Algebraic methods: Substitution method and Elimination method.

Subtracting the first equation from the second gives (10y=40), so (y=4). When (x)-coefficients are equal, subtract directly.

From the first equation, (x=22-4y). Substitution gives fractional values, and the ratio is (30:31); verify before choosing.

Adding the two equations eliminates \(y\): \(15x+7y+5x-7y=1+39\), so \(20x=40\) and hence \(x=2\). The nearby choices 1 and 3 would give \(20x=20\) and \(20x=60\), respectively, so they cannot satisfy the added equation. Exam tip: directly add or subtract equations when the coefficient of one variable has opposite signs.

Let the smaller number be (y) and the larger be (x=y+8). Substitution in (3x+2y=94) gives (5y+24=94), so (y=14).

Adding the two equations eliminates \(y\): \(18x=72\), so \(x=4\). Substituting into \(6x+11y=9\) gives \(24+11y=9\), hence \(y=-\frac{15}{11}\). Therefore, \(x-y=4-\left(-\frac{15}{11}\right)=\frac{59}{11}\). \(\frac{15}{11}\) is only the magnitude of \(y\), not the value of \(x-y\). Exam tip: add equations directly when a variable has opposite coefficients.

Adding gives (12x=96), so (x=8). Substituting in the first equation gives (32-9y=31), so (y=\frac{1}{9}).

Multiplying the first equation by (2) helps eliminate (y). After finding (x), substitute back carefully before evaluating (4x+y).

Let the tens digit be x and the units digit be y. Then x + y = 11. The original number is 10x + y, while the reversed number is 10y + x. Since the number decreases by 27 on reversal, (10x + y) − (10y + x) = 27, giving 9(x − y) = 27 and hence x − y = 3. Solving x + y = 11 and x − y = 3 gives x = 7 and y = 4. Therefore, the original number is 74. Option 83 has a digit sum of 11, but reversing it causes a decrease of 45, not 27. Exam tip: Represent a two-digit number as 10x + y before forming the equations.

Multiplying by (6) gives (3x+2y=42) and (2x+3y=48). Adding gives (5x+5y=90), so (x+y=18).

Multiplying the first equation by 20 gives \(5x+4y=120\), and multiplying the second by 20 gives \(4x-5y=20\). Multiply these equations by 5 and 4 respectively, then add them: \(41x=820\). Hence, \(x=20\). The nearby option \(18\) is incorrect because it does not satisfy both original equations. Exam tip: For linear equations containing fractions, first multiply by the LCM to remove the fractions.

Multiplying both equations by 10 gives 2x+3y=27 and 4x−y=11. From the second equation, y=4x−11. Substituting this into the first equation gives 2x+3(4x−11)=27, so 14x=60 and x=30/7. Therefore, y=4(30/7)−11=43/7. Hence, option A is correct. Exam tip: remove decimals carefully and verify the value in both original equations; an integer-looking distractor such as 5 or 6 is not necessarily correct.

Multiplying the first equation by 2 and the second by 4 removes the decimals, giving x-2y=3 and 4x+y=28. Multiply the first of these by 2 and add it to the second equation: 9x=59, so x=59/9. Substituting this into x-2y=3 gives y=16/9. Therefore, the solution is (59/9, 16/9). For example, option B satisfies the first equation, but in the second equation it gives 6+0.25(3/2)=6.375, not 7. Exam tip: When coefficients are decimals, first convert the equations to integer coefficients and then verify the final pair in both original equations.

Putting (x=4) in the second equation gives (y=\frac{5}{2}). Then (4a+5=17), so (a=3).

Putting (y=4) in the second equation gives (x=\frac{7}{2}). Then (14+4k=26), so (k=3).

Putting (y=5) in the second equation gives (x=3). Then (3p+5=14), so (p=3); match the option carefully.

Since the solution has \\(x=5\\), substitute it into the first equation: \\(2(5)+3y=13\\), so \\(10+3y=13\\) and \\(y=1\\). Substituting \\(x=5\\) and \\(y=1\\) into the second equation gives \\(5m-3=17\\), hence \\(5m=20\\) and \\(m=4\\). Therefore, option B is correct. Exam tip: When a solution coordinate is given, substitute the coordinates into both equations to determine the unknown parameter.

Let boat speed be (b) and stream speed be (s), so (b+s=15), (b-s=10). Adding gives (2b=25), so (b=12.5).

Let Ram's age be (r) and Shyam's be (s), so (r-s=4) and (r+s+10=44). Solving gives (r=19).

Let correct answers be (c) and wrong answers be (w), so (c+w=20) and (4c-w=55). Adding gives (5c=75), so (c=15).

Let the costlier ticket cost x rupees and the cheaper ticket cost y rupees. The equations are x+y=180 and x-y=40. Adding them eliminates y and gives 2x=220, so x=110 rupees. Therefore, option C is correct. In option B, the two prices would differ by 30 rupees, while in option D they would differ by 60 rupees; neither satisfies the condition. Exam tip: When the sum and difference are known, the larger quantity is (sum + difference) / 2.