Algebra/Chapter 2/Arithmetic

2.9: Algebraic Manipulation

---

Parentheses

Let's look at when it is OK to add or remove parentheses. The purpose of parentheses is to establish precedence. Precedence tells you which operation goes first. The operation rules for precedence say to evaluate the parentheses first (PEMDAS!). But, what can you do with ${\frac {x+1}{2}}={\frac {1}{2}}$ ?  $x+1$ doesn't have just one value; it has as many values as we choose to assign to $x$. This is where the distributive property shows its power. It allows us to rearrange the operations while maintaining precedence.

${\frac {x+1}{2}}={\frac {1}{2}}*(x+1)={\frac {1}{2}}x+{\frac {1}{2}}$

So we can change our equation to

${\frac {1}{2}}x+{\frac {1}{2}}={\frac {1}{2}}$

Add ${\frac {-1}{2}}$ to both sides

${\frac {1}{2}}x+{\frac {1}{2}}-{\frac {1}{2}}={\frac {1}{2}}-{\frac {1}{2}}$ ${\frac {1}{2}}x=0$

And multiply both sides by the inverse of ${\frac {1}{2}}$

${\frac {1}{2}}x*2=0*2$ $x=0$

Parentheses allow us to ensure that we treat expressions that have variables as if they were a value. For instance, if we want to know for which values the expression ${\frac {2}{x+1}}=3$ is true we need to use the properties of real numbers to place the variable $x$ by itself on one side of the equals sign. To do this we need to get $x+1$ out of the denominator of the fraction. We can do this by multiplying both sides of the equation by $(x+1)$. We don't know what the value of $(x+1)$ is, but it will always be the same thing on both sides of the equation so it doesn't change the notion of equality.

$(x+1)*{\frac {2}{x+1}}=3*(x+1)$

We use the inverse property to re-write ${\frac {2}{x+1}}$ as multiplication.

$(x+1)*2*{\frac {1}{x+1}}=3*(x+1)$

And the associative property to re-write the multiplication.

$2*(x+1)*{\frac {1}{x+1}}=3*(x+1)$

And the identity property to re-write $(x+1)*{\frac {1}{x+1}}$

$2*(1)=3*(x+1)$

Since 2 * (1) has no variables we can evaluate it. We use the distributive property to re-write 3 * (x+ 1)

$2=3*x+3*1$ $2=3*x+3$

We subtract 3 from both sides of the equation.

$2-3=3*x+3-3$ $-1=3*x$

And multiply both sides by ${\frac {1}{3}}$.

$-1*{\frac {1}{3}}=3*x*{\frac {1}{3}}$ $-{\frac {1}{3}}=x$

Using parentheses and the properties of real numbers and equality we were able to get x alone to determine the only number for which our initial statement is true.