Algebraic methods: Substitution method and Elimination method.

Given \(x=9-y\), substitute it into \(3x+y=19\): \(3(9-y)+y=19\). This gives \(27-3y+y=19\), or \(27-2y=19\). Hence, \(-2y=-8\) and \(y=4\). For example, \(y=5\) does not satisfy the equation. Exam tip: while expanding after substitution, write the negative sign in \(-3y\) carefully.

Subtracting the second equation from the first gives (4x=16), so (x=4). Equal (-y) terms can be removed by subtraction.

Given \(x+y=18\) and \(y=7\), substitute 7 for \(y\): \(x+7=18\). Therefore, \(x=18-7=11\), so option D is correct. If \(x=10\), the sum would be \(17\), not \(18\). Exam tip: When one variable is known, substitute its value into the equation to find the other variable.

Subtracting the second equation from the first gives (3y=15), so (y=5). Subtract to remove equal (x) terms.

The first equation becomes (x+y=9); adding it with (x-y=1) gives (2x=10). Divide by the common factor first.

Substituting (y=x-3) gives (3x-3=15), so (x=6). After substitution, solve the equation in one variable.

Subtracting the second equation from the first gives (3y=18), so (y=6). Subtraction is correct to remove equal (4x).

Substitute \(x=7\) in \(x+5y=32\): \(7+5y=32\). Thus, \(5y=32-7=25\), so \(y=5\). If \(y=4\), the left side becomes \(7+5(4)=27\), not 32. Exam tip: after substitution, isolate the term containing the unknown and divide by its coefficient.

On adding the two equations, y and -y cancel: 2x+y+2x-y=17+7, so 4x=24. Hence, x=6. If x were 5, then 4x would be 20, not 24. Exam tip: when variable terms have opposite signs, add the equations to eliminate that variable quickly.

Substituting (x=2y-4) gives (3y-4=14), so (y=6) and (x=8). Use the given expression for (x) directly.

Multiplying (2x+y=13) by (3) gives (6x+3y=39). Recognizing equivalent equations is also useful in exams.

Substituting (x=12-y) gives (12-y-2y=3), so (y=3). Simplify negative signs carefully in substitution.

Subtracting the second equation from the first gives (3x=15), so (x=5) and (y=5). Put the found variable into the smaller equation.

Substituting (y=3x-2) gives (4x-2=18), so (x=5) and (y=13). Find (x) first, then use the expression for (y).

The first equation becomes (x+y=10); adding it with (x-y=4) gives (2x=14). Reduce large coefficients first.

Substituting (x=y+2) gives (2y+4+4y=28), so (y=4) and (x=6). Choose a simple relation before substitution.

Given \(x=8\), substitute it in \(x+2y=18\): \(8+2y=18\). Thus, \(2y=10\), so \(y=5\). If \(y=4\), the left-hand side becomes \(16\), not \(18\). Exam tip: after substitution, transpose the constant term first and then divide by the coefficient of the variable.

Given \(3x-y=18\) and \(x=7\), substitute the value of \(x\): \(3(7)-y=18\), so \(21-y=18\). Hence, \(-y=-3\), giving \(y=3\). If \(y=4\), the left-hand side becomes \(21-4=17\), not 18. Exam tip: After substitution, take special care with signs when moving a negative term across the equation.

Subtracting the second equation from the first gives (2y=10), so (y=5). Eliminate the equal (2x) terms.

Given \(x-y=8\) and \(y=5\), substitute \(y=5\) into the first equation: \(x-5=8\). Adding 5 to both sides gives \(x=13\). For example, \(x=12\) gives \(12-5=7\), not 8. Exam tip: after substitution, carefully use inverse operations to isolate the unknown.