Algebraic methods: Substitution method and Elimination method.

The two equations represent the same straight line, so they do not give two separate conditions. When one equation is exactly a multiple of the other, every point satisfying one equation also satisfies the other. Therefore the pair does not select one particular ordered pair; it has infinitely many solutions. For example, the second equation can be written as y=14-3x, and every value of x gives a corresponding value of y.

To compare them, multiply 3x+y=14 by 2. This gives 6x+2y=28, which is exactly the first equation. Thus the equations are dependent and consistent, not contradictory. The correct choice is D, infinitely many solutions. Choice B would apply when the two lines meet at one point, while choice A would apply to distinct parallel lines.

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.

Subtract the second equation from the first: \((4x+7y)-(4x+3y)=41-25\). This gives \(4y=16\), so \(y=4\). If \(y=3\), then \(4y=12\), not 16. Exam tip: subtract equations directly when one variable has equal coefficients.

Adding the given equations gives \((9x-2y)+(3x+2y)=35+13\). The terms \(-2y\) and \(+2y\) cancel, so \(12x=48\), giving \(x=4\). Hence, option C is correct. If \(x=3\), the sum of the two equations would give \(12x=36\), not 48. Exam tip: add equations when a variable has equal coefficients with opposite signs.

The direct answer is option A: 63. Let the tens digit be x and the units digit be y. The digit sum gives x+y=9. “The tens digit is 3 more” gives x-y=3, not y-x=3. Add the equations: 2x=12, so x=6. Then y=9-6=3. A two-digit number with tens digit 6 and units digit 3 is 63. Option A is correct, and checking gives 6+3=9 and 6-3=3. Option B, 54, has digit sum 9, but 5 is only 1 more than 4. Option C, 72, has sum 9, but 7 is 5 more than 2. Option D, 81, has sum 9, but 8 is 7 more than 1. The common mistake is reversing the tens and units condition. Memory cue: tens digit is written first, so the number is 10x+y.

Using (x=y+1) gives (3y+3+2y=20), so (y=\frac{17}{5}) and (x=\frac{22}{5}). Then (3x-y=\frac{49}{5}).

Adding the two equations cancels \(3y\) and \(-3y\): \(7x=35\), so \(x=5\). Substituting this into \(2x+3y=25\) gives \(10+3y=25\), hence \(y=5\). Therefore, the solution is \((x,y)=(5,5)\). Option B satisfies the first equation, but in the second equation it gives \(5x-3y=3\), not \(10\). Exam tip: add equations directly when one variable has opposite coefficients.

Compare the equations after reducing the first one by dividing every term by 2. The first equation becomes \(3x+2y=20\), whereas the second equation is \(3x+2y=22\). Their left-hand sides are identical, but their constants differ, so the two equations represent distinct parallel lines. Parallel distinct lines never intersect; consequently, no ordered pair \((x,y)\) can satisfy both equations at the same time. Hence option B, “No solution,” is correct. Infinitely many solutions would require the two equations to be identical, while one solution would occur if the corresponding lines intersected. The phrase “two solutions” is not applicable to a consistent pair of linear equations in this form.

Let the number of boys be \(x\) and the number of girls be \(y\). Then \(x+y=45\) and \(x-y=9\). Adding the two equations gives \(2x=54\), so \(x=27\). Therefore, the number of boys is 27. Option 26 is close, but it would leave 19 girls, giving a difference of 7 rather than 9. Exam tip: for questions giving a total and a difference, add the equations to find the larger quantity directly.

On adding the two equations, \(-5y\) and \(+5y\) cancel: \(2x-5y+3x+5y=-1+31\). Thus, \(5x=30\), so \(x=6\). For example, \(x=5\) gives \(5x=25\), which does not satisfy the resulting equation \(5x=30\). Exam tip: Add equations directly when a variable has equal and opposite coefficients.

