Algebraic methods: Substitution method and Elimination method.

The intersection point must satisfy both equations. Subtracting the second equation from the first gives \(4y=24\), so \(y=6\). Substituting this into \(4x-y=10\) gives \(4x-6=10\), hence \(x=4\). Therefore, the intersection point is \((4,6)\). Option B merely reverses the coordinates and does not satisfy both equations. Exam tip: always substitute the obtained point into both original equations to verify it.

The coordinates of a point lying on a line must satisfy the equation of that line. Substituting \(\left(4,5\right)\) into the second line, \(kx+3y=23\), gives \(4k+3(5)=23\), or \(4k+15=23\). Thus, \(4k=8\) and \(k=2\). The first equation is also verified because \(4+5=9\). Exam tip: To find an unknown parameter when a point lies on a line, substitute the point’s coordinates directly into the line equation.

In elimination, adding equations cancels coefficients with equal magnitude and opposite signs, such as \(3x\) and \(-3x\), since their sum is \(0\). Equal constants alone do not eliminate a variable. Exam tip: first align one variable’s coefficients.

To eliminate y, multiply the first equation by 2 and the second equation by 3. This gives 4x+6y=36 and 9x-6y=15. Adding them produces 13x=51, so x=51/13, not x=3. Substitution then gives 2(51/13)+3y=18, hence 102/13+3y=234/13, so 3y=132/13 and y=44/13. Therefore none of the supplied original options is correct. To preserve one unambiguous answer, the corrected options should include (x=51/13, y=44/13) as option A, while the other three options remain distinct distractors. Direct checking confirms 2(51/13)+3(44/13)=234/13=18 and 3(51/13)-2(44/13)=65/13=5. The earlier integer key (3,4) is invalid because it gives 18 in the first equation but only 1 in the second.

Solve the first equation for y: y=22-4x. Substitute this into the second equation: 3x-2(22-4x)=1. Expanding gives 3x-44+8x=1, so 11x=45 and x=45/11. Now calculate y=22-4(45/11)=242/11-180/11=62/11. Hence the correct value is y=62/11, option A. A direct check confirms that 4(45/11)+62/11=242/11=22 and 3(45/11)-2(62/11)=135/11-124/11=1. The earlier option C, y=6, is not valid: the first equation would then give x=4, but the second equation would become 12-12=0 rather than 1. The fractional result is completely acceptable because the coefficients and constants do not force integer solutions.

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

The governing technique is elimination after removing decimal coefficients. Multiply both equations by 10 to obtain 4x + 3y = 31 and 2x − 5y = −11. Multiply the second transformed equation by 2: 4x − 10y = −22. Subtract the first equation from it: (4x − 10y) − (4x + 3y) = −22 − 31. Thus −13y = −53, so y = 53/13. Therefore option D is correct. The decimal scaling is valid because multiplying an entire equation by a nonzero number preserves its solution set. For verification, y = 53/13 gives x = 58/13, and substitution satisfies both original decimal equations. The other fractions do not.

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}).

The direct answer is option A, 84. Let x be the tens digit and y be the units digit. The statements give x+y=12 and x-y=4. Adding them gives 2x=16, so x=8. Then 8+y=12, so y=4. The number formed is 84. Option A is correct because its digit sum is 8+4=12 and its tens digit exceeds its units digit by 8-4=4. Option B, 75, has sum 12, but its difference is 2. Option C, 93, has sum 12, but its difference is 6. Option D, 66, has sum 12, but the tens digit is not 4 greater; the difference is zero. The order matters: the tens digit must be the larger digit. Exam cue: add sum and difference equations to get twice the tens digit.

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.

The direct answer is option D: \(y=\frac{62}{13}\). We need to solve both equations together. The second equation is \(2x-5y=-6\). Move \(-5y\) to the other side: \(2x=5y-6\), so \(x=\frac{5y-6}{2}\). Substitute this into \(4x+3y=50\): \(4\left(\frac{5y-6}{2}\right)+3y=50\). This becomes \(2(5y-6)+3y=50\), so \(10y-12+3y=50\), hence \(13y=62\), and \(y=\frac{62}{13}\). Option A, \(\frac{52}{13}\), does not satisfy the equations. Option B, \(\frac{56}{13}\), is also not the value obtained after simplification. Option C, \(\frac{58}{13}\), is another nearby but incorrect numerator. Option D is correct; substituting it gives the corresponding value of \(x\) and both original equations are satisfied. A useful check is to substitute the final value back into at least one equation. Keep signs carefully when moving a term across the equal sign.

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}).

Multiply the first equation by (2) and add the second. (x=\frac{58}{13}) and (y=\frac{47}{13}), so (x+y=\frac{105}{13}).

Clear denominators carefully. Multiplying the first equation by 12 gives 3x+2y=60. Multiplying the second by 6 gives 3x−2y=12. Adding these equations yields 6x=72, so x=12. Substituting into 3x+2y=60 gives 36+2y=60, hence y=12. Therefore option C is correct; the other ordered pairs fail at least one original equation.