Equivalent transformations 1

We discuss now how terms can be modified in such a way that the resulting term still equals the original term.

Subtraction can also be written as addition

\boxed{a-b=a+(-b)}

\boxed{a-(-b)=a+b}

Example 1

$3-4=3+(-4)$
$3-(-4)=3+4$
$a-b-3=a+(-b)+(-3)$
$a-2ab=a+(-2ab)$

The multiplication can be understood as an addition of equal numbers

\boxed{nb=n\cdot b=\underbrace{b+b+...+b}_{\text{n times a }"b"}}

where $n$ is a natural number

Example 2

$3\cdot 4=4+4+4$
$2b=b+b$
$3b=b+b+b$
$3(a+b)=(a+b)+(a+b)+(a+b)$
$3(-b)=(-b)+(-b)+(-b)=-b-b-b$

Commutative law

lat. commutare "to interchange"

If you add several variables (as usual from left to right) you can change the order of the variables. The same is true if you want to multiply several variables. So we have

\boxed{ab=ba\quad \text{and}\quad a+b=b+a}

$a+3+b+u=3+u+b+a$
$(ab+c)+abc=abc+(ab+c)$
$ab=ba$
$3abc=ba3c$
$a(a+b)(c+d+3)=(c+d+3)a(a+b)$

Note that subtraction and division are not commutative: $2-3\neq 3-2$ and $10:5\neq 5:10$ . But for subtraction, we have

\boxed{a-b=-b+a}

Why? Because $a-b=a+(-b)=(-b)+a=-b+a$

Associative law

lat. associare "to unite, connect, link, network", in the sense of forming groups.

If you want to add several variables, you can choose which group of variables to add first. You can simply go form left to right, or you can start with the middle two, or the last two, and so on. For example

\boxed{\underbrace{a+b+c+d}_{\text{left to right}} = \underbrace{a+(b+c)+d}_{\text{first the two in the middle}}= \underbrace{a+b+(c+d)}_{\text{first the last two}}}

The same is true if you multiply several variables (or subterms):

\boxed{\underbrace{abcd}_{\text{left to right}} = \underbrace{a(bc)d}_{\text{first the two in the middle}}= \underbrace{ab(cd)}_{\text{first the last two}}}

You can group variables into subterms using brackets as long as the operations between them are all $+$ , or all $\cdot$

Minus times Minus

It is

\boxed{(-a)b=-ab,\quad a(-b)=-ab, \quad (-a)(-b)=ab}

This follows form the commutative and associative laws. Uncollapse to see why.

Show

Indeed,

(-a)(-b)=((-1)a)((-1)b)=(-1)a(-1)b=(-1)(-1)ab=1ab=ab

and

(-a)b=((-1)a)b=(-1)ab=-ab

Exercise 1

What is correct, what not? Justify

$2xyz+3yzx+5zxy=3yxz+2xyz+5zyx$
$3k+12x = 12x+3k$
$c-d = c+(-d)$
$c-d = d-c$
$c-d = -d+c$
$c-d = d+(-c)$
$za = az$
$2x\cdot (-k) = -2xk$
$-3a\cdot (-2b) = -6ab$
$-3a\cdot (-2b) = 6ab$
$-x\cdot(-y)\cdot(-z) = xyz$
$-x\cdot(-y)\cdot(-z) = -xyz$
$3ab\cdot 5c=15abc$
$4x\cdot 5y \cdot (-x) = 20xy$
$(-a)(-b)=ab$

