subtract every term carefully and keep track of signs.

Given \(x-y=6\) and \(y=2\). Substituting 2 for \(y\) gives \(x-2=6\). Adding 2 to both sides gives \(x=8\). Therefore, option C is correct. The value 6 is only the right-hand side of the equation, not the value of \(x\). Exam tip: After substitution, apply the same operation to both sides to isolate the variable.

The first equation gives \(x=2\). Substituting this into \(y=x+5\) gives \(y=2+5=7\). Therefore, the ordered pair is \((x,y)=(2,7)\). In \((7,2)\), the values are reversed, so it is not correct. Exam tip: always write the value of \(x\) first and the value of \(y\) second in an ordered pair.

In option A, the coefficients of y are 3 and -3. Therefore, on adding the equations, \(3y+(-3y)=0\), so y is eliminated. In option B, adding eliminates x instead, while in option C the y-coefficients are 4 and -2, so y does not cancel directly. Exam tip: For elimination by addition, look for equal coefficients with opposite signs for the variable to be removed.

Given \(x=3\), substitute it into \(3x+2y=17\): \(3(3)+2y=17\), or \(9+2y=17\). Thus, \(2y=8\) and \(y=4\). If \(y=2\), the left-hand side becomes \(13\), not \(17\). Exam tip: substitute the given value first, then isolate the remaining variable step by step.

Given \(x=6\), substitute it into \(y=x-1\): \(y=6-1=5\). Therefore, \(y=5\) is correct. Choosing \(y=6\) is incorrect because 1 must be subtracted from the value of \(x\). Exam tip: after substitution, write each arithmetic step carefully.