Algebraic methods: Substitution method and Elimination method.

Adding the two equations gives (x+y=38) and subtracting gives (x-y=6). Therefore (x=22,\ y=16).

Multiply the first equation by (2) and add the second. (x=\frac{58}{13}) and (y=\frac{47}{13}), so (x+y=\frac{105}{13}).

Clear denominators carefully. Multiplying the first equation by 12 gives 3x+2y=60. Multiplying the second by 6 gives 3x−2y=12. Adding these equations yields 6x=72, so x=12. Substituting into 3x+2y=60 gives 36+2y=60, hence y=12. Therefore option C is correct; the other ordered pairs fail at least one original equation.

For infinitely many solutions, coefficients and constants must be in the same ratio. Since (4:6=10:15=2:3), (a=2).

Let the tens digit be \(x\) and the units digit be \(y\). Then \(x+y=9\). Taking the question to mean that the original number is 27 greater than its reversed number, \((10x+y)-(10y+x)=27\). Thus \(9(x-y)=27\), so \(x-y=3\). Solving the two equations gives \(x=6\) and \(y=3\). Therefore, the original number is \(63\). Although 54 has digit sum 9, its difference from 45 is only 9. Exam tip: represent a two-digit number as \(10x+y\) and its reverse as \(10y+x\).

Multiply the first equation by (2) and add the second. (x=\frac{58}{17}) and (y=\frac{69}{17}).

Substitute the given solution \(x=5,\ y=2\) into the first equation \(2x+ky=15\). This gives \(2(5)+k(2)=15\), or \(10+2k=15\). Hence \(2k=5\), so \(k=\frac{5}{2}\). If \(k=3\), the left-hand side becomes \(10+6=16\), not 15. Exam tip: To find a parameter from a given solution, substitute the coordinates into the equation containing that parameter.

Multiplying the first equation by 5 gives \(2x-3y=5\), and multiplying the second equation by 2 gives \(x+y=12\). From \(x+y=12\), write \(x=12-y\). Substituting gives \(2(12-y)-3y=5\), so \(24-5y=5\). Hence, \(-5y=-19\) and \(y=\frac{19}{5}\). The nearby distractor \(\frac{21}{5}\) does not satisfy both equations together. Exam tip: In linear equations containing fractions, first clear the denominators by multiplying both sides appropriately.

Multiplying the second equation by (3) gives (6y). It cancels with (-6y) in the first equation.

Subtracting the equations gives (10y=40), so (y=4). Then (x=\frac{25}{4}), hence (x-y=\frac{9}{4}).

Let the mother’s present age be \(x\) years and the daughter’s present age be \(y\) years. Then \(x+y=56\). The condition after 4 years gives \(x+4=3(y+4)\). Hence \(x=3y+8\). Substituting in the first equation, \(3y+8+y=56\), so \(4y=48\) and \(y=12\). Therefore, the daughter’s present age is 12 years. For example, 14 years does not satisfy the total-age condition. Exam tip: in age problems, add the stated number of years to both persons’ present ages.

Adding the two equations gives \((6x+5y)+(4x-5y)=2+18\), so \(10x=20\). Hence, \(x=2\). Substituting this into \(6x+5y=2\) gives \(12+5y=2\), so \(5y=-10\) and \(y=-2\). Therefore, the solution is \((2,-2)\). A close distractor such as \((3,-3)\) does not satisfy the first equation. Exam tip: when terms have opposite coefficients, adding the equations is a quick elimination method.

To make coefficients proportional, (3:6=a:8) must hold. This gives (a=4), and constants (7:20) are not in the same ratio.

Removing decimals gives (2x+5y=31) and (4x-y=13). The solution is (x=\frac{48}{11},\ y=\frac{49}{11}), so (x+y=\frac{97}{11}).

Adding the two equations eliminates the \(y\)-terms: \((8x-3y)+(2x+3y)=31+29\). Thus, \(10x=60\), so \(x=6\). Substituting this into \(2x+3y=29\) gives \(12+3y=29\), hence \(3y=17\) and \(y=\frac{17}{3}\). The value \(\frac{16}{3}\) does not satisfy the second equation. Exam tip: add equations directly when a variable has equal and opposite coefficients.

After clearing denominators, (2x+5y=70) and (4x-3y=12) are obtained. Elimination gives (x=\frac{135}{13}).

Let the length be
(l
) cm and the breadth be
(b
) cm. Since the perimeter is
80
cm,
2(l+b)=80
, so
l+b=40
. Also,
l-b=10
. Adding the two equations gives
2l=50
, hence
l=25
and
b=15
. Therefore, the area is
l imes b=25 imes15=375
square cm. Thus, option D is correct. 350 square cm would result from not using the correct length and breadth values in the product. Exam tip: first obtain
l+b
from the perimeter, then pair it with the given difference to form two linear equations.

The given solution \(x=4,\ y=7\) must satisfy the first equation \(px+2y=18\). Substituting gives \(4p+2(7)=18\), or \(4p+14=18\). Hence \(4p=4\), so \(p=1\). The second equation also checks out: \(3(4)-7=5\). Exam tip: In parameter-based questions, substitute the given solution into the equation containing the parameter.

On adding the two equations, \(3y\) and \(-3y\) cancel: \(7x=21\), so \(x=3\). Substituting \(x=3\) into \(5x+3y=14\) gives \(15+3y=14\). Hence \(3y=-1\) and \(y=-\frac{1}{3}\). \(\frac{1}{3}\) can result from a sign error. Exam tip: add equations first when the coefficients of one variable are opposites.

From the first equation, x=15-y. Substituting this into 2x-3y=10 gives 2(15-y)-3y=10, or 30-5y=10. Thus y=4 and x=11. Therefore, 3x+y=3(11)+4=37. Option 39 may seem close, but it is not obtained on substituting the correct values. Exam tip: after substitution, combine -2y and -3y carefully to get -5y.