Algebraic methods: Substitution method and Elimination method.

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.

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.

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.

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

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

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.

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.

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.

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.

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.

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

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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

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.