Algebraic methods: Substitution method and Elimination method.

Multiply the second equation by (3) to eliminate (x). In exams, making equal coefficients makes subtraction easier.

The direct answer is option C: the larger number is 30. Let the larger number be x and the smaller be y. “Nine more” gives x-y=9, and their sum gives x+y=51. Add the equations: 2x=60, so x=30. Then y=21, and 30-21=9 while 30+21=51. Option C satisfies both facts. Option A, 28, would make the smaller number 19, giving a difference of 9 but a sum of 47. Option B, 29, would pair with 20, giving sum 49. Option D, 31, would pair with 22, giving sum 53. The key is to assign the larger number to x, so subtraction produces a positive 9. Adding the equations is the quickest elimination method because the y terms cancel.

Let the tens digit be (x) and the units digit be (y), so (x-y=4). In exams, write the original number as (10x+y) and the reversed number as (10y+x).

From the second equation, put (y=6x-23) to get (x=6), (y=4). In exams, write the ratio in simplest form.

From (l-b=8) and (l+b=36), we get (l=22), (b=14). In exams, do not forget to use (l\times b) for area.

Multiply the two equations by 4 and 3 respectively to get \(8x+12y=72\) and \(15x-12y=3\). Adding them gives \(23x=75\), so \(x=\frac{75}{23}\). Substituting in \(2x+3y=18\) gives \(y=\frac{88}{23}\). Hence, \(x-2y=\frac{75}{23}-2\left(\frac{88}{23}\right)=-\frac{101}{23}\). Although \(-4\) is a nearby integer, it is not the exact value. Exam tip: when a fractional result is obtained, retain the exact fraction unless rounding is specifically asked.

Form (b+g=48) and (b-g=6), then subtract. In exams, such questions directly form sum and difference equations.

Multiply the first equation by (2) to eliminate (y) and get (x=3), (y=8). In exams, do not alter the original solution while evaluating the expression.

Let the numerator be (x) and denominator be (y), giving (y-x=3) and (\frac{x+2}{y+1}=\frac{3}{4}). In exams, solve the simple linear equations after cross multiplication.

Putting (x=6) gives (y=5). Then (24+5a=35) gives (a=\frac{11}{5}), so check option calculations carefully.

From (2x-y=7), put (y=2x-7) and solve the first equation. In exams, substitute the isolated variable into the correct equation.

Elimination gives (x=7), (y=2), so (xy=14). In exams, verify the solution before matching options.

Downstream speed is (15) and upstream speed is (7), so (b+s=15), (b-s=7). In exams, the stream speed is half the difference.

Form (12p+8q=184) and (5p+6q=88), then solve. In exams, take the prices of different items as separate variables.

Given \(y=5\), substitute it into \(3x+2y=25\): \(3x+10=25\), so \(x=5\). Now put \(x=5\) and \(y=5\) in \(mx-y=10\): \(5m-5=10\). Hence \(5m=15\), giving \(m=3\). Therefore, option B is correct. Exam tip: first substitute the given variable value into the equation that lets you find the other variable most directly.

The second equation is (\frac{3}{2}) times the first, so both represent the same line. In exams, if all ratios are equal, there are infinitely many solutions.

The ratio of variable coefficients is the same, but the constant ratio is different. In exams, such lines are parallel.

The direct answer is option B: the son is 12 years old now. Let the father’s present age be f and the son’s be s. Their present sum gives f+s=56. Four years ago, their ages were f-4 and s-4, and the father was five times the son: f-4=5(s-4). Expand: f-4=5s-20, so f=5s-16. Substitute into f+s=56: 5s-16+s=56, hence 6s=72 and s=12. Then f=44. Four years ago they were 40 and 8, and 40=5×8, so the result checks. Option A, 10, would make the father 46 and past ages 42 and 6, not five times. Option C, 14, gives father 42 and past ages 38 and 10. Option D, 16, gives father 40 and past ages 36 and 12. Always subtract four from both ages, not just one.

Multiply the first equation by (2) and add it to the second to get (x=4), (y=\frac{9}{2}). In exams, substitute fractional values carefully in the expression.

The direct answer is option D: x=68/7 and y=40/7. First clear the denominators in x/2+y/5=6 by multiplying every term by 10. This gives 5x+2y=60. From x-y=4, write x=y+4. Substitute: 5(y+4)+2y=60, so 7y+20=60, 7y=40, and y=40/7. Therefore x=y+4=40/7+28/7=68/7. Option D is correct. Option A has x=60/7,y=32/7; their difference is 4, but the first equation becomes 44/7, not 6. Option B has difference 4 but fails the first equation. Option C also has difference 4 but fails it. Direct checking of D gives x-y=4 and x/2+y/5=34/7+8/7=6. The memory cue is to multiply by the least common denominator before eliminating fractions.

Multiplying both equations by 10 gives \(3x+2y=27\) and \(5x-y=14\). From the second equation, \(y=5x-14\). Substituting this into the first equation gives \(3x+2(5x-14)=27\), so \(13x=55\). Hence, \(y=5\times\frac{55}{13}-14=\frac{93}{13}\). The value \(\frac{105}{13}\) does not satisfy both equations simultaneously. Exam tip: For linear equations containing decimals, first multiply by 10 or a suitable power of 10 to remove decimals.

Adding both equations gives (10x=100), so (x=10). Then (y=\frac{31}{7}), hence (x+2y=\frac{132}{7}), so no integer option is correct.

The given solution \(x=8, y=5\) must satisfy \(kx+5y=42\). Substituting gives \(8k+5(5)=42\), or \(8k+25=42\). Hence \(8k=17\), so \(k=\frac{17}{8}\). For example, \(\frac{19}{8}\) would not make the left-hand side equal to 42. Exam tip: when a solution is given, substitute it into the equation containing the unknown parameter.

For no solution, coefficients must be proportional and constants not proportional. Since (6:2=3), (a=9), and (18:11) is not the same ratio.

Adding the two equations gives \(7x-3y+2x+3y=22+23\), so \(9x=45\). Hence, \(x=5\). Substituting this into \(2x+3y=23\) gives \(10+3y=23\), and therefore \(y=\frac{13}{3}\). Thus, the correct ordered pair is \(\left(5,\frac{13}{3}\right)\). In option A, substituting \(x=4\) does not satisfy the first equation. Exam tip: When coefficients have opposite signs, add the equations first to eliminate that variable quickly.