Algebraic methods: Substitution method and Elimination method.

For a pair of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 to have infinitely many solutions, the three ratios a₁/a₂, b₁/b₂, and c₁/c₂ must be equal. Rewrite the equations as ax + 4y − 10 = 0 and 3x + 6y − 15 = 0. Since 4/6 = 10/15 = 2/3, we require a/3 = 2/3. Therefore a = 2, so option B is correct. This makes the first equation exactly two-thirds of the second: 2x + 4y − 10 = 0 after scaling, while the second is 3x + 6y − 15 = 0. Options A, C, and D fail the proportionality condition and would produce a unique or inconsistent relationship rather than infinitely many solutions.

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 \(10y+x\).

Use elimination to remove y. Multiply 7x + 2y = 32 by 2, obtaining 14x + 4y = 64. Add the second equation, 3x − 4y = −6. The y terms cancel, so 17x = 58 and x = 58/17. Substitute this value into the first equation: 7(58/17) + 2y = 32. Writing 32 as 544/17 gives 406/17 + 2y = 544/17, so 2y = 138/17 and y = 69/17. Therefore option A is correct. Substitution in the second equation also works: 3(58/17) − 4(69/17) = (174 − 276)/17 = −102/17 = −6. The other fractions arise from arithmetic or sign errors.

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.

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.

Clear the denominators carefully. Multiplying x/5 + y/2 = 7 by 10 gives 2x + 5y = 70. Multiplying x/3 − y/4 = 1 by 12 gives 4x − 3y = 12. To eliminate y, multiply the first equation by 3 and the second by 5: 6x + 15y = 210 and 20x − 15y = 60. Adding them gives 26x = 270, so x = 270/26 = 135/13. Therefore option B is correct. The other options arise from incorrect least common multiples or a sign error while combining the equations. Substitution of x = 135/13 into the two transformed equations produces the same y value and confirms the solution.

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.

The equations become (x+2y=24) and (2x-y=15). The solution is (x=\frac{54}{5},\ y=\frac{33}{5}), so (x-y=\frac{21}{5}).

The second equation is exactly twice the first: multiplying every term of \(2x+3y=11\) by 2 gives \(4x+6y=22\). Thus the two equations are not two different lines. They describe the same straight line, so every point on that line satisfies both equations. A pair of coincident linear equations has infinitely many solutions, not one or two.

To see this algebraically, divide the second equation by 2. It becomes \(2x+3y=11\), identical to the first equation. There is therefore only one independent condition for two variables. For example, choosing any suitable value of x determines a corresponding y, and all such pairs lie on the same line. Hence option A, infinitely many solutions, is correct. “No solution” would apply when parallel distinct lines are represented.

Multiply the second equation, \(2a+5b=250\), by 2 to get \(4a+10b=500\). Subtracting \(4a+3b=276\) gives \(7b=224\), so \(b=32\). Substituting this into \(2a+5b=250\) gives \(2a+160=250\), hence \(a=45\). For example, \(a=40\) does not satisfy both equations together. Exam tip: in elimination, first make the coefficients of one variable equal.

Adding the two equations gives \(7x-2y+3x+2y=39+21\), so \(10x=60\) and hence \(x=6\). Substituting this in \(3x+2y=21\) gives \(18+2y=21\), so \(2y=3\). Therefore, \(x+2y=6+3=9\). Option 8 would require \(2y=2\), which does not satisfy the given equations. Exam tip: add equations directly when terms with opposite coefficients cancel.

Substitute \(x=3y-4\) into \(2x+5y=37\): \(2(3y-4)+5y=37\). Thus, \(6y-8+5y=37\), so \(11y=45\). Therefore, \(y=\frac{45}{11}\). For example, \(\frac{42}{11}\) would give \(11y=42\), not \(11y=45\). Exam tip: while substituting, distribute the coefficient \(2\) to both terms inside the bracket.

Let the numbers of boys and girls be x and y respectively. Then x+y=68 and 2x+y=112. Subtracting the first equation from the second gives x=44. Therefore, y=68-44=24. Hence, the correct answer is 24. If 26 were chosen, the total would become 70, so it cannot be correct. Exam tip: In such questions, subtracting the equations often eliminates one variable directly.

Dividing the first equation by 2 gives \(x+y=17\), and dividing the second equation by 3 gives \(x-y=5\). Adding these equations gives \(2x=22\), so \(x=11\). Substituting in \(x+y=17\) gives \(y=6\). Hence, the correct solution is \(x=11,\ y=6\). In option \(x=10,\ y=7\), the sum is 17, but the difference is not 5. Exam tip: first divide by any common factor to simplify the equations before eliminating a variable.

From (5x-y=11), put (y=5x-11) in the first equation and solve. In exams, combine terms carefully after substitution.

Multiply the second equation by (2) to eliminate (y), then find (x). In exams, eliminate one variable first and then calculate the required expression.

Multiply the first equation by (2) to eliminate (x) and get (y=5). In exams, watch the signs after multiplication.