Algebraic methods: Substitution method and Elimination method.

From the second equation \(2x-y=8\), we get \(y=2x-8\). Substituting this into the first equation gives \(4x+3(2x-8)=26\), so \(10x=50\) and hence \(x=5\). Then \(y=2(5)-8=2\). Therefore, the solution is \(x=5,\ y=2\). Option B satisfies the second equation, but in the first equation \(4(4)+3(3)=25\neq26\). Exam tip: Always verify the ordered pair in both original equations.

The first equation becomes (x+2y=5), while the second is (x+2y=7). Same left side with different right side gives no solution.

Using elimination, multiply the first equation by 3 and the second equation by 2: (6x+9y=930) and (6x+4y=680). Subtracting gives (5y=250), so (y=50). Substituting this into (3x+2y=340) gives (3x+100=340), hence (x=80). Therefore, the correct answer is ₹80. A value such as ₹70 does not satisfy both equations together. Exam tip: In elimination, first make the coefficients of one variable equal and then subtract the equations.

The direct answer is option C, 74. Let the tens digit be x and the units digit be y. The two facts become x+y=11 and x-y=3, because the tens digit is 3 greater. Add the equations: 2x=14, so x=7. Substituting in x+y=11 gives 7+y=11, hence y=4. A two-digit number with tens digit 7 and units digit 4 is 74. Option A, 65, has digit sum 11 but 6 is only 1 greater than 5, not 3. Option B, 83, has sum 11, but 8 is 5 greater than 3. Option C works because 7+4=11 and 7-4=3. Option D, 92, has sum 11, but 9-2=7. Memory cue: for digit questions, write sum and difference equations, then add them to find the tens digit.

Substitute \(x=4\) and \(y=4\) into \(kx+2y=20\): \(4k+2(4)=20\). Thus, \(4k+8=20\), so \(4k=12\) and \(k=3\). The other equation is also satisfied because \(4+4=8\). Exam tip: to find a parameter, substitute the given solution into the equation containing that parameter.

The coefficient of y in the first equation is 4, while it is -2 in the second equation. Multiplying the second equation by 2 gives 10x-4y=2. On adding it to 3x+4y=17, +4y and -4y cancel. Multiplying by 1 would leave -2y, so y would not be eliminated. Exam tip: In elimination, make the coefficients of one variable equal in magnitude and opposite in sign.

Use (y=2x-6) from the first equation. Substitution gives (7x=31), so (x=\frac{31}{7},\ y=\frac{20}{7}).

The direct answer is option C: 49/5. From x-y=1, add y to both sides to get x=y+1. Substitute into 3x+2y=25: 3(y+1)+2y=25, so 5y+3=25, 5y=22, and y=22/5. Then x=22/5+1=27/5. Therefore x+y=27/5+22/5=49/5. Option C is correct. Option A, 8, is not the calculated sum and does not come from the equations. Option B, 9, is also incorrect. Option D, 51/5, is different from the required numerator sum 27+22=49. A decimal check gives x=5.4,y=4.4, whose sum is 9.8=49/5; the original equations give 25 and 1. Keep denominators common when adding fractions, and do not confuse x+y with either variable alone.

The coefficient of \(7x\) is the same in both equations. Subtract the second equation from the first: \((7x+4y)-(7x-y)=45-15\), which gives \(5y=30\). Hence, \(y=6\). If \(y=7\), then \(5y=35\), so it cannot satisfy the result. Exam tip: Add or subtract equations to eliminate a variable when its coefficients are equal or opposites.

The governing concept is solving a pair of simultaneous linear equations by elimination. Let the father’s age be x years and the son’s age be y years. The information gives x+y=50 and x−y=30, because the father is older. Adding the two equations eliminates y: (x+y)+(x−y)=50+30, so 2x=80 and x=40. Therefore the father is 40 years old, so option A is correct. Substitution confirms the result: the son’s age is 50−40=10, and 40−10=30, exactly as required. Options B, C, and D do not satisfy both the stated sum and the stated difference simultaneously.

A pair of linear equations can be solved by eliminating one variable. Here the coefficients of y are 2 and 3, so multiply the first equation by 3 and the second by 2. This gives \(9x+6y=390\) and \(4x+6y=240\). The y terms now match, so subtraction removes them and leaves \(5x=150\), giving \(x=30\).

Substitute this value into either original equation. Using \(3x+2y=130\), we get \(3(30)+2y=130\), so \(90+2y=130\). Hence \(2y=40\) and \(y=20\). Checking in the second equation gives \(2(30)+3(20)=60+60=120\), so the result satisfies both equations. Therefore option C is correct.

