Algebraic methods: Substitution method and Elimination method.

Adding (x+y=14) and (x-y=4) gives (2x=18), so (x=9) and (y=5). In word problems, form equations first.

From the second equation, 4x-y=5 ⇒ y=4x-5. Substituting this into the first equation gives 2x+3(4x-5)=13. Thus, 14x-15=13, so 14x=28 and x=2. Now, y=4(2)-5=3. Therefore, the solution is (2, 3). Option (3, 2) does not satisfy the second equation because 4(3)-2=10, not 5. Exam tip: Always verify the obtained pair in both original equations.

Subtracting gives (2x=12), so (x=6) and (y=2). When (y)-coefficients are equal, subtraction is easy.

From (x-y=2), (x=y+2), so (\frac{y+2}{2}+y=7) and (y=4). In fraction questions, try to clear denominators first.

Putting (x=10-y) gives (3(10-y)+2y=24), so (y=6) and (x=4). A sum equation makes substitution easier.

Subtracting the first equation from the second gives (x=3), then (2x+5y=26) gives (y=4). Substitute back after elimination.

Subtract the first equation from the second: \(3x+2y-(x+2y)=25-11\). Thus, \(2x=14\), so \(x=7\). Substituting \(x=7\) into \(x+2y=11\) gives \(2y=4\), hence \(y=2\). Therefore, the solution is \((7, 2)\). Option B, \((5,3)\), does not satisfy the second equation. Exam tip: when a variable has equal coefficients in both equations, subtract the equations to eliminate it directly.

Putting (y=2x+1) gives (3x+1=10), so (x=3) and (y=7). If (y=) form is given, substitute directly.

From (3y-1+y=11), (4y=12), so (y=3) and (x=8). Keeping the expression in brackets is safer.

Putting (x=13-y) gives (2(13-y)+3y=34), so (y=8) and (x=5). Keep the formed equations organized.

From the second equation, \(2x-y=9\), we get \(y=2x-9\). Substituting this into the first equation gives \(4x+2(2x-9)=30\), so \(8x=48\) and \(x=6\). Then \(y=2(6)-9=3\). Hence, the solution is \((6,3)\). The close option \((5,5)\) does not satisfy the second equation because \(2(5)-5=5\), not 9. Exam tip: substitute the final values into both original equations to verify the solution.

Adding both equations gives (14x=98), so (x=7) and (y=4). Add opposite (y) terms to eliminate them.

The second equation is (2) times the first, so both represent the same line. Such a pair has infinitely many solutions.

Multiplying the first equation by (2) gives the same left side but constant (16). Since it conflicts with (20), there is no solution.

Subtracting the first from the second gives (3x=12), so (x=4) and (y=4). After elimination, substitute into any original equation.

From the first equation, \(x=2y+1\). Substituting this into \(3x-y=18\) gives \(3(2y+1)-y=18\). Thus, \(6y+3-y=18\), so \(5y=15\) and \(y=3\). Hence, \(x=2(3)+1=7\), making \((7,3)\) the correct pair. Option \((5,2)\) satisfies the first equation but not the second one. Exam tip: verify the obtained values in both original equations.

Putting (3x-5) in place of (y) gives (5x-5=20), so (x=5) and (y=10). Replace the entire expression carefully.

If the father's age is (x) and the son's age is (y), then (x+y=50) and (x=4y). This gives (5y=50), so the ages are (40) and (10).

Dividing (50x+30y=500) by (10) gives (5x+3y=50). Solving with (x+y=12) gives (x=7) and (y=5).

On adding the two equations, the y-terms cancel: (x+2y)+(3x-2y)=13+23. Thus, 4x=36 and x=9. Substituting x=9 into x+2y=13 gives 9+2y=13, so y=2. Therefore, the solution is (9,2). Option (8,3) gives 14 in the first equation, so it is not correct. Exam tip: In elimination, add or subtract equations to cancel terms with opposite coefficients.