Solution

$2xyz+3yzx+5zxy=3yxz+2xyz+5zyx$ correct, you can rearrange each $x, y$ and $z$ , and also the terms between the plus signs(commutative law).
$3k+12x = 12x+3k$ correct, commutative law
$c-d = c+(-d)$ correct, subtraction can be written as an addition
$c-d = d-c$ wrong, e.g. insert $c=1$ and $d=2$
$c-d = -d+c$ correct, we can write $c-d=c+(-d)=(-d)+c=-d+c$
$c-d = d+(-c)$ wrong, e.g. insert $c=1$ and $d=2$
$za = az$ correct, commutative law
$2x\cdot (-k) = -2xk$ correct, we can write $2x\cdot (-k)=2x(-1k)=2x(-1)k=(-1)2xk=-2xk$ . We use the associative and the commutative laws.
$-3a\cdot (-2b) = -6ab$ wrong, e.g. $a=1$ and $b=1$
$-3a\cdot (-2b) = 6ab$ correct, $-3a(-2b)=-3a(-2)b=-3(-2)ab=6ab$ . We use the associative and the commutative laws.
$-x\cdot(-y)\cdot(-z) = xyz$ wrong
$-x\cdot(-y)\cdot(-z) = -xyz$ correct
$3ab\cdot 5c=15abc$ correct
$4x\cdot 5y \cdot (-x) = 20xy$ wrong
$(-a)(-b)=ab$ correct

Distributive law

This law tells you what happens if you want to get rid of the brackets in a term like $a(b+c)$ :

\boxed{a(b+c)=ab+ac}

So the multiplication distributes over the addition, which we can illustrate by using the dot notation for the multiplication:

\boxed{a{\color{red}\cdot}(b+c)=a{\color{red}\cdot} b+a{\color{red}\cdot }c}

Example 3

$3\cdot (4+5) = 3\cdot 4 + 3\cdot 5$
$2a(x+y) = 2ax+2ay$
$5k(3t+2b) = 5k\cdot 3t+5k\cdot 2b = 15kt+10kb$

Because addition can be traced back to subtraction, the distributive law applies accordingly:

a(b-c) = a\cdot \left(b+(-c)\right) = ab+a\cdot(-c) = ab+(-ac)=ab-ac

So we also have the following rule:

\boxed{a(b-c) = ab-ac}

Example 4

$3\cdot (4-5) = 3\cdot 4 - 3\cdot 5$
$2a(x-y) = 2ax-2ay$
$5k(3t-2b) = 5k\cdot 3t-5k\cdot 2b = 15kt-10kb$

It is important to note that the distributive law applies only to the case where you multiply a variable with a bracket which contains a plus or minus. So the following is wrong:

a(bc)=abac \quad \text{WRONG}!!

Actually the associative law deals this case and tells us how to get rid of the brackets:

a(bc)=abc \quad \text{CORRECT}!!

Expanding and factoring out

"Ausmultiplizieren und Ausklammern."

If the distributive law is applied from left to right, therefore the parentheses are resolved, and two terms are introduced with are added or subtracted, we call this to expand (or multiply out) the term. For example

$3(a+b)=3a+3b$
$3(a-b)=3a-3b$

If the law is applied from right to left, therefore factors are formed (an multiplication is introduced), we call this to factor out the term.

Example 5

$3a+3b=3(a+b)$
$3a-3b=3(a-b)$
$1.5a+2.1a=a(1.5+2.1)=3.6a$

What is the rule for factoring out? We need two terms, which we add or subtract, such as

3abx+3xc

Then we identify the common numbers and variables in each term

{\color{red}3}ab{\color{red}x}+{\color{red}3x}c

and take those to the front and wrap the rest of the two terms into brackets

{\color{red}3x}(ab+c)

A tricky situation occurs in the following case:

{\color{red}3}a+{\color{red}3}

If we take the three to the front, and wrap the rest into brackets, we get

{\color{red}3}(a)

But clearly this is wrong: ${\color{red}3}a+{\color{red}3}\neq {\color{red}3}(a)$ . For example, use $a=1$ . So what went wrong? Well, the three can be written as $3\cdot 1$ , so we get

{\color{red}3}a+{\color{red}3}={\color{red}3}a+{\color{red}3}\cdot 1={\color{red}3}(a+1)

and this is now correct, which can easily be verified by expanding the term $3(a+1)$ .

