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.

Let the tens digit be (x) and units digit be (y). From (x+y=11) and (x-y=3), (x=7,\ y=4), so the number is (74).

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}).

Using (x=y+1) gives (5y+3=25), so (y=\frac{22}{5}) and (x=\frac{27}{5}). Hence (x+y=\frac{49}{5}).

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.

Let the father’s age be (x) and the son’s age be (y). Adding (x+y=50) and (x-y=30) gives (x=40).

The governing concept is solving a pair of simultaneous linear equations by elimination. Multiply the first equation by 3: 9x+6y=390. Multiply the second by 2: 4x+6y=240. Subtracting the second new equation from the first eliminates y and gives 5x=150, so x=30. Substitute this into 2x+3y=120: 2(30)+3y=120, hence 3y=60 and y=20. Therefore option C is correct. Options A, B and D do not satisfy both original equations; substituting y=20 with x=30 verifies 90+40=130 and 60+60=120.

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.

Twice the first equation is (4x+6y=24), but the second is (4x+6y=25). Therefore there is no solution.

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.

Let the larger number be (x) and the smaller be (y). Adding (x+y=29) and (x-y=7) gives (2x=36), so (x=18).

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}).