Algebraic methods: Substitution method and Elimination method.

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

Solving gives (x=4) and (y=0). In exams, check the whole expression even when one value is zero.

Form (x-y=9) and (x+y=51), then add them. In exams, taking the larger number as (x) simplifies the solution.

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.

Form (f+s=56) and (f-4=5(s-4)), then solve. In exams, apply addition or subtraction correctly for past and future ages.

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.