Formulate at least three equivalent equations for each of the followings:
- $$\dfrac{x}{2}-7=\dfrac{1}{2}$$
- $$3x-5=10$$
Development:
- We have to move and operate terms of the equations to obtain equivalents.
For example, a first step in the first equation is to multiply it by $$2$$ in order to eliminate the denominators:
$$$2 \Big[ \dfrac{x}{2}-7=\dfrac{1}{2}\Big] \Rightarrow x-14=1$$$
The equation that is obtained is equivalent to the initial.
Now we can express $$x$$ as $$3x-2x$$, but moving $$2x$$ to the second term, so its sign changes:
$$$3x-2x-14=1 \Rightarrow 3x-14=2x+1$$$
Finally, if we move a unit of the term $$14$$ to the other side, we obtain:
$$$3x-13=2x+2$$$
Then we can extract common factor in the second member, after which we introduce brackets:
$$$3x-13=2(x+1)$$$
- In the second equation we can move the number $$10$$:
$$$3x-5-10=0$$$
Also, the independent terms can be operated:
$$$3x-15=0$$$
Now, the equation can be simplified if it is divided by $$3$$:
$$$[3x-15=0]/3 \Rightarrow x-5=0$$$
With the latter step we have already obtained the three equivalent equations, but we might obtain many others, it is enough breaking down terms and moving them from side to side of the equality.
Solution:
- $$x-14=1; 3x-14=2x+1; 3x-13=2(x+1)$$
- $$3x-5-10=0; 3x-15=0; x-5=0$$