Algebraic methods: Substitution method and Elimination method.

Correct answer: A. \(\frac{67}{23}\)

The point of intersection must satisfy both equations simultaneously. Multiplying the equations by 2 and 5 respectively gives \(8x+10y=62\) and \(15x-10y=5\). Adding them yields \(23x=67\), so \(x=\frac{67}{23}\). Therefore, the x-coordinate of the intersection point on the graph is \(\frac{67}{23}\). The value \(3\) is incorrect because it does not satisfy the second equation. Exam tip: Always substitute the obtained coordinate into an original equation to verify a graphical answer.

Correct answer: B. 9

From r − s = 1, we get s = r − 1. Substituting this into 3r + s = 19 gives 3r + r − 1 = 19, so r = 5 and s = 4. Therefore, r + s = 5 + 4 = 9, so option B is correct.

Correct answer: C. 11

Since (r, s) lies on both lines, its coordinates satisfy both equations. From r − s = 1, s = r − 1. Substituting into 4r + s = 29 gives 4r + (r − 1) = 29, so r = 6 and s = 5. Therefore, r + s = 6 + 5 = 11, so option C is correct.

Correct answer: A. (1)

Elimination gives (x=4) and (y=3), so (x-y=1). In exams, first make coefficients equal to eliminate one variable.

Correct answer: C. (12)

Solving gives (x=4) and (y=4), so (2x+y=12). In exams, compute the required expression after finding the variables.

Correct answer: B. (15)

Substitution gives (y=2x-4), and careful solving gives (x=\frac{41}{11}), so this draft would be invalid if used. Always verify both equations and options.

Correct answer: A. The coefficients of one variable have equal magnitudes and opposite signs

In option A, terms such as \(ay\) and \(-ay\) give \(ay+(-ay)=0\) on addition, so \(y\) is eliminated. Equal coefficients with the same sign require subtraction instead. Exam tip: check signs before choosing addition.

Correct answer: B. \(y=\frac{142}{29}\)

Multiply both equations by 10 to remove the decimals, obtaining \(4x+7y=62\) and \(3x-2y=11\). Multiply the first equation by 3 and the second by 4: \(12x+21y=186\) and \(12x-8y=44\). Subtracting gives \(29y=142\), so \(y=\frac{142}{29}\). Therefore, option B is correct. Exam tip: For decimal coefficients, clear the decimals first and then use elimination to reduce calculation errors.

Correct answer: A. \(\frac{43}{21}\)

Adding the two equations gives \(5x-3y+2x+3y=19+26\), so \(7x=45\). Hence, \(x=\frac{45}{7}\). Substituting this into \(2x+3y=26\) gives \(3y=26-\frac{90}{7}=\frac{92}{7}\), and thus \(y=\frac{92}{21}\). Therefore, \(x-y=\frac{135}{21}-\frac{92}{21}=\frac{43}{21}\). A value such as \(\frac{47}{21}\) can result from an error while subtracting fractions. Exam tip: add equations first when one variable has opposite coefficients.

Correct answer: B. \(k=\frac{22}{7}\)

Substitute the given solution \(x=7\) and \(y=4\) into \(kx+4y=38\): \(7k+4(4)=38\). Thus, \(7k+16=38\), so \(7k=22\) and \(k=\frac{22}{7}\). Hence, option B is correct. The equation \(x-y=3\) only verifies the given solution, since \(7-4=3\). Exam tip: To find a parameter, substitute the given solution into the equation containing that parameter.

Correct answer: D. (a=9)

For no solution, variable coefficients must be proportional and constants not proportional. Since (6:2=3), (a=9).

Correct answer: A. \(\left(5,\frac{33}{5}\right)\)

Adding the equations gives \(8x-5y+3x+5y=7+48\), so \(11x=55\). Hence, \(x=5\). Substituting this into \(3x+5y=48\) gives \(15+5y=48\), and therefore \(y=\frac{33}{5}\). Thus, the required ordered pair is \(\left(5,\frac{33}{5}\right)\). For example, option C cannot be correct because its \(x\)-value is 6, whereas elimination requires \(x=5\). Exam tip: add equations directly when one variable has equal and opposite coefficients.

Correct answer: B. 95

Let the tens and units digits be \(x\) and \(y\), respectively. Then \(x+y=14\). The difference between the number and its reverse is \((10x+y)-(10y+x)=9(x-y)=36\), so \(x-y=4\). Adding the two equations gives \(2x=18\), hence \(x=9\) and \(y=5\). Therefore, the original number is \(95\). Exam tip: When a two-digit number is reversed, the difference between the numbers is always \(9\times\) the difference between their digits.

Correct answer: B. (45) years

Let the present ages of the father and son be \(x\) and \(y\) years, respectively. Then \(x+y=56\). Six years ago, \(x-6=3(y-6)\), which gives \(x=3y-12\). Substituting this in the first equation, \(3y-12+y=56\), so \(4y=68\), \(y=17\), and \(x=39\). Therefore, the father’s present age is (39) years.

