Algebraic Method

Algebraic Method is a general term for various tools and techniques that will help us solve more complex exercises in the future. It is mostly concern about using algebraic operations to isolate variables and solve equations. This approach is fundamental for solving equations in various mathematical contexts.

Distributive Property

This property helps us to clear parentheses and assists us with more complex calculations. Let's remember how it works. Generally, we will write it like this:

Z×(X+Y)=ZX+ZY Z\times(X+Y)=ZX+ZY

Z×(XY)=ZXZY Z\times(X-Y)=ZX-ZY

Extended Distributive Property

The extended distributive property is very similar to the distributive property, but it allows us to solve exercises with expressions in parentheses that are multiplied by other expressions in parentheses.
It looks like this:

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

Factoring

The factoring method is very important. It will help us move from an expression with several terms to one that includes only one by taking out the common factor from within the parentheses.
For example:
2A+4B2A + 4B

This expression consists of two terms. We can factor it by reducin by the greatest common factor. In this case, it's the 2 2 .
We will write it as follows:

2A+4B=2×(A+2B) 2A+4B=2\times(A+2B)

Algebraic Method

In this article, we’ll explain each of these topics in detail, But each of these topics will be explained even more in detail in their respective articles.

Practice Algebraic Technique

Examples with solutions for Algebraic Technique

Exercise #1

(a+b)(c+d)= (a+b)(c+d)=

Video Solution

Step-by-Step Solution

Let's simplify the given expression, open the parentheses using the distributive property:

(a+b)(c+d)=ac+ad+bc+bd (\textcolor{red}{a}+\textcolor{blue}{b})(c+d)=\textcolor{red}{a}c+\textcolor{red}{a}d+\textcolor{blue}{b}c+\textcolor{blue}{b}d Therefore, the correct answer is option A.

Answer

ac+ad+bc+bd \text{ac+ad}+bc+bd

Exercise #2

(a+4)(c+3)= (a+4)(c+3)=

Video Solution

Step-by-Step Solution

When we encounter a multiplication exercise of this type, we know that we must use the distributive property.

Step 1: Multiply the first factor of the first parentheses by each of the factors of the second parentheses.

Step 2: Multiply the second factor of the first parentheses by each of the factors of the second parentheses.

Step 3: Group like terms.

 

a * (c+3) =

a*c + a*3

4  * (c+3) =

4*c + 4*3

 

ac+3a+4c+12

 

There are no like terms to simplify here, so this is the solution!

Answer

ac+3a+4c+12 ac+3a+4c+12

Exercise #3

It is possible to use the distributive property to simplify the expression?

If so, what is its simplest form?

(a+c)(4+c) (a+c)(4+c)

Video Solution

Step-by-Step Solution

We simplify the given expression by opening the parentheses using the extended distributive property:

(x+y)(t+d)=xt+xd+yt+yd (\textcolor{red}{x}+\textcolor{blue}{y})(t+d)=\textcolor{red}{x}t+\textcolor{red}{x}d+\textcolor{blue}{y}t+\textcolor{blue}{y}d Keep in mind that in the distributive property formula mentioned above, we assume that the operation between the terms inside the parentheses is an addition operation, therefore, of course, we will not forget that the sign of the term's coefficient is ery important.

We will also apply the rules of multiplication of signs, so we can present any expression within parentheses that's opened with the distributive property as an expression with addition between all the terms.

In this expression we only have addition signs in parentheses, therefore we go directly to opening the parentheses,

We start by opening the parentheses:

(x+c)(4+c)x4+xc+c4+cc4x+xc+4c+c2 (\textcolor{red}{x}+\textcolor{blue}{c})(4+c)\\ \textcolor{red}{x}\cdot 4+\textcolor{red}{x}\cdot c+\textcolor{blue}{c}\cdot 4+\textcolor{blue}{c} \cdot c\\ 4x+xc+4c+c^2 To simplify this expression, we use the power law for multiplication between terms with identical bases:

aman=am+n a^m\cdot a^n=a^{m+n}

In the next step like terms come into play.

We define like terms as terms in which the variables (in this case, x and c) have identical powers (in the absence of one of the variables from the expression, we will refer to its power as zero power, this is because raising any number to the power of zero results in 1).

We will also use the substitution property, and we will order the expression from the highest to the lowest power from left to right (we will refer to the regular integer as the power of zero),

Keep in mind that in this new expression there are four different terms, this is because there is not even one pair of terms in which the variables (different) have the same power. Also it is already ordered by power, therefore the expression we have is the final and most simplified expression:4x+xc+4c+c2c2+xc+4x+4c \textcolor{purple}{4x}\textcolor{green}{+xc}\textcolor{black}{+4c}\textcolor{orange}{+c^2 }\\ \textcolor{orange}{c^2 }\textcolor{green}{+xc}\textcolor{purple}{+4x}\textcolor{black}{+4c}\\ We highlight the different terms using colors and, as emphasized before, we make sure that the main sign of the term is correct.

We use the substitution property for multiplication to note that the correct answer is option A.

Answer

Yes, the meaning is 4x+cx+4c+c2 4x+cx+4c+c^2

Exercise #4

It is possible to use the distributive property to simplify the expression below?

What is its simplified form?

(ab)(cd) (ab)(c d)

Video Solution

Step-by-Step Solution

Let's remember the extended distributive property:

(a+b)(c+d)=ac+ad+bc+bd (\textcolor{red}{a}+\textcolor{blue}{b})(c+d)=\textcolor{red}{a}c+\textcolor{red}{a}d+\textcolor{blue}{b}c+\textcolor{blue}{b}d Note that the operation between the terms inside the parentheses is a multiplication operation:

(ab)(cd) (a b)(c d) Unlike in the extended distributive property previously mentioned, which is addition (or subtraction, which is actually the addition of the term with a minus sign),

Also, we notice that since there is a multiplication among all the terms, both inside the parentheses and between the parentheses, this is a simple multiplication and the parentheses are actually not necessary and can be remoed. We get:

(ab)(cd)=abcd (a b)(c d)= \\ abcd Therefore, opening the parentheses in the given expression using the extended distributive property is incorrect and produces an incorrect result.

