Algebraic methods: Substitution method and Elimination method.

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.

The answer is A, 4y=12. Subtract the complete second equation from the first: (x+3y)−(x−y)=13−1. When the minus sign is distributed through the parentheses, x−y becomes −x+y in the subtraction. Thus the left side is x+3y−x+y. The x terms cancel, and 3y+y=4y. The right side is 13−1=12, so the resulting equation is 4y=12. Option B would incorrectly treat the second y-term as if its sign did not change. Option C has an incorrect right-hand side. Option D is wrong because x must be eliminated when the equations are subtracted. The safest habit is to use parentheses before subtracting an entire equation.

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.

The correct answer is option D: (3, 2). In elimination, look for a variable with equal and opposite coefficients. Adding the equations 3x + y = 11 and 3x − y = 7 cancels y: 6x = 18. Divide by 6 to get x = 3. Substitute into the first equation: 3(3) + y = 11, so 9 + y = 11 and y = 2. Therefore the ordered pair is (3, 2). Option A, (2, 5), gives 3x + y = 11 but gives 6 − 5 = 1, not 7, in the second equation. Option B, (4, −1), gives 11 in the first equation but 13 in the second. Option C, (1, 8), gives 11 in the first but −5 in the second. Option D satisfies both: 9 + 2 = 11 and 9 − 2 = 7. Memory cue: add equations when opposite terms cancel immediately, then substitute to find the remaining variable.

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.

The correct answer is option B: (4, 12). The first equation already gives y = 3x, so substitution is the quickest method. Put this into x + y = 16: x + 3x = 16, hence 4x = 16 and x = 4. Now y = 3x = 3 × 4 = 12. Therefore the solution is (4, 12). Option A, (3, 13), has y = 13 rather than 3x = 9, so it fails the first equation. Option B satisfies both: 12 = 3 × 4 and 4 + 12 = 16. Option C, (5, 11), fails because 11 is not 15 and because 5 + 11 = 16 only satisfies the second equation. Option D, (6, 10), fails the first equation since 10 is not 18, although its sum is 16. Always test both equations, not just one. Memory cue: when one equation gives a variable directly, substitute it into the other equation immediately.

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 \(y=2\).

An equation remains equivalent when every term on both sides is divided by the same non-zero number. Divide each term of 2x + 2y = 14 by 2: (2x)/2 + (2y)/2 = 14/2. Each coefficient and the constant simplify, giving x + y = 7. Therefore option B is correct. The operation must be applied to every term, not just one part of the equation. Option A leaves 14 unchanged, while option C changes only the coefficient of x and option D changes only the coefficient of y. The simplified equation has exactly the same solutions as the original because division was by the non-zero number 2.

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.