Algebraic methods: Substitution method and Elimination method.

Substitute \(x=2y+1\) into the second equation: \(5(2y+1)-3y=33\). Thus, \(10y+5-3y=33\), so \(7y=28\) and \(y=4\). Now \(x=2(4)+1=9\). Therefore, the correct solution is \(x=9,\ y=4\). The close distractor \(x=7,\ y=3\) satisfies the first equation but gives \(5x-3y=26\), not 33. Exam tip: verify the ordered pair in both original equations.

Adding both equations gives (11x=55), so (x=5). From the second equation (15+5y=26), hence (y=\frac{11}{5}).

The given solution \(x=5, y=4\) must satisfy \(3x+my=23\). Substituting gives \(3(5)+m(4)=23\), or \(15+4m=23\). Hence \(4m=8\) and \(m=2\). If \(m=3\), the left-hand side becomes \(27\), not \(23\). Exam tip: To find a parameter from a given solution, substitute the solution into the relevant equation.

The first equation is (5) times the second. Therefore both are the same equation and have infinitely many solutions.

Dividing the first equation by 3 gives \(2x-3y=4\), whereas the second equation is \(2x-3y=5\). Their left-hand sides are identical but their constants differ, so they represent distinct parallel lines and have no common solution. Infinitely many solutions would occur only if both equations represented the same line. Exam tip: identical variable expressions with different constants indicate an inconsistent pair.

Let the larger number be (x) and the smaller be (y). From (x+y=64) and (x-y=18), (2y=46), so (y=23).

Multiplying the first equation by (6) gives (3x+2y=30). Using (x=y+3) gives (y=\frac{21}{5}) and (x=\frac{36}{5}).

Multiplying the first equation by (6) gives (2x+3y=42). Using (x=y+3) gives (5y=36), so (y=\frac{36}{5}).

From the second equation, \(2x+y=13\), we get \(y=13-2x\). Substituting this into the first equation gives \(4x-7(13-2x)=-19\), so \(18x=72\). Hence, \(x=4\). If \(x=3\), the two equations do not give the same value of \(y\). Exam tip: while substituting, distribute the negative sign in \(-7(13-2x)\) carefully.

From the first equation, \(y=22-3x\). Substituting this into \(x+2y=19\) gives \(x+2(22-3x)=19\), so \(-5x=-25\) and hence \(x=5\). Then \(y=7\), giving \(x-y=5-7=-2\). The distractor \(-1\) can result from an error while subtracting the values. Exam tip: find and verify both variable values before evaluating the required expression.

Let the length be \(l\) and the breadth be \(b\). The statements give \(l+b=28\) and \(l-b=6\). Subtracting the second equation from the first gives \(2b=22\), so \(b=11\). Hence, 11 is the correct answer. If the breadth were 12, the length would be 18 and their sum would be 30, not 28. Exam tip: For sum-and-difference word problems, first form separate equations for the sum and the difference.

Adding the two equations eliminates \(y\): \(9x+3x=37+11\), so \(12x=48\). Hence, \(x=4\). If \(x=3\), then \(12x=36\), so it cannot satisfy the resulting equation. Exam tip: add equations when one variable has equal coefficients with opposite signs.

For infinitely many solutions, both equations must be proportional. The ratio of (4) and (2) is (2), so (a=6).

Multiplying \(x+y=5\) by 3 gives \(3x+3y=15\). On adding it to \(2x-3y=7\), the terms \(+3y\) and \(-3y\) cancel, so \(y\) is eliminated. Multiplying \(x-y=5\) by 3 gives \(-3y\), which would not cancel the given \(-3y\) on addition. Exam tip: for elimination by addition, the selected variable must have equal coefficients with opposite signs.

From \(x-y=3\), we get \(x=y+3\). Substituting this into \(3x+2y=19\) gives \(3(y+3)+2y=19\), so \(5y=10\) and \(y=2\). Hence, \(x=5\), and the ordered pair is \((5,2)\). Although \((6,3)\) satisfies \(x-y=3\), it gives \(3x+2y=24\ne19\). In an exam, verify the final pair in both equations.

Subtract \(5x-y=14\) from \(5x+4y=44\). This gives \(5y=30\), so \(y=6\). The nearby option \(y=4\) does not satisfy both equations simultaneously. Exam tip: when the coefficients of one variable are already equal, subtract the equations to eliminate that variable quickly.

From the first equation, y=14-2x. Substituting this into x+2y=16 gives x+2(14-2x)=16, so -3x=-12 and x=4. Hence y=6, and xy=4×6=24. A value such as 22 can result from an error while handling the coefficient 2 during substitution. Exam tip: After finding x and y, verify them in either original equation.

Substitute \(x=3\) and \(y=2\) into \(2x+ky=16\): \(2(3)+k(2)=16\). Thus, \(6+2k=16\), so \(2k=10\) and \(k=5\). For example, \(k=4\) gives a left-hand side of \(14\), not \(16\). Exam tip: while substituting an ordered pair, place the values of \(x\) and \(y\) in their correct terms.

Twice the first equation is (6x+8y=20), but the second is (6x+8y=21). Therefore there is no solution.

The first equation is (7) times the second. Therefore both are identical equations and have infinitely many solutions.