Algebraic methods: Substitution method and Elimination method.

Correct answer: B. \(y=\frac{19}{7}\)

Multiply the first equation by 2 to get \(10x+4y=48\). Subtract \(3x+4y=22\) from it: \(7x=26\), so \(x=\frac{26}{7}\). Substituting this into \(5x+2y=24\) gives \(2y=24-\frac{130}{7}=\frac{38}{7}\), hence \(y=\frac{19}{7}\). The distractor \(y=2\) does not satisfy the original equations. Exam tip: make one variable’s coefficients equal before subtracting to eliminate it quickly.

Correct answer: A. \(\frac{52}{7}\)

From the second equation, \(y=3x-8\). Substituting this into \(x+2y=11\) gives \(x+2(3x-8)=11\), so \(7x=27\) and \(x=\frac{27}{7}\). Then \(y=3\times\frac{27}{7}-8=\frac{25}{7}\). Hence, \(x+y=\frac{27}{7}+\frac{25}{7}=\frac{52}{7}\). The option \(8\) may result from assuming integer values, but the given equations have fractional solutions. Exam tip: when adding fractions with the same denominator, add only their numerators.

Correct answer: B. \(\left(\frac{47}{13},\frac{23}{13}\right)\)

From the second equation, \(y=9-2x\). Substituting this into \(7x-3y=20\) gives \(7x-3(9-2x)=20\), so \(13x=47\). Hence, \(x=\frac{47}{13}\) and \(y=9-2\times\frac{47}{13}=\frac{23}{13}\). Although \((3,3)\) satisfies the second equation, it gives \(7(3)-3(3)=12\neq20\) in the first equation. Exam tip: always verify the ordered pair in both original equations.

Correct answer: D. (y=5)

Multiply the second equation by (5) and add with the first. Checking the options in both equations gives the correct value (y=5).

Correct answer: C. \(x=5\)

The coefficients of y are +5 and -5, so y is eliminated when the equations are added: \(6x+5y+4x-5y=39+11\). Thus, \(10x=50\), giving \(x=5\). If \(x=4\), then \(10x=40\), so it cannot satisfy the added equation. Exam tip: Before eliminating, look for terms with equal coefficients and opposite signs.

Correct answer: A. 1

Adding the two equations eliminates the y-terms: 8x=40, so x=5. Substituting x=5 into 3x+2y=23 gives 15+2y=23, hence y=4. Therefore, x-y=5-4=1. Option 3 is incorrect because it does not match the difference of the correct values of x and y. Exam tip: Add equations when a variable has equal coefficients with opposite signs.

Correct answer: B. -2

Multiplying the first equation by -2 gives (-4x-6y=-26). On adding it to (4x-5y=1), we get (-4x+4x=0), so x is eliminated. Multiplying by 2 would produce (4x), which would not cancel with the (4x) in the second equation when added. Exam tip: for elimination by addition, make the coefficients equal in magnitude and opposite in sign.

Correct answer: C. (x=4)

Use (y=2x-3) from the second equation. Substitution and option checking show (x=4) is correct.

Correct answer: C. There are infinitely many solutions

Compare the two equations after multiplying the first equation by 2. The result is 4x+6y=32, which is exactly the second equation. Thus both equations represent the same straight line, not two different intersecting lines. Every point satisfying 2x+3y=16 also satisfies the second equation, so there are infinitely many ordered pairs in the solution set. For example, x=2 gives y=4, while x=5 gives y=2; both are different solutions. Therefore option C is correct. A unique solution would require two distinct lines meeting at one point. No solution would occur for parallel distinct lines, where the coefficient ratios agree but the constant ratio differs. Option D is also false: x=0 gives y=16/3, but it is only one of infinitely many solutions, not the only one.

Correct answer: B. There is no solution

The first equation becomes (x+2y=6), while the second is (x+2y=8). Same left side with different right side gives no solution.

Correct answer: C. 13

Let the larger number be \(x\) and the smaller number be \(y\). Then \(x+y=21\) and \(x-y=5\). Adding the equations gives \(2x=26\), so \(x=13\). Therefore, the larger number is 13. If 14 were chosen, the other number would be 7 and their difference would be 7, not 5. Exam tip: In sum-and-difference questions, add the two equations to find the larger number directly.

Correct answer: D. (16)

Use (y=2x-9) from the first equation. Solving gives (x=\frac{31}{5}) and (y=\frac{17}{5}), so (2x+y=\frac{79}{5}).

