This post follows on from this post where the logic for the below is discussed. I am not going to define here what easy means!

Here is the strategy/guiding principle:

Fundamental Principle of Solving ‘Easy’ Equations

Identify what is difficult or troublesome about the equation and get rid of it. As long as you do the same thing to both numbers (the “Lhs” and the “Rhs”), the equation will be replaced by a simpler equation with the same solution.

Inverses

For the moment think of a function as an object f that associates to an input number x an output number f(x):

x\mapsto f(x),

which could be read as x is sent to f(x).

Examples of functions include:

  • the add-three function,  e.g. 4\mapsto 7.
  • the subtract-seven function, e.g. 26\mapsto 19.
  • the multiply-by-eight function, e.g. 7\mapsto 56.
  • the divide-by-four function, e.g. 44\mapsto 11.
  • the squaring function, e.g. 5\mapsto 25.
  • the square root function, e.g. 100\mapsto 10.
  • a power function, e.g. 2\mapsto 2^4=16.
  • a root function, e.g. 27\mapsto \sqrt[3]{27}=3.

Some functions have inverses f^{-1} that undo the action of f.

x\mapsto f(x)\mapsto f^{-1}(f(x))=x.

Call f(x) by y:

y=f(x)\mapsto f^{-1}(f(x))=x\mapsto f(x)=y,

that is if f^{-1} is an inverse of f then, usually, f is an inverse of $f^{-1}$.

For example, we have inverses of the above:

  • the inverse of the plus three function is the subtract three function, e.g. 4\mapsto 7\mapsto 4.
  • the inverse of the subtract seven function is the add seven function, e.g 26\mapsto 19\mapsto 26.
  • the inverse of the times eight function function is the divide-by-eight function, e.g. 7\mapsto 56\mapsto 7
  • the inverse of the divide-by-four is the times four function, e.g. 44\mapsto 11\mapsto 44.

 

So… this “get rid of it” mentioned in the above principle is “apply the appropriate inverse”. In the below when I write \underset{f^{-1}}{\Rightarrow} it means I applied the inverse function f^{-1} to both of the numbers “Lhs” and “Rhs”.

Examples

Solve each of the following for x. As per the previous post the first step is the implicit assumption that x makes the equation true.

  • x-3=-10.

What is troublesome here is we want to conclude that x= some number but here we have the subtract 3 function:

x\mapsto x-3,

messing things up. Apply the inverse function:

\displaystyle \underset{+3}{\Rightarrow}x=-7.

  • \frac{x}{5}=1.

What is troublesome here is we want to conclude that x= some number but here we have the divide-by-five function:

\displaystyle x\mapsto \frac{x}{5}.

Apply the inverse function:

\displaystyle \underset{\times 5}{\Rightarrow}x=5.

  • \sqrt{x}=2.

Going a bit faster now. The square root is in the way: apply the inverse:

\displaystyle \underset{x^2}{\Rightarrow}x=4.

  • -3x=12.

The multiply-by-(-3) is in the way: apply the inverse:

\displaystyle \underset{\div(-3)}{\Rightarrow} x=-4.

It might us take a number of implications before we can conclude the value of x. As long as you follow the principle of doing the same thing to both numbers as explained in the previous post, there can be a number of ways to get to the correct conclusion about the value of x. For example, consider the equation:

  • 2x+3=11.

Solution 1: The plus three is in the way: apply the inverse:

\displaystyle \underset{-3}{\Rightarrow}2x=8.

The multiply-by-two is in the way: apply the inverse:

\displaystyle \underset{\div 2}{\Rightarrow}x=4.

Solution 2: The multiply-by-two is in the way: apply the inverse:

\displaystyle \underset{\div 2}{\Rightarrow}x+\frac{3}{2}=\frac{11}{2}.

Here we had to use the definition of division, multiplying out and commutativity to properly divide a sum:

\displaystyle(a+b)\div c=(a+b)\times\frac{1}{c}=\frac{1}{c}(a+b)

\displaystyle \frac{1}{c}\times a+\frac{1}{c}\times b

=a\div c+b\div c.

The plus 3/2 is in the way: apply the inverse:

\displaystyle \underset{-3/2}{\Rightarrow}x=4.

Experience will tell us it is easier to deal with the +3 before the times two.

  • 6x-7=2x+13

Here there is a big problem. You want to conclude that x must equal some number but here x appears in both numbers. This is bad. The 2x shouldn’t be there: apply the inverse. The subtract seven is also in the way: apply that inverse too:

\displaystyle \underset{-2x+7}{\Rightarrow} 4x=20.

Can you finish this off?

  • \displaystyle \frac{2x}{3}+\frac{x}{4}=\frac{11}{6}.

Here the difficulty is that we have fractions. We don’t really want the division by three, four and six: if we apply the inverses of the first two of these — multiply-by-three and multiply-by-four — we end up multiplying by twelve. This will also deal with the six we don’t want:

\displaystyle \underset{\times 12}{\Rightarrow} 12\left(\frac{2x}{3}+\frac{x}{4}\right)=12\frac{11}{6}

\displaystyle \Rightarrow 4(2x)+3(x)=2(11)

\Rightarrow 8x+3x=22,

where we used ‘multiplying out’. Can you finish this off?

  • \displaystyle 4=\frac{5}{x}.

Here the problem is we want x= some number but instead we have division-by-x: apply the inverse:

\underset{\times x}{\Rightarrow}4x=5.

Can you finish this off?

In the rest of this mini-series of posts we will talk about some subtleties to do with inverses and also solving equations with a number of variables.

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