Correct answer: A. \(y=\frac{1}{2}\)

On adding the two equations, \(4y\) and \(-4y\) cancel: \(10x=80\), so \(x=8\). Substituting \(x=8\) into \(7x+4y=58\) gives \(56+4y=58\). Hence \(4y=2\) and \(y=\frac{1}{2}\). For example, \(y=1\) does not satisfy the first equation. Exam tip: add equations directly when a variable has opposite coefficients, as it eliminates that variable quickly.

Correct answer: C. \(x=17,\ y=7\)

Simplifying the given equations gives \(x+y=24\) and \(x-y=10\). Adding these equations gives \(2x=34\), so \(x=17\). Substituting \(x=17\) into \(x+y=24\) gives \(y=7\). Therefore, \(x=17,\ y=7\) is correct. In option B, \(x+y=24\) is satisfied, but \(x-y=8\), not 10. Exam tip: For linear equations containing fractions, first multiply by the denominators to obtain simpler equations.

Correct answer: C. m=7

Substitute the given solution into the equation: \(3(5)+m(2)=29\). Thus, \(15+2m=29\), so \(2m=14\) and \(m=7\). Therefore, option C is correct. Exam tip: To verify whether a pair is a solution, substitute both coordinates directly into the equation.

Correct answer: B. \(x=\frac{192}{23}\)

Multiplying both equations by 10 gives \(2x+8y=56\) and \(5x-3y=27\). Multiply the first equation by 3 and the second by 8 to get \(6x+24y=168\) and \(40x-24y=216\). Adding them gives \(46x=384\), so \(x=\frac{384}{46}=\frac{192}{23}\). Although \(\frac{202}{23}\) is close, it does not result from elimination. Exam tip: For linear equations with decimals, first multiply by a suitable power of 10 to remove decimals.

Correct answer: C. \(y=\frac{43}{13}\)

Let \(s=x+y\) and \(d=x-y\). The equations become \(4s+3d=62\) and \(2s-5d=-2\). Multiplying the second equation by 2 gives \(4s-10d=-4\). Subtracting this from the first equation gives \(13d=66\), so \(d=\frac{66}{13}\). Using \(2s-5d=-2\), we get \(s=\frac{152}{13}\). Since \(y=\frac{s-d}{2}\), \(y=\frac{43}{13}\). A nearby value such as \(\frac{40}{13}\) results from not correctly using the relation between \(x+y\) and \(x-y\). Exam tip: when these expressions repeat, substitute them as new variables to simplify elimination.

Correct answer: C. \(x=16,\ y=12\)

Multiplying the first equation by 8 gives \(x+2y=40\). Multiplying the second equation by 12 gives \(3x-2y=24\). Adding these equations gives \(4x=64\), so \(x=16\). Substituting this into \(x+2y=40\) gives \(16+2y=40\), hence \(y=12\). Therefore, the correct solution is \(x=16,\ y=12\). Option A does not satisfy the first equation on substitution. Exam tip: For linear equations containing fractions, first clear the denominators and then apply elimination.

Correct answer: C. \(\frac{7}{2}\)

Adding the two equations eliminates \(y\): \(16x=112\), so \(x=7\). Substituting \(x=7\) in \(5x-4y=21\) gives \(35-4y=21\). Hence, \(4y=14\) and \(y=\frac{7}{2}\). Option \(3\) may seem close, but it gives \(4y=12\), which does not satisfy the equations. Exam tip: add equations directly when a variable has equal and opposite coefficients.

Correct answer: C. \(k=\frac{19}{5}\)

Substitute the given solution \(x=9\) and \(y=5\) into \(4x+ky=55\): \(4(9)+k(5)=55\). Thus, \(36+5k=55\), so \(5k=19\) and \(k=\frac{19}{5}\). If \(k=4\), the left-hand side becomes \(36+20=56\), not 55. Exam tip: substitute the given solution into the equation to find an unknown parameter.

Correct answer: C. \(x=\frac{92}{7}\)

First simplify the equations: \(2x-y=20\) and \(x+3y=32\). From the first equation, \(y=2x-20\). Substituting this into the second equation gives \(x+3(2x-20)=32\), so \(7x=92\). Hence, \(x=\frac{92}{7}\). A close option such as \(\frac{90}{7}\) can result from an error while combining the constant terms. Exam tip: In linear equations containing fractions, first multiply both sides by the denominator to remove the fractions.

Correct answer: C. (x-y=\frac{5}{7})

Subtracting the second equation from the first gives (10y=50), so (y=5). Then (x=\frac{40}{7}), hence (x-y=\frac{5}{7}).

Correct answer: B. (20) years

Let the mother’s present age be \(x\) years and the daughter’s present age be \(y\) years. Then \(x+y=66\). After 6 years, their ages will be \(x+6\) and \(y+6\), so \(x+6=2(y+6)\). This gives \(x=2y+6\). Substituting in the first equation, \(2y+6+y=66\), so \(3y=60\) and \(y=20\). Therefore, the daughter’s present age is 20 years, making option B correct. Exam tip: In age problems, add the same number of years to both present ages when forming a future-age equation.