Algebraic methods: Substitution method and Elimination method.

Substituting \(x=3\) gives \(2(3)-y=4\), or \(6-y=4\). Thus, \(-y=-2\), so \(y=2\). In the exam, handle the negative sign carefully; \(6\) is the value of \(2x\), not of \(y\).

The governing concept is preserving an equation while reducing it to an equivalent simpler form. Every term on both sides must be divided by the same non-zero common factor. In 6x + 3y = 12, the common factor is 3. Dividing each term by 3 gives 6x/3 + 3y/3 = 12/3, hence 2x + y = 4. Therefore option B is correct. Option A divides the coefficients but leaves the constant 12 unchanged, option C changes the coefficient of x incorrectly, and option D leaves the original y coefficient and constant. The simplified equation has exactly the same solutions as the original because the same non-zero operation was applied to both sides.

The direct answer is option A: \((x,y)=(3,2)\). Add the equations \(x+y=5\) and \(x-y=1\). The \(y\) terms cancel: \(2x=6\), so \(x=3\). Substitute into \(x+y=5\): \(3+y=5\), hence \(y=2\). Option A satisfies both equations: \(3+2=5\) and \(3-2=1\). Option B, \((2,3)\), gives sum 5 but difference \(-1\), not 1. Option C, \((4,1)\), gives sum 5 but difference 3. Option D, \((1,4)\), gives sum 5 but difference \(-3\). Thus checking both original equations confirms A. When coefficients of one variable are opposites, addition is usually the quickest elimination method.

Given \(x=2\), substitute it in \(2x+y=7\): \(2(2)+y=7\), or \(4+y=7\). Hence, \(y=3\). If \(y=5\), the left side becomes \(2(2)+5=9\), not 7. Exam tip: after substitution, solve the resulting simple linear equation carefully.

Substituting (y=x+1) in (x+y=7) gives (2x+1=7), so (x=3) and (y=4). Use the already isolated variable in exams.

Subtracting the second equation from the first gives (2x=6), so (x=3). Subtract equal like terms when their signs are the same.

Adding both equations gives (2x=14), so (x=7) and (y=3). After finding one variable, substitute it in any equation.

Adding both equations gives (3x=9), so (x=3) and (y=5). Variables with opposite signs cancel quickly in elimination.

The governing method is substitution: use the expression given for one variable in the other equation. Since x=2y, replace x in x+y=9 by 2y. This gives 2y+y=9, so 3y=9 and y=3. Substituting this value back into x=2y gives x=2(3)=6. Therefore option B is correct. A direct check confirms the pair (x,y)=(6,3): it satisfies x=2y because 6=2×3, and it satisfies x+y=9 because 6+3=9. Option A is the value of y rather than x, option C merely repeats the total x+y, and option D results from an incorrect multiplication or addition. The substitution method is appropriate because one equation already expresses x directly in terms of y.

Substituting (y=3x) gives (4x=8), so (x=2) and (y=6). Substitution is easiest when one variable is already expressed as a multiple.

The two equations describe relationships between the same two unknown numbers, x and y. Because the coefficient of y is 1 in both equations, subtracting one equation from the other removes y completely. This leaves an equation containing only x, which can be solved directly. This is the elimination method, because one variable is eliminated by combining the equations.

Subtract the second equation from the first: \(4x+y-(x+y)=14-5\). The y terms cancel, so \(3x=9\). Dividing both sides by 3 gives \(x=3\). Substitution can check it: if x is 3, then \(3+y=5\), so y is 2, and \(4(3)+2=14\). Therefore option B is correct.

From the second equation (x=y+2); substituting gives (3y+2=11), so (y=3) and (x=5). Substitute the isolated variable carefully.

The appropriate algebraic method is substitution. From the simpler equation x+y=5, isolate x to obtain x=5-y. Substitute this into the second equation: 2(5-y)+3y=12. Expanding gives 10-2y+3y=12, so y=2. Substituting y=2 into x=5-y gives x=3. Therefore the solution is (x,y)=(3,2), and option B is correct. Verification is important: 3+2=5 and 2(3)+3(2)=6+6=12, so both original equations hold. Option A reverses the two values and gives 2+3=5 but 2(2)+3(3)=13, so it fails the second equation. Options C and D also satisfy neither equation pair simultaneously. Thus only B is unambiguous.

