Algebraic methods: Substitution method and Elimination method.

From the second equation, \(2x+y=13\), we get \(y=13-2x\). Substituting this into the first equation gives \(4x-7(13-2x)=-19\), so \(18x=72\). Hence, \(x=4\). If \(x=3\), the two equations do not give the same value of \(y\). Exam tip: while substituting, distribute the negative sign in \(-7(13-2x)\) carefully.

From the first equation, \(y=22-3x\). Substituting this into \(x+2y=19\) gives \(x+2(22-3x)=19\), so \(-5x=-25\) and hence \(x=5\). Then \(y=7\), giving \(x-y=5-7=-2\). The distractor \(-1\) can result from an error while subtracting the values. Exam tip: find and verify both variable values before evaluating the required expression.

Let the length be \(l\) and the breadth be \(b\). The statements give \(l+b=28\) and \(l-b=6\). Subtracting the second equation from the first gives \(2b=22\), so \(b=11\). Hence, 11 is the correct answer. If the breadth were 12, the length would be 18 and their sum would be 30, not 28. Exam tip: For sum-and-difference word problems, first form separate equations for the sum and the difference.

Adding the two equations eliminates \(y\): \(9x+3x=37+11\), so \(12x=48\). Hence, \(x=4\). If \(x=3\), then \(12x=36\), so it cannot satisfy the resulting equation. Exam tip: add equations when one variable has equal coefficients with opposite signs.

For infinitely many solutions, both equations must be proportional. The ratio of (4) and (2) is (2), so (a=6).

Multiplying \(x+y=5\) by 3 gives \(3x+3y=15\). On adding it to \(2x-3y=7\), the terms \(+3y\) and \(-3y\) cancel, so \(y\) is eliminated. Multiplying \(x-y=5\) by 3 gives \(-3y\), which would not cancel the given \(-3y\) on addition. Exam tip: for elimination by addition, the selected variable must have equal coefficients with opposite signs.

From \(x-y=3\), we get \(x=y+3\). Substituting this into \(3x+2y=19\) gives \(3(y+3)+2y=19\), so \(5y=10\) and \(y=2\). Hence, \(x=5\), and the ordered pair is \((5,2)\). Although \((6,3)\) satisfies \(x-y=3\), it gives \(3x+2y=24\ne19\). In an exam, verify the final pair in both equations.

Subtract \(5x-y=14\) from \(5x+4y=44\). This gives \(5y=30\), so \(y=6\). The nearby option \(y=4\) does not satisfy both equations simultaneously. Exam tip: when the coefficients of one variable are already equal, subtract the equations to eliminate that variable quickly.

From the first equation, y=14-2x. Substituting this into x+2y=16 gives x+2(14-2x)=16, so -3x=-12 and x=4. Hence y=6, and xy=4×6=24. A value such as 22 can result from an error while handling the coefficient 2 during substitution. Exam tip: After finding x and y, verify them in either original equation.

Substitute \(x=3\) and \(y=2\) into \(2x+ky=16\): \(2(3)+k(2)=16\). Thus, \(6+2k=16\), so \(2k=10\) and \(k=5\). For example, \(k=4\) gives a left-hand side of \(14\), not \(16\). Exam tip: while substituting an ordered pair, place the values of \(x\) and \(y\) in their correct terms.

Use the consistency test for two linear equations. Multiplying 3x+4y=10 by 2 gives 6x+8y=20, whereas the second equation states 6x+8y=21. Since identical left-hand sides cannot have two different right-hand values, no ordered pair can satisfy both equations. Equivalently, the coefficient ratios are equal, 3/6=4/8, but the constant ratio 10/21 is different; this represents two distinct parallel lines. Hence option A, no solution, is correct. Option B would be true only if the second equation were 6x+8y=20. Option C is impossible for parallel distinct lines, and substituting x=2,y=1 gives 10 in the first equation but 20, not 21, in the second.

The coefficients of \(y\) are \(-3\) and \(+3\), so adding the equations eliminates \(y\): \((2x-3y)+(4x+3y)=-4+22\). Thus, \(6x=18\), giving \(x=3\). For example, \(x=4\) would give \(6x=24\), which does not satisfy \(6x=18\). Exam tip: Add equations directly when a variable has equal and opposite coefficients.

From the first equation, \(x=26-4y\). Substituting this into \(3x-2y=8\) gives \(3(26-4y)-2y=8\), or \(78-14y=8\). Hence, \(14y=70\) and \(y=5\). Substituting \(y=5\) into \(x+4y=26\) gives \(x=6\). Therefore, the correct solution is \(x=6,\ y=5\). Option B merely interchanges the values and does not satisfy the first equation. Exam tip: verify the ordered pair in both original equations.

Direct answer: D, y=4. Use substitution. From 2x−y=5, rearrange to y=2x−5. Substitute this expression in the first equation: 6x+5(2x−5)=47. Thus 6x+10x−25=47, so 16x=72 and x=9/2. Therefore y=2(9/2)−5=9−5=4. Verification gives 6(9/2)+5(4)=27+20=47 and 2(9/2)−4=9−4=5, so both equations are satisfied. A, y=2, does not produce a common solution in both equations. B, y=3, also fails when the corresponding x is tested in both equations. C, y=5 gives an incorrect value after substitution and does not satisfy the pair simultaneously. D, y=4, satisfies both equations and is correct. A useful warning is that changing 2x−y=5 to y=5−2x would reverse the sign incorrectly; carefully moving terms prevents this common error.