Adding the two equations eliminates \(2y\) and \(-2y\): \(8x=40\), so \(x=5\). Substituting \(x=5\) into \(5x+2y=29\) gives \(25+2y=29\), hence \(2y=4\) and \(y=2\). For example, \(y=3\) does not satisfy the first equation. Exam tip: add equations directly when a variable has opposite coefficients.

Substitute \(y=3x-7\) into \(2x+y=18\): \(2x+3x-7=18\). Thus, \(5x=25\), so \(x=5\). Then \(y=3(5)-7=8\). Hence, the correct pair is \(x=5,\ y=8\). Option C satisfies the first equation, but it gives \(2x+y=23\), not 18. Exam tip: Always verify the obtained pair in both equations.

Subtract the second equation from the first: \((3x+4y)-(3x-y)=38-13\). Thus, \(5y=25\), so \(y=5\). If \(y=4\), then \(5y=20\), which does not satisfy the resulting equation. Exam tip: subtract equations when a variable has the same coefficient to eliminate it quickly.

Use (y=4x-11) from the second equation. Substitution gives (14x-33=27), so (x=\frac{30}{7}).

Adding the two equations gives \((x+2y)+(3x-2y)=13+7\), so \(4x=20\). Hence, \(x=5\). Substituting this into \(x+2y=13\) gives \(5+2y=13\), and therefore \(y=4\). Thus, the solution is \(x=5,\ y=4\). In option C, \(x=6\) does not satisfy the first equation. Exam tip: adding equations is a quick elimination method when terms have opposite coefficients.

From the second equation, 2x-y=3, solve for y to obtain y=2x-3. Substitute this into x+3y=19: x+3(2x-3)=19. Simplifying gives x+6x-9=19, so 7x=28 and x=4. Then y=2(4)-3=5. The required expression is x+2y=4+2(5)=14, so option D is correct. Verification in the original equations gives 4+15=19 and 8-5=3. The distractors 11, 12, and 13 may result from using only one variable, subtracting instead of adding the final terms, or making an arithmetic error, but none equals the correctly evaluated expression. The governing idea is to solve the pair first, then substitute the ordered values into the requested expression rather than confusing it with either original equation.

Substituting the given values in the equation gives \(2k+4(3)=22\). Thus, \(2k+12=22\), so \(2k=10\) and \(k=5\). If \(k=4\), the left-hand side becomes \(20\), not \(22\). Exam tip: In parameter questions, substitute the given values of \(x\) and \(y\) carefully before solving.

From the second equation, \(4x-y=3\), we get \(y=4x-3\). Substituting this into the first equation gives \(2x+3(4x-3)=19\), so \(14x=28\) and \(x=2\). Hence, \(y=4(2)-3=5\), and the solution is \(x=2,\ y=5\). Option A does not satisfy the second equation. Exam tip: always verify the obtained values in both original equations.

Adding the two equations eliminates the y-terms: 5x + 2y + 3x - 2y = 28 + 4, so 8x = 32. Hence, x = 4. Substituting x = 4 into 3x - 2y = 4 gives 12 - 2y = 4, so y = 4. Therefore, x + y = 4 + 4 = 8. Option 7 is not correct because the simultaneous solution is x = 4 and y = 4. Exam tip: when variable terms have opposite coefficients, add the equations to eliminate that variable quickly.

From the second equation, \(x=9-y\). Substituting this into \(3x+5y=31\) gives \(3(9-y)+5y=31\), or \(27+2y=31\). Hence \(y=2\) and \(x=9-2=7\). If \(x=6\), then the second equation gives \(y=3\), but the first equation becomes \(33\), not \(31\). Exam tip: verify the obtained values in both original equations.

From the first equation, \(y=6x-21\). Substituting this into \(2x+3y=17\) gives \(2x+3(6x-21)=17\), so \(20x=80\). Hence, \(x=4\) and \(y=3\), making option C correct. For example, \(x=5, y=1\) gives \(13=17\) in the second equation, so it is not a solution. Exam tip: Always verify the ordered pair in both original equations.