Add the equations \(x+y=6\) and \(3x-y=6\). The \(y\)-terms cancel, giving \(4x=12\), so \(x=3\). Hence, option C is correct. For example, \(x=2\) cannot satisfy both equations together. Exam tip: add two equations when a variable has equal coefficients with opposite signs.

Adding both equations gives (3x=12), so (x=4) and (y=1). Eliminate first, then find the other variable.

Given \(x=5-y\), substitute it into \(2x+y=8\): \(2(5-y)+y=8\). This gives \(10-2y+y=8\), or \(10-y=8\). Hence, \(y=2\). If \(y=3\), then \(x=2\) and \(2x+y=7\), so it does not satisfy the given equation. Exam tip: while expanding after substitution, write the negative term \(-2y\) carefully.

The governing concept is the elimination method for a pair of linear equations in two variables. The coefficient of y is −1 in both equations, so subtracting the second equation from the first eliminates y directly: (5x − y) − (2x − y) = 9 − 3. Simplifying gives 5x − 2x − y + y = 6, hence 3x = 6. Dividing both sides by 3 yields x = 2, so option B is correct. A check confirms the result: when x = 2, the first equation gives 10 − y = 9, so y = 1; the second gives 4 − 1 = 3, which is also true. The other values of x cannot produce one common value of y satisfying both original equations, so they are not valid solutions.

Substitute \(x=7\) into \(x+y=12\) to get \(7+y=12\). Subtracting 7 from both sides gives \(y=5\). Therefore, option C is correct. If \(y=6\), the sum would be 13, so it does not satisfy the given equation. Exam tip: When the value of one variable is given, substitute it into the equation first.

The governing concept is solving a pair of simultaneous linear equations by elimination. Because the coefficient of x is 1 in both equations, subtracting the second equation from the first removes x directly: (x + 3y) − (x + y) = 13 − 7. The left side becomes x + 3y − x − y = 2y, while the right side is 6. Thus 2y = 6, and division by 2 gives y = 3. Therefore option B is correct. For verification, substitute y = 3 into x + y = 7, giving x = 4. In the first equation, 4 + 3(3) = 4 + 9 = 13, so the pair satisfies both equations. The other proposed values do not produce one common x in both equations.

The first equation becomes (x+y=8); adding it with (x-y=2) gives (2x=10). Simplifying an equation first makes solving easier.

From the first equation, \(y=x+4\). Substituting this into \(2x+y=10\) gives \(2x+(x+4)=10\). Hence, \(3x+4=10\), so \(3x=6\) and \(x=2\). For example, \(x=3\) does not satisfy the second equation. Exam tip: After substitution, remove brackets carefully and combine like terms.

When two linear equations contain the same term with the same coefficient, subtraction can eliminate that term. This is the elimination method. Removing one variable leaves a simple equation in the other variable. After finding that variable, the result can be substituted back if the value of the second variable is also needed. Careful subtraction of every term is essential, including the signs on the right-hand sides.

The equations are 
m 3x+2y=16
m and 
m 3x-y=7
m. Subtract the second equation from the first: 
m (3x+2y)-(3x-y)=16-7
m. The 
m 3x
m terms cancel, and 
m 2y-(-y)=3y
m, so 
m 3y=9
m. Dividing by 3 gives 
m y=3
m. Therefore option C is correct. The other choices do not satisfy the difference between the two equations.

Given \(x=2\), substitute it into \(x+4y=18\): \(2+4y=18\). Thus, \(4y=16\), so \(y=4\). If \(y=3\), the left-hand side becomes \(2+12=14\), not 18. Exam tip: after substitution, transpose the constant term first and then divide by the coefficient of \(y\).

On adding the two equations, \(y\) and \(-y\) cancel: \((2x+y)+(2x-y)=11+5\), so \(4x=16\). Hence, \(x=4\). The nearby option \(x=3\) would give \(4x=12\), not the required \(4x=16\). Exam tip: add the equations when a variable has opposite coefficients so that it is eliminated.

The governing concept is the substitution method for simultaneous linear equations. One variable, x, is already expressed in terms of y as x = 3y − 1, so this expression can replace x in the second equation. Substitution gives (3y − 1) + y = 11. Combining like terms produces 4y − 1 = 11; adding 1 to both sides gives 4y = 12, and division by 4 gives y = 3. Now substitute y = 3 into x = 3y − 1: x = 3(3) − 1 = 9 − 1 = 8. Hence the solution is x = 8 and y = 3, so option A is correct. Checking gives 8 = 3(3) − 1 and 8 + 3 = 11. Each other option fails at least one original equation.