REMOVING BRACKETS, SIMPLIFYING

Simplifying Algebraic Expressions

When we do Algebra, we use variables (letters) to represent Real numbers, so when we organize and operate on them, we use the same laws and properties that apply to Real numbers. The Distributive, Associative and Commutative properties apply to variables just as they do to Real Numbers. So does the Order of Operations (BEDMAS)

Example 1:

Simplify 2(x + 3) – 4 + 5x

Solution:

The distributive property says that 2(x + 3) = 2(x) + 2(3), so we get 2x + 6 from that.
Now we have 2x + 6 – 4 + 5x -- and the associative property says that's the same as
2x + 5x + 6 – 4 when we rearrange the order of the terms.

So, 2(x + 3) – 4 + 5x = 2x + 5x + 6 – 4 = 7x + 2

Brackets: Positives and Negatives

Let's look at an example,then we'll state the rules:

Example 2:

Simplify 2(x + 3) – 4( 5x – 6)

Solution:

As in example 1, the first bracket = 2x + 6.
The 2nd bracket is multiplied by negative 4 -- so we apply the rules of multiplication for signed numbers (integers).

– 4 × 5x = – 20x ......... and .........– 4 × – 6 = + 24
So, 2(x + 3) – 4( 5x – 6) = 2x + 6 – 20x + 24
the same as 2x – 20x + 6 + 24 = – 18x + 30

 When we MULTIPLY or DIVIDE,the result of 2 opposite signs is negative.The result of 2 identical signs is positive.

Beware!! When we see a negative sign before a bracket, it means the bracket is multiplied by negative ONE -- but we don't write the 1 -- we just write something like – (x + 3). We have to change both signs in the brackets when we remove them because we're multiplying by – 1.

Example 3:

– (2x + 3) = – 2x – 3 ..... and ..... – 4( – 2x – 3) = + 8x + 12

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Collecting Like Terms:

Once we remove brackets, we have to collect like terms to complete the simplification. So what are like terms? Again -- as always in math -- the words say it all.

in like terms the variable and its powers are identical.

So, x³ , 17 x³ , 9 x³ and 4267 x³ are all like terms

When I collect them I will get 4276 x³

Whereas x³, 17x² , – 9x 7 , 2ax – 8 and 4267x 0. 357 are not like terms

because though the variable, x, is the same, the powers or exponents are different.

If we have 3x 5y + 7x - 2y + x ², we would collect like terms as follows:

3x and 7x are like terms = 10x

– 5y and 2y are like terms = – 7y

x² has no buddies, so he stands alone.

3x – 5y + 7x – 2y + x² = 10x – 7y + x 2 .

One of the greatest sources of confusion for new students of Algebra when they start collecting terms is the fact that we don't write a one (1) when we talk about one x, or one y.

We simply write x or y not 1x, or 1y.

In math, it's pretty obvious -- if we write x, we mean one x. If we meant five x, that's what we'd write -- like this -- 5x.

So, if we have 4x + x , we have 5x :

and x – 7x = – 6x
and since we divide x by itself and we get 1 (in the numerator).

We also don't write a one (1) when we talk about x to the first power. We just write x.

When we learn the laws of exponents, we learn that to multiply terms in a given variable that has been raised to various powers or exponents, we add the powers.

So, since x = x 1 , .......... then .......... x 2 · x = x 2 + 1 = x ³

Example 4:

Simplify....... 2x (6x – 3) – 4x ( 5x – 1)

Solution:

2x (6x – 3) – 4x ( 5x – 1) = 12x² – 6x – 20x² + 4x =
12x² – 20x² – 6x + 4x = – 8x² – 2x

Now get a pencil, an eraser and a note book, copy the questions,
do the practice exercise(s), then check your work with the solutions.
If you get stuck, review the examples in the lesson, then try again.

Practice Exercises

1) Simplify by removing brackets and collecting like terms.

 a) 3x – ( x + 3 ) b) 4d – 2( c + d ) c) 2r – s – 3( r + 2s ) d) 4n – x + ( n – 2x ) e) 3x – ( x – 4 ) – x f) 2r – ( 5r – 7 ) + 3 g) – 4( – 3x + 2) – 5x h) 1 – (6x – 5) + 9x

2) Simplify by removing brackets and collecting like terms.

 a) 3a(5a – 7) – 2a(a – 3) b) 4x(5 – 3x) – 2(3x – 4) c) – 3x(5 – 7x) + 2 (3 – 12x) d) – 2( – x² + 9) – (6 – 5 x²) e) – 5x²(x – 9) + 3(x³ – x²) f) 3x – (4x + 4) – ( – 2 + 7x)

3) if a = 4 and x = – 2, evaluate (find the number value for) the answers for #2.

4)

a) write an algebraic expression for the perimeter of this rectangle.

b) write an algebraic expression for the area of this rectangle.

c) If x = 2.7 cm, find the value of the perimeter and area.

Solutions

1) Simplify by removing brackets and collecting like terms.

 a) 3x – ( x + 3 ) = 3x – x – 3 = 2x – 3 b) 4d – 2( c + d ) = 4d – 2c – 2d = 2d – 2c c) 2r – s – 3( r + 2s ) = 2r – s – 3r – 6s = – r – 7s d) 4n – x + ( n – 2x ) = 4n – x + n – 2x = 5n – 3x e) 3x – ( x – 4 ) – x = 3x – x + 4 – x = x + 4 f) 2r – ( 5r – 7 ) + 3 = 2r – 5r + 7 + 3 = – 3r + 10 g) – 4( – 3x + 2) – 5x = 12x – 8 – 5x = 7x – 8 h) 1 – (6x – 5) + 9x = 1 – 6x + 5 + 9x = 3x + 6

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2) Simplify by removing brackets and collecting like terms.

 a) 3a(5a – 7) – 2a(a – 3) = 15a² – 21a – 2a² + 6a = 13a² – 15a b) 4x(5 – 3x) – 2(3x – 4) = 20x – 12x² – 6x + 8 = – 12x² + 14x + 8 c) – 3x(5 – 7x) + 2(3 – 12x) = – 15x + 21x² + 6 – 24x = 21x² – 39x + 6 d) – 2( – x² + 9) – (6 – 5 x²) = 2x² – 18 – 6 + 5 x² = 7x² – 24 e) – 5x²(x – 9) + 3(x³ – x²) = – 5x³ + 45x² + 3x³ – 3x² = – 2x³ + 42x² f) 3x – (4x + 4) – ( – 2 + 7x) = 3x – 4x – 4 + 2 – 7x = – 8x – 2

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3) if a = 4 and x = – 2

 a) 13a² – 15a = 13(4)² – 15(4) = 13(16) – 60 = 148 b) – 12x² + 14x + 8 = – 12(– 2)² + 14(– 2) + 8 = – 68 c) 21x² – 39x + 6 = 21(– 2)² – 39(– 2) + 6 = 168 d) 7x² – 24 = 7 (– 2)² – 24 = 4 e) – 2x³ + 42x² = –2(– 2)³ + 42 (–2)² = 184 f) – 8x – 2 = – 8(–2) – 2 = 14.

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4)

a) perimeter = 2 (4x + 2x + 7) = 2 (6x + 7) = 12x + 14
b) area = 4x(2x + 7) = 8x² + 28x
c) If x = 2.7 cm perimeter = 12(2.7) + 14 = 46.4 cm
area = 8(2.7)² + 28(2.7) = 133.92 cm²

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