Therefore, the correct answer is option d.

Answer

No, abcd abcd .

Exercise #5

(3+20)×(12+4)= (3+20)\times(12+4)=

Video Solution

Step-by-Step Solution

Simplify this expression paying attention to the order of arithmetic operations. Exponentiation precedes multiplication whilst division precedes addition and subtraction. Parentheses precede all of the above.

Therefore, let's first start by simplifying the expressions within the parentheses. Then we can proceed to perform the multiplication between them:

(3+20)(12+4)=2316=368 (3+20)\cdot(12+4)=\\ 23\cdot16=\\ 368 Therefore, the correct answer is option A.

Answer

368

Exercise #6

(12+2)×(3+5)= (12+2)\times(3+5)=

Video Solution

Step-by-Step Solution

Simplify this expression by paying attention to the order of arithmetic operations which states that exponentiation precedes multiplication, division precedes addition and subtraction and that parentheses precede all of the above.

Thus, let's begin by simplifying the expressions within the parentheses, and following this, the multiplication between them.

(12+2)(3+5)=148=112 (12+2)\cdot(3+5)= \\ 14\cdot8=\\ 112 Therefore, the correct answer is option C.

Answer

112

Exercise #7

(35+4)×(10+5)= (35+4)\times(10+5)=

Video Solution

Step-by-Step Solution

We begin by opening the parentheses using the extended distributive property to create a long addition exercise:

We then multiply the first term of the left parenthesis by the first term of the right parenthesis.

We multiply the first term of the left parenthesis by the second term of the right parenthesis.

Now we multiply the second term of the left parenthesis by the first term of the left parenthesis.

Finally, we multiply the second term of the left parenthesis by the second term of the right parenthesis.

In the following way:

(35×10)+(35×5)+(4×10)+(4×5)= (35\times10)+(35\times5)+(4\times10)+(4\times5)=

We solve each of the exercises within parentheses:

350+175+40+20= 350+175+40+20=

We solve the exercise from left to right:

350+175=525 350+175=525

525+40=565 525+40=565

565+20=585 565+20=585

Answer

585

Exercise #8

Break down the expression into basic terms:

8y2 8y^2

Step-by-Step Solution

To break down the expression 8y2 8y^2 , we identify the basic components. The expression y2 y^2 is a shorthand fory×y y \times y . Therefore, 8y2 8y^2 can be decomposed as 8yy 8 \cdot y \cdot y .

Answer

8yy 8\cdot y\cdot y

Exercise #9

Break down the expression into basic terms:

3a3 3a^3

Step-by-Step Solution

To break down the expression 3a3 3a^3 , we recognize that a3 a^3 means a×a×a a \times a \times a . Therefore, 3a3 3a^3 can be decomposed as 3aaa 3 \cdot a\cdot a\cdot a .

Answer

3aaa 3 \cdot a\cdot a\cdot a

Exercise #10

Break down the expression into basic terms:

8y 8y

Step-by-Step Solution

To break down the expression 8y 8y , we can see it as the multiplication of 8 8 and y y :

8y=8y 8y = 8 \cdot y

This shows the expression as a product of two factors, 8 8 and y y .

Answer

8y 8\cdot y

Exercise #11

Break down the expression into basic terms:

5m 5m

Step-by-Step Solution

To break down the expression 5m 5m , we recognize it as the product of 5 5 and m m :

5m=5m 5m = 5 \cdot m

This expression can be seen as a multiplication of the constant 5 5 and the variable m m .

Answer

5m 5\cdot m

Exercise #12

Break down the expression into basic terms:

5x2 5x^2

Step-by-Step Solution

To break down the expression 5x2 5x^2 into its basic terms, we identify each component in the expression:

5is a constant multiplier 5 \, \text{is a constant multiplier}

x2 x^2 means xx x \cdot x

Therefore, 5x2 5x^2 can be rewritten as 5xx 5 \cdot x \cdot x .

Answer

5xx 5\cdot x\cdot x

Exercise #13

Break down the expression into basic terms:

3y3 3y^3

Step-by-Step Solution

To break down the expression 3y3 3y^3 into its basic terms, we understand the components of the expression:

3is a constant multiplier 3 \, \text{is a constant multiplier}

y3 y^3 can be rewritten as yyy y \cdot y \cdot y

Thus, 3y3 3y^3 can be decomposed into 3yyy 3 \cdot y \cdot y \cdot y .

Answer

3yyy 3\cdot y\cdot y \cdot y

Exercise #14

Break down the expression into basic terms:

4a2 4a^2

Step-by-Step Solution

To break down the expression 4a2 4a^2 into basic terms, we need to look at each factor:

4is a constant multiplier 4 \, \text{is a constant multiplier}

a2 a^2 means aa a \cdot a

Hence, 4a2 4a^2 is equivalent to 4aa 4 \cdot a \cdot a .

Answer

4aa 4\cdot a\cdot a

Exercise #15

Break down the expression into basic terms:

6b2 6b^2

Step-by-Step Solution

To break down the expression 6b2 6b^2 into its fundamental parts, we analyze each element:

6is a constant multiplier 6 \, \text{is a constant multiplier}

b2 b^2 represents bb b \cdot b

Therefore, 6b2 6b^2 is decomposed as 6bb 6 \cdot b \cdot b .

Answer

6bb 6\cdot b\cdot b