Algebraic methods: Substitution method and Elimination method.

Adding \(3x-5y=-1\) and \(2x+5y=21\) eliminates the \(y\)-terms: \(5x=20\). Hence, \(x=4\), so option B is correct. The value of \(x\) cannot be obtained directly from either equation alone because each also contains \(y\). Exam tip: add two equations when one variable has equal and opposite coefficients.

Use (y=3x-11) from the second equation. Substitution gives (x=\frac{27}{5},\ y=\frac{16}{5}), so (2x+y=14).

Clear the denominators using the LCM first. The solution is (x=\frac{138}{13},\ y=\frac{66}{13}).

Multiply both equations by (10) to remove decimals. Then elimination gives (y=\frac{53}{13}).

Multiply the first equation by (2) and eliminate (y). (x=\frac{64}{11}) and (y=\frac{58}{11}), so (x-y=\frac{6}{11}).

After clearing denominators, the equations become (8x+3y=60) and (3x-2y=6). Elimination gives (x=\frac{138}{25},\ y=\frac{132}{25}).

The given solution \(x=5, y=3\) must satisfy \(kx+3y=25\). Thus, \(5k+3(3)=25\), so \(5k+9=25\). Hence \(5k=16\) and \(k=\frac{16}{5}\). For instance, using \(k=\frac{14}{5}\) makes the left-hand side \(23\), not \(25\). Exam tip: when a solution is given, substitute its values directly into the original equation.

For no solution, coefficients of (x) and (y) are proportional but constants are not. Since (4:2=2), (a=6) is correct.

Use (y=3x-10) from the second equation. Substitution gives (11x=61), so (x=\frac{61}{11},\ y=\frac{73}{11}).

Let the tens digit be (x) and units digit be (y). From (x+y=12) and (x-y=4), (x=8,\ y=4), so the number is (84).

Let the present ages of the father and son be \(x\) years and \(y\) years respectively. Then \(x+y=50\). The condition after 5 years gives \(x+5=2(y+5)\), which simplifies to \(x-2y=5\). Solving this with \(x+y=50\) gives \(y=15\) and \(x=35\). Therefore, the father's present age is 35 years. If the father were 40 years old, the son would be 10; after 5 years, 45 would not be twice 15. Exam tip: in age problems, add the stated number of years to both ages for a future condition.

Adding the two equations eliminates \(-5y\) and \(+5y\): \(7x-5y+2x+5y=4+41\). Thus, \(9x=45\), so \(x=5\). If \(x=4\), then \(9x=36\), not 45. Exam tip: add equations directly when the coefficients of one variable are opposites.

Use (x=\frac{5y-6}{2}) from the second equation. Substitution gives (13y=62), so (y=\frac{62}{13}).

Multiplying the first equation by 2 gives \(x+y=18\), and multiplying the second equation by 3 gives \(x-y=6\). Adding these equations gives \(2x=24\), so \(x=12\). Substituting \(x=12\) into \(x+y=18\) gives \(y=6\). In option B, \(x-y=2\), so it does not satisfy the second equation. Exam tip: For linear equations involving fractions, first clear the denominators and then use elimination.

Using elimination, multiply the first equation by 5 and the second equation by 2: \(15p+10q=925\) and \(4p+10q=430\). Subtracting gives \(11p=495\), so \(p=45\). Substituting in \(3p+2q=185\), we get \(135+2q=185\), hence \(2q=50\) and \(q=25\). Therefore, the correct answer is ₹25. Substituting ₹20 does not satisfy both equations together. Exam tip: in elimination, make the coefficients of one variable equal before subtracting the equations.

The given solution \(x=7,\ y=4\) must satisfy \(2x+my=34\). Substituting gives \(2(7)+m(4)=34\), or \(14+4m=34\). Hence, \(4m=20\) and \(m=5\). If \(m=4\), the left-hand side becomes \(30\), not \(34\). Exam tip: To find an unknown parameter from a given solution, substitute the values directly into the equation.

The first equation becomes (2x+3y=6). When (c=8), the left side is the same but the right side is different, so there is no solution.

From the second equation, \(x=y+6\). Substituting this into the first equation gives \(\frac{y+6}{3}+\frac{y}{5}=4\). Multiplying by 15, \(5y+30+3y=60\), so \(8y=30\) and \(y=\frac{15}{4}\). Hence, \(x=\frac{15}{4}+6=\frac{39}{4}\). If \(x=\frac{37}{4}\), then \(x-y=6\) gives \(y=\frac{13}{4}\), which does not satisfy the first equation. Exam tip: after substitution in equations containing fractions, multiply by the LCM to clear the denominators.

Multiplying both equations by 10 gives \(5x-2y=19\) and \(3x+4y=26\). Multiplying the first equation by 2 gives \(10x-4y=38\). Adding it to the second equation gives \(13x=64\), so \(x=\frac{64}{13}\). Substituting this into \(5x-2y=19\) gives \(y=\frac{73}{26}\). A nearby value such as \(\frac{68}{26}\) can result from an error in elimination or substitution. Exam tip: For linear equations with decimals, first multiply by 10 or 100 to remove the decimals.

Let (x+y=s) and (x-y=d). Solving gives (s=\frac{57}{7}) and (d=\frac{23}{7}), so (x=\frac{40}{7}).