Let the number of girls be \(g\). Then the number of boys is \(g+8\). Using the total, \(g+(g+8)=52\), so \(2g=44\). Hence, \(g=22\). If the number of girls were 23, the number of boys would be 31 and the total would become 54, so that option is incorrect. Exam tip: In such questions, represent the smaller number by a variable and write the larger number as variable plus the given difference.

Adding \(3x-5y=-1\) and \(2x+5y=21\) eliminates the \(y\)-terms: \(5x=20\). Hence, \(x=4\), so option B is correct. The value of \(x\) cannot be obtained directly from either equation alone because each also contains \(y\). Exam tip: add two equations when one variable has equal and opposite coefficients.

Use (y=3x-11) from the second equation. Substitution gives (x=\frac{27}{5},\ y=\frac{16}{5}), so (2x+y=14).

The answer is D, 22. Treat x and y as the two numbers. Add the equations: (2x+3y)+(3x+2y)=92+98, giving 5x+5y=190 and therefore x+y=38. Now subtract the first equation from the second: (3x+2y)−(2x+3y)=98−92, so x−y=6. Add x+y=38 and x−y=6: 2x=44, hence x=22. Then y=38−22=16. Since 22 is greater than 16, the larger number is 22. Verification gives 2(22)+3(16)=92 and 3(22)+2(16)=98. Option A is the smaller number; 18 and 20 do not satisfy the required sum and difference. The key idea is to form simple equations for x+y and x−y.

The elimination principle requires the coefficients of the variable to be eliminated to have equal magnitudes and opposite signs. In the first equation, the coefficient of y is −6. In the second equation it is +2, so multiplying the entire second equation by 3 gives 9x + 6y = 27. The +6y then cancels the −6y when the equations are added, leaving an equation only in x. Therefore option D, 3, is correct. Multiplying by 1 leaves +2y, which cannot cancel −6y directly. Multiplying by 2 gives +4y, and multiplying by −2 gives −4y; neither produces the required +6y. The same multiplier must be applied to every term of the second equation.

The equal coefficients of x make elimination immediate. Subtract the second equation from the first: (4x + 7y) − (4x − 3y) = 53 − 13. The x terms cancel and the y terms give 7y + 3y = 10y, so 10y = 40 and y = 4. Substitute y = 4 into 4x − 3y = 13: 4x − 12 = 13, hence 4x = 25 and x = 25/4. Therefore x − y = 25/4 − 4 = 25/4 − 16/4 = 9/4, making option A correct. Substitution verifies both original equations. The other choices arise from errors in subtraction, sign handling, or putting the result over the wrong denominator.

The governing concept is the consistency condition for a pair of linear equations. Multiplying the first equation by 2 gives 4x + 6y = 22, whereas the second equation is 4x + 6y = 21. The left-hand sides are identical, but the constants are different, so the same expression cannot equal both 22 and 21. Geometrically, the two equations represent distinct parallel lines, which never intersect. Therefore the pair is inconsistent and has no common ordered pair (x, y). Option C is correct. Option A would require different slopes, option B would require the equations to be proportional including their constants, and option D is impossible for a linear pair.

The answer is B, y=17/3. The y-coefficients are 9 and −3, so multiply the second equation by 3: 15x−9y=24. Add it to 4x+9y=71. The y terms cancel, giving 19x=95, so x=5. Substitute x=5 into 5x−3y=8: 25−3y=8. Hence −3y=−17, and y=17/3. Check the first equation: 4(5)+9(17/3)=20+51=71. Therefore B is correct. The other fractions are distractors because substituting any of them with the corresponding x does not satisfy both original equations. A useful memory cue is to multiply equations so that one pair of coefficients becomes opposites before adding.

The governing concept is elimination followed by evaluation of an expression. Adding the equations is useful because -4y and +4y are opposites: (3x - 4y) + (5x + 4y) = -2 + 34, which gives 8x = 32 and x = 4. Substitute x = 4 into 3x - 4y = -2: 12 - 4y = -2, so -4y = -14 and y = 7/2. Now calculate the requested expression: 2x + y = 2(4) + 7/2 = 8 + 7/2 = 16/2 + 7/2 = 23/2. Therefore option A is correct. The other choices can arise from sign mistakes, incorrect division, or adding the final terms incorrectly. Both original equations verify x = 4 and y = 7/2.

The governing concept is translating verbal conditions into a pair of linear equations and solving by substitution. Let the first number be x and the second be y. “The first is 3 less than twice the second” gives x = 2y - 3, while their sum gives x + y = 42. Substitute the first relation into the sum: (2y - 3) + y = 42, so 3y - 3 = 42, 3y = 45, and y = 15. Then x = 2(15) - 3 = 27. Since 15 is smaller than 27, the smaller number is 15, so option C is correct. A check gives 27 + 15 = 42 and 27 is exactly 3 less than 30, twice 15. The other options fail at least one condition.

The governing concept is solving a pair of simultaneous linear equations by elimination. Multiply 6x - 7y = -8 by 5 to obtain 30x - 35y = -40. Multiply 4x + 5y = 38 by 7 to obtain 28x + 35y = 266. Adding these equations eliminates y and gives 58x = 226, so x = 113/29. Substitute this value into 4x + 5y = 38: 4(113/29) + 5y = 38, which gives 5y = 650/29 and y = 130/29. Therefore x + y = 113/29 + 130/29 = 243/29. Option A is correct. The other numerical options are not equal to the value obtained from the simultaneous equations. The corrected options now contain one unambiguous answer.