Exercise 2

Expand or factor out:

$3b(2k+z)$
$n(3-p)$
$ax-bx$
$-2u(5r-4p)$
$2a-2$
$2ab-xab$
$-(x+2)$
$2+2x$
$2x+2.3x$
$1.6 abc-0.5 abc+0.1 abc$
$\frac{2}{3}a +\frac{1}{5}a$

Solution

$3b(2k+z)=3b2k+3bz=6bk+3bz$
$n(3-p)=n\cdot 3-pn=3n-pn$
$ax-bx=x(a-b)$
$-2u(5r-4p)=-2u5r+2u4p=-10ur+8up$
$2a-2=2(a-1)$
$2ab-xab=ab(2-x)$
$-(x+2)=-x-2$
$2+2x=2\cdot 1+2\cdot x=2(1+x)$
$2x+2.3x=x(2+2.3)=x(4.3)=4.3x$
$1.6 abc-0.5 abc+0.1 abc=abc(1.6-0.5+0.1)=abc(1.2)=1.2\cdot abc$
$\frac{2}{3}a +\frac{1}{5}a= a(\frac{2}{3}+\frac{1}{5})=a\cdot \frac{13}{15}=\frac{13}{15}a$

A minus in front of a bracket

If you have a minus in from of a bracket, and you want to resolve the bracket, then

take the minus away,
change all signs in the bracket, and
remove the bracket

Example 6

$-(3x+5y) = -(+3x+5y)=-3x-5y$
$-(9a-2b) = -(+9a-2b)-9a+2b$
$7k+3b-(2a+4b) = 7k+3b-(+2a+4b)= 7k+3b-2a-4b = 7k-2a-b$
$3x+8y-(x-4y) = 3x+8y-(+x-4y) = 3x+8y-x+4y = 2x+12y$
$-(k-p) = -(+k-p)= -k+p$
$-(-3x-2y) =3x+2y$
$-(12a-9b) = -(+12a-9b) = -12a+9b$

Why is that? Uncollapse to see an explanation.

Show

We already know that $-a=-1\cdot a$ . The same is true for parentheses. A minus for the bracket is nothing else than the bracket multiplied by $-1$ :

-(a+b)=-1\cdot (a+b)

Expanding this term, we get

-(a+b)=-a-b

and similar

-(a-b)=-a+b=b-a

Exercise 3

Resolve the brackets and if possible simplify the term

$-(w+z)$
$-(2i+3j)$
$-(7k-9s)$
$-(21uv-4x)$
$18p-4d-(3u+2a)$
$9x+2y-(3a-6x)$
$2a-3b-(6b-2a)$

Solution

$-(w+z)=-w-z$
$-(2i+3j)=-2i-3j$
$-(7k-9s)=-7k+9s$
$-(21uv-4x)=-21uv+4x$
$18p-4d-(3u+2a)=18p-4d-3u-2a$
$9x+2y-(3a-6x)=9x+2y-3a+6x=15x+2y-3a$
$2a-3b-(6b-2a)=2a-3b-6b+2a=4a-9b$

Products of sums and differences

To expand the term

(a+b)(c+d)

do the following: Form all possible pairs of variables, where one variable is from the first and the other variable from the second bracket. Multiply those variables, and add everything up.

(a+b)(c+d)=ac+ad+bc+bd

If there is a minus involved, the same rule procedure applies, but think of the minus sign stuck to the variable.

(a-b)(c+d)=ac+ad-bc-bd

(a+b)(c-d)=ac-ad+bc-bd

(a-b)(c-d)=ac-ad-bc+bd

Example 7

$(w+x)(y+z) = wy+wz+xy+xz$
$(2p+n)(k+3r) = 2pk+6pr+nk+3nr$
$(4a+3b)(5x+2y) = 20ax+8ay+15bx+6by$
$(a+b+c)(x+y) = ax+bx+cx+ay+by+cy$
$(2r+s)(3w+8q+m) = 6rw+16rq+2rm+3sw+8sq+sm$
$(v-n)(g-q) = vg-vq-ng+nq$
$(7x-2t)(4y-6z) = 28xy-42xz-8ty+12tz$
$(2w-az-4r)(3p-7j) = 6wp-14wj-3azp+7azj-12rp+28rj$

Why is that? Uncollapse to see an explanation.

