Algebraic methods: Substitution method and Elimination method.

Multiply both equations by 10 to remove decimals: \(5x+4y=61\) and \(3x-2y=17\). Multiply the second equation by 2 and add it to the first equation: \(11x=95\), so \(x=\frac{95}{11}\). Substituting into \(3x-2y=17\) gives \(y=\frac{49}{11}\). Therefore, \(x+y=\frac{95}{11}+\frac{49}{11}=\frac{144}{11}\). A value such as \(\frac{139}{11}\) results from an incorrect addition or calculation. Exam tip: for linear equations with decimals, first clear the decimals to make elimination easier.

Adding the two equations eliminates y because −3y and +3y cancel: 12x = 84, so x = 7. Substituting x = 7 into 2x + 3y = 23 gives 14 + 3y = 23, hence 3y = 9 and y = 3. Therefore, option B is correct. Verification gives 10(7) − 3(3) = 61 and 2(7) + 3(3) = 23. Exam tip: When the coefficients of one variable are equal in magnitude and have opposite signs, add the equations to eliminate that variable directly.

Multiplying the first equation by 6 gives \(x+2y=36\), and multiplying the second equation by 4 gives \(2x-y=20\). Multiply the second equation by 2: \(4x-2y=40\). Adding it to the first equation gives \(5x=76\), so \(x=\frac{76}{5}\). A value such as \(\frac{72}{5}\) does not satisfy both equations on substitution. Exam tip: In linear equations involving fractions, first clear the denominators to simplify elimination.

Let the length be \(l\) cm and the breadth be \(b\) cm. From the perimeter, \(2(l+b)=112\), so \(l+b=56\). Also, \(l-b=16\). Adding the two equations gives \(2l=72\), hence \(l=36\) and \(b=20\). Therefore, the area is \(l\times b=36\times20=720\ \text{cm}^2\). A choice such as \(700\ \text{cm}^2\) is not the product of the correct length and breadth. Exam tip: In rectangle-perimeter questions, first convert \(2(l+b)\) carefully into \(l+b\).

Substitute the given solution \(x=6,\ y=1\) into the first equation \(px+5y=43\). This gives \(6p+5=43\), so \(6p=38\) and hence \(p=\frac{38}{6}=\frac{19}{3}\). The second equation also checks out because \(3(6)-1=17\). If \(p=\frac{20}{3}\), the left-hand side of the first equation becomes \(45\), not \(43\). Exam tip: to find a parameter, substitute the given values of \(x\) and \(y\) into the equation containing that parameter.

Using the elimination method, add the two equations: \((9x+2y)+(3x-2y)=10+14\), which gives \(12x=24\), so \(x=2\). Substituting \(x=2\) into \(9x+2y=10\) gives \(18+2y=10\), hence \(2y=-8\) and \(y=-4\). Therefore, option B is correct. The nearby distractor \(y=-3\) is incorrect because it makes the first equation equal to \(18-6=12\), not 10. Exam tip: when coefficients of one variable have opposite signs, add the equations to eliminate that variable directly.

From the first equation, x=31-y. Substituting this into the second equation gives 4(31-y)-3y=19, or 124-7y=19. Thus y=15 and x=16. Therefore, 2x-y=2(16)-15=17, so option C is correct. Exam tip: Verify the values of x and y in both original equations before selecting the answer.

Multiplying the given equations by 5 and 4 gives \(x+4y=50\) and \(3x-y=28\). From the second equation, \(y=3x-28\). Substituting this into the first equation gives \(x+12x-112=50\), so \(13x=162\). Hence, \(x=\frac{162}{13}\) and \(y=\frac{122}{13}\). Therefore, \(x-y=\frac{40}{13}\). Exam tip: For linear equations containing fractions, first clear the denominators to obtain simpler integer-coefficient equations.

Twice the first equation is (8x+14y=62), but the second is (8x+14y=65). Therefore there is no solution.

Multiply the second equation by (3) and add it to the first. Solving gives (y=\frac{119}{15}).

Multiply the first equation by (2) and the second by (3), then subtract. This gives (b=\frac{412}{11}).

Adding both equations gives (12x=72), so (x=6). Then (y=\frac{12}{5}), hence (x+2y=\frac{54}{5}).

From the first equation, \(x=5y-8\). Substituting this into \(4x+3y=61\) gives \(4(5y-8)+3y=61\). Thus, \(20y-32+3y=61\), so \(23y=93\). Therefore, \(y=\frac{93}{23}\), which is option C. Taking \(\frac{88}{23}\) would give \(23y=88\), which does not satisfy the given equation. Exam tip: after substitution, multiply the number outside the bracket by every term inside it.

The first equation becomes (3x+5y=15). At (k=3), the second becomes (3x+5y=18), so there is no solution.

Adding the equations gives (8x=40), so (x=5), then (y=2). In exams, add directly when coefficients are opposite.

From the second equation, put (y=19-4x), giving (x=4) and (y=3). In exams, isolate the easier variable first.

Adding the two equations eliminates y because the terms 4y and -4y cancel: 10x = 20, so x = 2. Substituting x = 2 into 7x + 4y = 2 gives 14 + 4y = 2, hence y = -3. Therefore, x - y = 2 - (-3) = 5. Exam tip: Add equations directly when one variable has equal and opposite coefficients.

This question needs careful substitution after elimination; careless cancellation gives a wrong value. Check each obtained value in both equations before marking.

Adding the two equations gives (10x=50), so (x=5). In exams, eliminate terms with opposite coefficients first.

Substitute (x=11-y) carefully in the first equation; incorrect simplification changes the answer. Always check the obtained values in both equations.