Algebraic methods: Substitution method and Elimination method.

The given equation is \(x+y=20\). Substituting \(y=8\) gives \(x+8=20\). Subtracting 8 from both sides gives \(x=12\). If \(x=10\), the sum would be 18, so it is incorrect. Exam tip: Substitute the given value into the equation and simplify to find the unknown.

Adding both equations gives (6x=36), so (x=6). It is easy to eliminate opposite (y) terms by addition.

Substituting (x=14-3y) gives (14-2y=8), so (y=3). After substitution, keep variable terms on one side.

Subtracting the second equation from the first gives (3x=18), so (x=6). Subtract to remove equal (-y) terms.

Substitute \(y=2x\) into \(2x+y=16\): \(2x+2x=16\), so \(4x=16\). Hence \(x=4\) and \(y=2\times4=8\). For example, \(x=3,\ y=6\) gives \(2x+y=12\), not 16. Exam tip: verify the obtained values in both original equations.

From the second equation, \(y=6\). Substituting it into \(x+3y=23\) gives \(x+3(6)=23\), or \(x+18=23\). Hence, \(x=5\). The close distractor \(x=6\) is incorrect because it makes the left-hand side \(24\), not \(23\). Exam tip: Verify the obtained value by substituting it back into the original equation.

Substituting (y=x+2) gives (4x+2=18), so (x=4) and (y=6). Find (x) first, then substitute for (y).

Subtracting the second equation from the first gives (2x=12), so (x=6). Use elimination immediately on equal (y) terms.

From the second equation (x=y+4); substituting gives (5y+4=19), so (y=3) and (x=7). Choose the simpler equation for substitution.

Subtracting the second equation from the first gives (5x=30), so (x=6) and (y=5). After finding one variable, find the other immediately.

Putting (x=4) gives (12-2y=4), so (y=4). Check signs while solving a term with a negative coefficient.

Substituting (x=3y+2) gives (4y+2=18), so (y=4) and (x=14). Put the given relation directly into the main equation.

Subtracting the second equation from the first gives (4y=16), so (y=4). Subtract to remove equal (2x).

Given y=8, substitute it into x+2y=21: x+2(8)=21, so x+16=21. Subtracting 16 from both sides gives x=5. Option 4 may seem close, but substituting it gives x+2y=20, not 21. Exam tip: after substitution, multiply first and then isolate the unknown term.

The first equation becomes (2x+y=19), and substituting (y=x-5) gives (3x-5=19). Simplifying the equation first saves time.

From the first equation (y=15-3x); substituting gives (x+45-9x=13), so (x=4) and (y=3). Isolate the simpler variable in substitution.

Subtracting the equations gives (x-y=-2), so (x=y-2); substituting gives (y=6) and (x=4). Put the relation obtained by elimination into an original equation.

From (x-y=1), (y=x-1), so (2x+x-1=11) and the solution is ( (4,3) ). In exams, isolate the easier variable first.

From the second equation, \(x-y=2\), so \(x=y+2\). Substituting this into \(2x+3y=19\) gives \(2(y+2)+3y=19\), or \(5y=15\). Hence, \(y=3\), and then \(x=3+2=5\). Therefore, the solution is \((5,3)\). For example, \((6,2)\) gives \(x-y=4\), so it is not correct. Exam tip: verify the ordered pair in both original equations.

Subtracting the second equation from the first gives (3y=12), so (y=4) and (x=5). Equal (x)-coefficients make elimination faster.