Show

We want to multiply out the term $(a+b)(c+d)$ . To do this, we apply the distributive law twice

(a+b)(c+d)=(a+b)c+(a+b)d=c(a+b)+d(a+b)=ca+cb+da+db=ac+ad+bc+bd

Similar for the term $(a-b)(c+d)$ : we write the subtraction as an addition $(a+(-b))(c+d)$ and apply the above rule. We get

(a-b)(c+d)=(a+(-b))(c+d)=ac+ad+(-b)c+(-b)d=ac+ad-bc-bd

Finally, there may be more than two parentheses, e.g.

(a+b)(n+k)(x+y)

Then it is advisable to first resolve only two brackets:

(a+b)(n+k)(x+y) = (an+ak+bn+bk)(x+y)

And then form all possible pairs to expand this term:

(an+ak+bn+bk)(x+y)=anx+akx+bnx+bkx+any+aky+bny+bky

Exercise 4

Expand and if possible simplify

$(d+i)(p+n)$
$(2x+z)(3y+u)$
$(t+s)(p-a)$
$(3z-f)(2e+k)$
$(c-x)(d-y)$
$(3k-2x)(9y-8z)$
$3p(8z+4f)(y-4e)$
$-2i(h-3z)(4t+9a)$
$(q+r)(t+z)(m+v)$
$(2a+u)(e-p)(3x+y)$
$(2a-b)(a+b)$
$-(3x-2y)(x+y)$
Is this equation correct? $(x-y)(a-b)=-(y-x)(b-a)$

Solution

$(d+i)(p+n)=d n + d p + i n + i p$
$(2x+z)(3y+u)=2 u x + u z + 6 x y + 3 y z$
$(t+s)(p-a)=-a s - a t + p s + p t$
$(3z-f)(2e+k)=-f k - 2 f e + 3 k z + 6 e z$
$(c-x)(d-y)=c d - c y - d x + x y$
$(3k-2x)(9y-8z)=27ky-24kz-18xy+16xz$
$3p(8z+4f)(y-4e)=3p(8yz-32ez+4fy-16ef)=24pyz-96pez+12pfy-48pef$
$-2i(h-3z)(4t+9a)=-2i(4ht+9ah-12tz-27az)\\=-8iht-18iah+24itz+54iaz$
$(q+r)(t+z)(m+v)=(q+r)(m t + m z + t v + v z)\\=m q t + m q z + m r t + m r z + q t v + q v z + r t v + r v z$
$(2a+u)(e-p)(3x+y)=(2a+u)(-3 p x - p y + 3 e x + e y)\\=-6 a p x - 2 a p y + 6 e a x + 2 e a y - 3 p u x - p u y + 3 e u x + e u y$
$(2a-b)(a+b)=2a^2+2ab-ab-b^2=2a^2+ab-b^2$
$-(3x-2y)(x+y)=-(3x^2+3xy-2xy-2y^2)=-(3x^2+xy-2y^2) = -3x^2-xy+2y^2$
Is this statement correct? $(x-y)(a-b)=-(y-x)(b-a)$ No, because the left side is $(x-y)(a-b)=ax-bx-ay+by$ and the right side is $-(y-x)(b-a)=-(by-ay-bx+ax)\\=-by+ay+bx-ax=-ax+bx+ay-by$ Clearly these terms are not the same.

