Algebraic methods: Substitution method and Elimination method.

Subtracting gives (10y=50), so (y=5), and substitution gives (x=\frac{40}{11}). Fractional values are valid if both equations satisfy them.

The direct answer is option B, 15. Let the greater number be x and the smaller number be y. The given facts are x+y=23 and x-y=7. From x+y=23, x=23-y. Substitute this in the second equation: 23-y-y=7, so 23-2y=7, hence 2y=16 and y=8. Therefore x=23-8=15. Option A, 14, has sum 22 with 8, not 23. Option B, 15, gives sum 15+8=23 and difference 15-8=7, so it is correct. Option C, 16, would give the paired number 7 if the sum were 23, but the difference would be 9, not 7. Option D, 17, would give the paired number 6 and difference 11. Memory cue: for sum S and difference D, the greater number is (S+D)/2; here (23+7)/2=15. The supplied answer and explanation are correct.

Let the denominator be (y), so the numerator is (y-3). From (\frac{y-1}{y+2}=\frac{4}{5}), (y=13), so the fraction is (\frac{10}{13}).

Subtracting the first equation from the second gives (3x=21), so (x=7). Then (y=\frac{13}{3}), so the ratio is (21:13).

Adding gives (18x=54), so (x=3) and (y=\frac{17}{5}). Hence (3x-y=\frac{28}{5}); do not guess from options.

Adding the two equations gives \((4x+7y)+(8x-7y)=1+35\), so \(12x=36\). Hence, \(x=3\). Substituting \(x=3\) into \(4x+7y=1\) gives \(12+7y=1\), so \(7y=-11\) and \(y=-\frac{11}{7}\). Therefore, the correct ordered pair is \((3,-\frac{11}{7})\). Option B does not satisfy both equations when \(x=2\). Exam tip: When coefficients of one variable have opposite signs, add the equations to eliminate that variable quickly.

From the second equation, (y=6x-27). Substitute carefully; expert questions may have fractional answers.

Adding gives (15x=45), so (x=3) and (y=2). Therefore (y-x=-1); sign reversal can change the answer.

Let the length be l cm and the breadth be b cm. From the perimeter, 2(l+b)=64, so l+b=32. The second condition gives l-b=6. Adding the two equations yields 2l=38, hence l=19 cm. Therefore, option C is correct. Exam tip: In perimeter problems, first form and simplify the equation 2(l+b). Option B is incorrect because if the length were 18 cm, the breadth would be 14 cm, giving a perimeter of 64 cm—but this conflicts with the required length difference of 6 cm.

From the first equation (y=2x-9). Substitution gives (5x+4x-18=12), so (x=\frac{10}{3}); simplify carefully.

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.

Word problems become easier when the unknown quantities are represented by variables. Since one number is 8 greater than the other, the larger number can be written as the smaller number plus 8. The statement about three times one number and twice the other gives a second equation. Substituting the first relationship into that equation reduces the problem to one variable. The wording is consistent with taking the coefficient 3 for the larger number, as used in the supplied solution.

Let the smaller number be 
m y
m. Then the larger number is 
m x=y+8
m. The condition gives 
m 3x+2y=94
m. Substitute: 
m 3(y+8)+2y=94
m, so 
m 3y+24+2y=94
m and 
m 5y=70
m. Hence 
m y=14
m. The larger number is 22, and 
m 3(22)+2(14)=94
m, so option B is correct.

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.

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 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 (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 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.

Let \(u=\frac{1}{x}\) and \(v=\frac{1}{y}\). The equations become \(2u+3v=13\) and \(3u-2v=4\). Multiply the first equation by 2 and the second by 3 to get \(4u+6v=26\) and \(9u-6v=12\). Adding them gives \(13u=38\), so \(u=\frac{38}{13}\). Hence, \(\frac{1}{x}=\frac{38}{13}\). The value \(\frac{31}{13}\) is for \(v=\frac{1}{y}\), not for \(\frac{1}{x}\). Exam tip: For reciprocal terms, substitute \(u=\frac{1}{x}\) and \(v=\frac{1}{y}\) first to form linear equations.

Let u=x-1 and v=y+2. The equations become 2u+3v=25 and 4u-3v=5. Adding them eliminates v and gives 6u=30, so u=5. Substituting into the first equation gives 10+3v=25, hence v=5. Therefore, x=6 and y=3, so x+y=9; option B is correct. Exam tip: when the coefficients of one variable have opposite signs, add the equations to eliminate that variable directly.