Correct answer: A. (x=\frac{28}{5},\ y=\frac{21}{5})

Use (y=2x-7) from the first equation. Substitution gives (5x=28), so (x=\frac{28}{5}) and (y=\frac{21}{5}).

Correct answer: C. \(x=5\)

The term \(3y\) is the same in both equations, so subtract the second equation from the first: \((5x+3y)-(2x+3y)=31-16\). This gives \(3x=15\), hence \(x=5\). For example, \(x=4\) would give \(3x=12\), not 15. Exam tip: In elimination, subtract equations directly when one variable already has equal coefficients.

Correct answer: B. (y=2)

Adding both equations gives (8x=36), so (x=\frac{9}{2}). Then (x-2y=3) gives (y=\frac{3}{4}).

Correct answer: C. \(x=3,\ y=7\)

From the second equation, \(x=10-y\). Substituting this into \(3x+5y=44\) gives \(3(10-y)+5y=44\), so \(30+2y=44\). Hence, \(y=7\) and \(x=10-7=3\). In option A, \(x+y=10\), but \(3x+5y=42\), not 44. Exam tip: always verify the obtained pair in both original equations.

Correct answer: B. 9 rupees

Let the price of one pen be ₹\(x\) and that of one pencil be ₹\(y\). Then \(x+y=18\) and \(2x+3y=45\). Multiplying the first equation by 2 gives \(2x+2y=36\). Subtracting this from the second equation gives \(y=9\). Hence, \(x=18-9=9\), so one pen costs 9 rupees. If the pen cost 10 rupees, the pencil would cost 8 rupees, which does not satisfy the second condition. Exam tip: In price-based word problems, assign separate variables to the prices and form equations first.

Correct answer: C. (x=5)

From the second equation use (y=2x-4). Correct substitution gives (4x+6x-12=34), so (x=\frac{23}{5}).

Correct answer: C. 4

Subtract the second equation from the first so that the x-terms cancel:
(2x+7y)−(2x+3y)=36−20.
Thus, 4y=16, so y=4. Option 3 is incorrect because for y=3, the difference of the left-hand sides would be 12, not 16. Exam tip: In the elimination method, add or subtract equations to eliminate a variable with equal coefficients.

Correct answer: D. (10)

Adding both equations gives (8x=32), so (x=4). Then (3x+4y=30) gives (y=\frac{9}{2}), so (x+y=\frac{17}{2}).

Correct answer: C. \(y=4\)

From the first equation, \(x=2y+3\). Substituting this into \(3x-4y=17\) gives \(3(2y+3)-4y=17\). Therefore, \(6y+9-4y=17\), so \(2y=8\) and hence \(y=4\). For example, \(y=3\) does not satisfy the second equation. Exam tip: after substitution, multiply \(3\) by both terms inside the bracket.

Correct answer: B. \(-x+y=2\)

Multiplying \(-x+y=2\) by 3 gives \(-3x+3y=6\). Adding this to \(3x+2y=16\) cancels \(3x\) with \(-3x\), so \(x\) is eliminated. If option A is multiplied by 3, it produces \(3x\), which will not cancel the given \(3x\) on addition. Exam tip: for elimination, make the coefficients of one variable equal in magnitude and opposite in sign.

Correct answer: A. \(x=4,\ y=3\)

From the second equation, \(y=11-2x\). Substituting this into \(4x+5y=31\) gives \(4x+5(11-2x)=31\), so \(-6x=-24\). Hence, \(x=4\) and \(y=3\). The close distractor \(x=5,\ y=1\) satisfies the second equation, but it gives \(4(5)+5(1)=25\), not 31, in the first equation. Exam tip: always verify the ordered pair in both original equations.

Correct answer: A. \(\frac{21}{5}\)

The coefficients of \(y\) are \(-3\) and \(+3\), so adding the equations eliminates \(y\): \((8x-3y)+(2x+3y)=25+17\). Thus, \(10x=42\), giving \(x=\frac{42}{10}=\frac{21}{5}\). Option \(4\) is close, but solving \(10x=42\) does not give an integer. Exam tip: in elimination, first check whether the coefficients of a variable are additive inverses.

Correct answer: B. 50

Write the equations as \(2x+y=140\) and \(x+2y=130\). Multiplying the second equation by 2 gives \(2x+4y=260\). Subtracting the first equation gives \(3y=120\), so \(y=40\). Substituting \(y=40\) into \(2x+y=140\) gives \(2x=100\), hence \(x=50\). Option 40 is the value of \(y\), not \(x\). Exam tip: verify the obtained values in both original equations.