Algebraic methods: Substitution method and Elimination method.

To make coefficients proportional, (4:8=a:10) must hold. This gives (a=5), while constants are not in the same ratio.

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.

The direct answer is option C: \(x+2y=\frac{54}{5}\). Add the equations \(9x-5y=42\) and \(3x+5y=30\). The \(y\)-terms cancel: \(12x=72\), so \(x=6\). Substitute this into \(3x+5y=30\): \(18+5y=30\), hence \(5y=12\) and \(y=\frac{12}{5}\). Now calculate the requested expression, not just one variable: \(x+2y=6+2\left(\frac{12}{5}\right)=\frac{30}{5}+\frac{24}{5}=\frac{54}{5}\). Option A, \(\frac{44}{5}\), is too small and does not result from the solved values. Option B, \(\frac{49}{5}\), is also incorrect. Option C matches the exact calculation. Option D, \(\frac{59}{5}\), is too large. A useful habit is to add equations when opposite coefficients, such as \(-5y\) and \(+5y\), appear; then substitute and finally evaluate the expression asked. Do not stop after finding \(x\), because the question asks for \(x+2y\).

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.

Two linear equations have no solution when they represent distinct parallel lines. For equations \(a_1x+b_1y+c_1=0\) and \(a_2x+b_2y+c_2=0\), this occurs when the coefficients of \(x\) and \(y\) are proportional but the constants are not proportional. Such lines have the same slope but different positions.

Divide the first equation by 3: \(3x+5y=15\). If \(k=3\), the second equation becomes \(3x+5y=18\). The left sides are identical, but the right sides differ, so one pair cannot satisfy both equations. The lines are distinct and parallel, and there is no common solution. Therefore option B is correct.

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.

Add the equations: \((6x+5y)+(4x-5y)=43+7\Rightarrow 10x=50\Rightarrow x=5\).

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

Adding gives (7x=35), so (x=5) and (y=3). Therefore (x+y=8); substitute back before choosing the option.

From the second equation, (y=2x-1). Substitute carefully and verify with both equations before using (xy).

Adding the two equations eliminates y: 12x = 36, so x = 3. Substituting x = 3 into 3x + 4y = 25 gives 9 + 4y = 25, hence y = 4. Therefore, x - y = 3 - 4 = -1. In the exam, check the sign carefully; 1 would be y - x, not x - y.

Subtracting the second equation from the first gives (12y=36), so (y=3). When coefficients are equal, subtraction is efficient.

Adding gives (8x=24), so (x=3) and (y=\frac{9}{2}). Thus (2x+y=\frac{21}{2}); solve fully before comparing options.

Substituting (x=2y+1) gives (3(2y+1)-y=17). Always simplify brackets carefully and verify the option.

Adding the two equations gives \((8x-3y)+(2x+3y)=13+17\), so \(10x=30\). Hence, \(x=3\). Substituting this into \(2x+3y=17\) gives \(6+3y=17\), and therefore \(y=\frac{11}{3}\). Option B does not satisfy the first equation when \(x=4\) and \(y=3\) are substituted. Exam tip: add equations first when a variable has opposite coefficients, so that variable is eliminated directly.