Algebraic methods: Substitution method and Elimination method.

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.

The first equation is (2) times the second. Therefore both equations are identical and give infinitely many solutions.

Twice the first equation is (4x+6y=24), but the second is (4x+6y=30). Same left side with different constants means no solution.

From the first equation, \(x=12-y\). Substituting this into \(2x-3y=9\) gives \(2(12-y)-3y=9\), or \(24-5y=9\). Hence, \(-5y=-15\) and \(y=3\). For example, \(y=4\) does not satisfy both equations together. Exam tip: while expanding \(2(12-y)\), keep the negative sign in \(-2y\) correct.

Use (y=13-2x) from the second equation. Option checking shows (x=5,\ y=3) satisfies both equations.

The given solution \(x=4,\ y=3\) must satisfy the first equation \(2x+ky=18\). Substituting gives \(2(4)+k(3)=18\), or \(8+3k=18\). Hence \(3k=10\), so \(k=\frac{10}{3}\). If \(k=3\), the left-hand side becomes \(17\), so it is not correct. Exam tip: substitute the given solution into the equation containing the unknown parameter.

Substitute \(x=2\) and \(y=5\) into \(ax+2y=16\): \(2a+2(5)=16\). Thus, \(2a+10=16\), so \(2a=6\) and \(a=3\). The second equation is also satisfied because \(2+5=7\). If \(a=2\), the first equation gives \(14=16\), which is false. Exam tip: always verify the given ordered pair in both equations.