The binomial formulas

If we expand term $(a+b)(c+d)$ for special cases, we obtain the binomial formulas:

\begin{array}{llllll} (a+b)^2 & =& (a+b)(a+b) &= & a^2+2ab+b^2 &\quad \text{ 1st binomial form}\\ (a-b)^2 & = & (a-b)(a-b) &=& a^2-2ab+b^2& \quad \text{ 2nd binomial form}\\ && (a+b)(a-b)& =& a^2-b^2 & \quad \text{ 3rd binomial form} \end{array}

The binomial formulas can be expressed in words nicely:

"square of the first, plus twice the product of both, plus square of the second summand"
"square of the first, minus twice the product of both, plus square of the second summand"
"square of the first minus square of the second summand".

The binomial formulas are obtained by simply expanding the terms:

$(a+b)^2=(a+b)(a+b)=aa+ab+ba+bb=a^2+2ab+b^2$
$(a-b)^2=(a-b)(a-b)=aa-ab-ba+bb=a^2-2ab+b^2$
$(a+b)(a-b)=aa-ab+ba-bb=a^2-b^2$

Example 8

$(x+y)^2=x^2+2xy+y^2$
$(2x+y)^2 =(2x)^2+2\cdot 2xy+y^2 = 4x^2+4xy+y^2$
$(3x+5y)^2 = (3x)^2+2\cdot 3x5y+(5y)^2=9x^2+30xy+25y^2$
$(x-y)^2 = x^2-2xy+y^2$
$(2x-y)^2= (2x)^2-2\cdot 2xy+y^2= 4x^2-4xy+y^2$
$(3x-5y)^2= (3x)^2-2\cdot 3x5y+(5y)^2= 9x^2-30xy+25y^2$
$(x+y)(x-y)= x^2-y^2$
$(2x+y)(2x-y) = (2x)^2-y^2=4x^2-y^2$
$(3x+5y)(3x-5y)= (3x)^2-(5y)^2= 9x^2-25y^2$

It is useful to know the binomial formulas by heart, especially when it comes to factorizing a term. For example, write the term

s^2+2st+t^2

as the product of two factors. Knowing that this is 1st binomial form, we can write

s^2+2st+t^2 =(s+t)^2

Writing terms as the product of two factors is often useful for solving equations, or simplifying fractions, as we will see later.

Exercise 5

Expand or factor out using the binomial formulas:

$(n+p)^2$
$(z-w)^2$
$(3p+q)^2$
$(i-8u)^2$
$(c+d)(c-d)$
$(e-p)(e+p)$
$(5x+y)(5x-y)$
$(m+b)(b-m)$
$x^2+y^2+2xy$
$p^2-2pt+t^2$
$4x^2+y^2+4xy$
$25a^2+36b^2-60ab$
$r^2-t^2$
$64c^2-81d^2$

Solution

$(n+p)^2 = n^2+2np+p^2$
$(z-w)^2=z^2-2zw+w^2$
$(3p+q)^2=(3p)^2+2(3p)q+q^2=9p^2+6pq+q^2$
$(i-8u)^2=i^2-16iu+(8u)^2=i^2-16iu+64u^2$
$(c+d)(c-d)=c^2-d^2$
$(e-p)(e+p)=e^2-p^2$
$(5x+y)(5x-y)=(5x)^2-y^2=25x^2-y^2$
$(m+b)(b-m)=(b+m)(b-m)=b^2-m^2$
$x^2+y^2+2xy=x^2+2xy+y^2=(x+y)^2$
$p^2-2pt+t^2=(p-t)^2$
$4x^2+y^2+4xy=(2x)^2+4xy+y^2$ . Candidate: $(2x+y)^2$ . Expand to verify. Correct!
$25a^2+36b^2-60ab=(5a)^2-60ab+(6b)^2$ . Candidate: $(5a-6b)^2$ . Expand to verify. Correct!
$r^2-t^2=(r-t)(r+t)$
$64c^2-81d^2=(8c)^2-(9d)^2=(8c-9d)(8c+9d)$