What is the Transposing Method (Shortcut) in solving linear equations?

Oct 8, 2015

It is a popular world wide algebra solving process that performs by moving (transposing) algebraic terms from one side to the other side of an equation, while keeping the equation balanced.

Explanation:

Some advantages of the Transposing Method.
1. It proceeds faster and it helps avoid the double writing of terms (variables, numbers, letters) on both sides of the equation in every solving step.
Exp 1. Solve: 5x + a - 2b - 5 = 2x - 2a + b - 3
5x - 2x = -2a + b - 3 - a + 2b + 5
3x = - 3a + 3b + 2
$x = - a + b + \frac{2}{3}$
2. The "smart move" of the Transposing Method allows students to smartly avoid doing operations such as cross multiplication and distributive multiplication that are sometimes unnecessary.
Exp 2. Solve $\frac{3 t}{t - 1} = \frac{5}{x - 7} .$
Don't proceed cross multiplication and distributive multiplication.
$\left(x - 7\right) = \frac{5 \left(t - 1\right)}{3 t}$
$x = 7 + \frac{5 \left(t - 1\right)}{3 t}$
3. It easily helps transform math and science formulas.
Exp 3. Transform $\frac{1}{f} = \frac{1}{d 1} + \frac{1}{d 2}$ to get d2 in terms of others.
$\frac{1}{d 2} = \frac{1}{f} - \frac{1}{d 1} = \frac{d 1 - f}{f d 1}$
$d 2 = \frac{f d 1}{d 1 - f}$

Apr 10, 2016

Transposing Method is a world wide solving process that should be taught at Algebra 1 level. This method will greatly improve students' math skills.

Explanation:

The balancing method looks simple, reasonable, easy to understand, at the beginning of learning equation solving.
Students are taught to do in the right side what they did to the left side.
However, when the equation get more complicated at higher levels, the abundant double writing of algebra terms, on both sides of the equation, takes too much time. It also makes students confused and easily committed mistakes.
Here is an example of disavantage of the balancing method.
Solve: $\frac{m + 1}{m - 1} = \frac{2 m}{x - 5}$. Cross multiply:
$\left(m + 1\right) \left(x - 5\right) = 2 m \left(m - 1\right)$
$\left(m + 1\right) x - 5 \left(m + 1\right) = 2 m \left(m - 1\right)$
+ 5(m + 1) = + 5(m + 1)
(m + 1)x = 2m(m - 1) + 5(m + 1)
:(m + 1) = :(m + 1)
$x = \frac{2 m \left(m - 1\right)}{m + 1} + 5$
Compare to solving by transposing method:
$\left(x - 5\right) = \frac{\left(2 m\right) \left(m - 1\right)}{m + 1}$
$x = 5 + \frac{\left(2 m\right) \left(m - 1\right)}{m + 1}$