Algebraic methods: Substitution method and Elimination method.

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.

Elimination gives (p=\frac{340}{7}). Substituting it in either equation gives (q=\frac{200}{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.

The first equation becomes (2x+3y=9). When (c=10), the left side is the same but the right side is different.

Multiply the first equation by (30) to get (6x+5y=210). Using (x=y+6) gives (x=\frac{240}{11}).

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.

Subtracting the first equation from the second directly gives (x-y=2). In such questions, subtraction saves time.

Adding both equations gives (10x=80), so (x=8). Then (y=\frac{23}{7}), hence (x+y=\frac{79}{7}).

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.

Let the tens digit be (x) and units digit be (y). From (10x+y=7(x+y)) and (x-y=3), the number is (63).

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.

Multiplying the second equation by (3) gives (15y). It cancels with (-15y) in the first equation.

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.

Adding both equations gives (12x=24), so (x=2). Substituting in the first equation gives (y=-\frac{12}{5}).

To make coefficients proportional, (4:8=a:10) must hold. This gives (a=5), while constants are not in the same ratio.