From the second equation, \(3x-y=17\), we get \(y=3x-17\). Substituting this into \(2x+5y=0\) gives \(2x+5(3x-17)=0\), so \(17x=85\) and \(x=5\). Then \(y=3(5)-17=-2\). Therefore, the correct answer is \(-2\). The value \(-1\) does not satisfy both equations together. Exam tip: verify the obtained values in both original equations.

The governing concept is the consistency condition for a pair of linear equations. Multiply the first equation, 2x+3y=12, by 2. It becomes 4x+6y=24. The second equation has exactly the same left-hand side but equals 25, so the pair would require the same expression to be both 24 and 25, which is impossible. Geometrically, the equations represent distinct parallel lines: their coefficients are proportional, but their constants are not in the same proportion. Hence there is no common ordered pair (x,y), and option A is correct. Infinite solutions would require the second equation to be 4x+6y=24, while one or two solutions do not apply to this inconsistent pair.

For no solution, coefficients must be proportional while constants are not. At (k=4), the left sides are proportional but (12) and (9) are not in the same ratio.

On adding the two equations, \(-3y\) and \(+3y\) cancel: \(5x-3y+2x+3y=2+19\). Thus, \(7x=21\), so \(x=3\). If \(x=2\), then \(7x=14\), which does not equal the obtained sum \(21\). Exam tip: Add equations when a variable has equal coefficients with opposite signs.

From the second equation, \(2x-y=9\), we get \(y=2x-9\). Substituting this into the first equation gives \(4x+5(2x-9)=39\), so \(14x=84\) and \(x=6\). Then \(y=2(6)-9=3\). Hence, the solution is \(x=6,\ y=3\). For instance, option B gives \(2x-y=6\), not 9. Exam tip: always verify the obtained values in both original equations.

From the first equation, \(y=23-2x\). Substituting this into \(x+3y=19\) gives \(x+3(23-2x)=19\), so \(-5x=-50\), hence \(x=10\) and \(y=3\). Therefore, \(x-2y=10-2(3)=4\). Option \(5\) would be obtained for \(x-y\), not for the required expression. Exam tip: verify the values in both original equations after solving.

The direct answer is option C: 18. Let the larger number be x and the smaller number be y. Their sum gives x+y=29. Their difference gives x-y=7. Add the equations: 2x=36, so x=18. Then y=29-18=11, and 18-11=7, confirming the interpretation. Option C is correct because 18 is the larger number and the pair has sum 29 and difference 7. Option A, 15, would pair with 14 to make sum 29, but the difference is only 1. Option B, 16, would pair with 13, giving difference 3. Option D, 20, would pair with 9, giving sum 29 but difference 11. The formula shortcut is useful: larger number = (sum+difference)/2 = (29+7)/2 = 18. Memory cue: add sum and difference to get twice the larger number.

Substitute \(x=2y+1\) into the second equation: \(5(2y+1)-3y=33\). Thus, \(10y+5-3y=33\), so \(7y=28\) and \(y=4\). Now \(x=2(4)+1=9\). Therefore, the correct solution is \(x=9,\ y=4\). The close distractor \(x=7,\ y=3\) satisfies the first equation but gives \(5x-3y=26\), not 33. Exam tip: verify the ordered pair in both original equations.

Adding both equations gives (11x=55), so (x=5). From the second equation (15+5y=26), hence (y=\frac{11}{5}).

The given solution \(x=5, y=4\) must satisfy \(3x+my=23\). Substituting gives \(3(5)+m(4)=23\), or \(15+4m=23\). Hence \(4m=8\) and \(m=2\). If \(m=3\), the left-hand side becomes \(27\), not \(23\). Exam tip: To find a parameter from a given solution, substitute the solution into the relevant equation.

Dividing the first equation by 3 gives \(2x-3y=4\), whereas the second equation is \(2x-3y=5\). Their left-hand sides are identical but their constants differ, so they represent distinct parallel lines and have no common solution. Infinitely many solutions would occur only if both equations represented the same line. Exam tip: identical variable expressions with different constants indicate an inconsistent pair.

This is a word problem involving a pair of simultaneous linear equations. Let the larger number be x and the smaller number be y. Then x+y=64 and x−y=18. Adding these equations gives 2x=82, so x=41. Subtracting the second equation from the first gives 2y=46, or directly y=64−41=23. Therefore the smaller number is 23, so option C is correct. A quick check confirms both conditions: 41+23=64 and 41−23=18. Options A, B, and D fail at least one condition; for example, 24 would require the larger number to be 40, whose difference from 24 is only 16.

Multiplying the first equation by (6) gives (3x+2y=30). Using (x=y+3) gives (y=\frac{21}{5}) and (x=\frac{36}{5}).

Multiplying the first equation by (6) gives (2x+3y=42). Using (x=y+3) gives (5y=36), so (y=\frac{36}{5}).