Algebraic methods: Substitution method and Elimination method.

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.

Subtracting the second equation from the first gives (3x=6), so (x=2). Equal (y) terms can be removed by subtraction.

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.

Subtracting the second equation from the first gives (2y=6), so (y=3). When (x) is equal, subtraction is easiest.

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.

Subtracting the second equation from the first gives (3y=9), so (y=3). Subtract to remove the common (3x).

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.

Substituting (x=3y-1) gives (4y-1=11), so (y=3) and (x=8). Use the given expression for (x) directly.

The first equation becomes (2x+y=10); subtracting (x+y=7) gives (x=3). Divide by common factors first.

From the second equation (x=10-y); substituting gives (10-y-2y=1), so (y=3). Watch brackets and signs in substitution.

The second given equation directly states (x+y=9), so the value of (x+y) is 9. There is no need to solve the first equation. Options such as 8 or 10 do not match the stated sum. Exam tip: If the required expression is already given directly in an equation, avoid unnecessary calculations.

Subtracting the second equation from the first gives (2x=10), so (x=5) and (y=4). Subtraction is correct to remove equal (y) terms.

Substitution gives (x+2x-1=8), so (x=3) and (y=5). Find (x) first, then use the expression for (y).

The first equation is (x+y=5); adding it with (x-y=1) gives (2x=6). Reduce large coefficients first.

The first equation is (2(x+2y)=18), so it is (2) times the second. Recognizing proportional equations is also important.

Given \(x=4\), substitute it into \(x+2y=14\): \(4+2y=14\). Thus, \(2y=10\), so \(y=5\). If \(y=4\), the left-hand side becomes \(4+2(4)=12\), not 14. Exam tip: When the value of one variable is given directly, substitute it into the other equation to find the remaining variable.

Given \(x=5\), substitute it into \(2x-y=8\): \(2(5)-y=8\), so \(10-y=8\). Subtracting 10 from both sides gives \(-y=-2\), hence \(y=2\). For example, if \(y=3\), the left side becomes \(10-3=7\), not 8. Exam tip: verify the obtained value by substituting it back into the original equation.

Subtracting the second equation from the first gives (2y=8), so (y=4). Eliminate the equal (2x) terms.