Algebraic methods: Substitution method and Elimination method.

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.

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.

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.

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

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.

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.

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

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.

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

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.

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}).

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

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

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.

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.

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

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.

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

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.

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.

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.

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.

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.