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Get to learn how to factorize the expressions of the form a³ + b³ + c³, a + b + c=0, or can be put in the form a³ + b³ + c³, a + b + c=0 by going through this page. Check out the step-by-step process listed for solving the factorization of expressions of the form a³ + b³ + c³, a + b + c=0, and apply the related knowledge while solving similar problems. Make the most out of the problems available and get a good hold of the concept of factorization of expressions.

Also, See:

• Factorization of Expressions of the Form a^3 + b^3 + c^3 – 3abc
• Factorization of Expressions of the Form a^3 – b^3
• Factorization of Expressions of the Form a^3 + b^3

Proof of a³ + b³ + c³ = 3abc when a + b  + c = 0

Given Statement is a³ + b³ + c³ ; a+b+c=0
We can rewrite the expression a³ + b³ + c³ = a³ + b³ – (-c)³
= a³ + b³ – (a + b)³, [Since, a + b  + c = 0]
= a³ + b³ – {a³ + b³ + 3ab(a + b)}
= -3ab(a + b)
= -3ab(-c)
= 3abc

Solved Examples on Factorization of Expressions of the form a^3 + b^3 + c^3, a + b + c=0

Example 1.
Factorize (x + y)³ + (z – y)³ – (x + z)³?
Solution:
Given (x + y)³ + (z – y)³ – (x + z)³
= (x + y)³ + (z – y)³ +(-(x + z)³)  where (x+y)+(z-y)+(-(x+z)) =0
Therefore, given expression = 3(x+y)(z-y)(-(x+z))
=3(x+y)(y-z)(x+z)

Example 2.
Factorize u³ + v³ + w³?
Solution:
Given Expression = u³ + v³ + w³
=u³ + v³-(-w)³
=u³ + v³-(u+v)³
= u³ + v³– (u + v)³, [Since, u +v  + w = 0]
=u³ + v³-{u³ + v³ + 3uv(u + v)}
= u³ + v³ – {u³ + v³ + 3uv(u + v)}
= -3uv(u + v)
= -3uv(-w)
= 3uvw

Example 3.
Factorize (u + v)³ + (w – v)³ – (u + w)³?
Solution:
Given Expression = (u + v)³ + (w – v)³ – (u + w)³
=(u + v)³ + (w – v)³ – (u + w)³
=(u + v)³ + (w – v)³ +(-(u + w)³)  where (u+v)+(w-v)+(-(u+w)) =0
Therefore, given expression = 3(u+v)(w-v)(-(u+w))
=3(u+v)(v-w)(u+w)

Example 4.
Factorize the Expression (x + y)³ + (z + y)³ + (x + z)³
Solution:
Given Expression = (x + y)³ + (z + y)³ + (x + z)³
= (x + y)³ + (z + y)³ + (x + z)³ where (x + y)+ (z + y) + (x + z)= 0
=3(x+y)(z+y)(x+z)

Learn the process of factorization of algebraic expressions of the form a³ + b³ + c³ – 3abc by going through the entire article. We have got you covered with everything like how to factorize expressions of the form a³ + b³ + c³ – 3abc by giving enough examples. Get to know the derivation and formula of the expression a³ + b³ + c³ – 3abc and use it while solving problems on the topic easily. Step by Step Explanations provided for all the questions on Factorization of Expressions of the Form a³ + b³ + c³ – 3abc make it easy for you to master the concept.

Identity a³ + b³ + c³ – 3abc =  (a + b + c)(a² + b² + c² – bc – ca – ab)

Read More:

• Factorization of Expressions of the Form a^3 + b^3
• Factorization of Expressions of the Form a^3 – b^3

Factoring Expressions of the Form a³ + b³ + c³ – 3abc Examples

Example 1.
Factorize  u³ + v³ – 3uv + 1?
Solution:
Given Expression = u³ + v³ – 3uv + 1
= u³ + v³ + 1³ – 3 ∙ u ∙ v ∙ 1
= (u + v + 1)(u² + v² + 1² – v ∙ 1 – 1 ∙ u – uv)
= (u + v + 1)(u² +v² – uv – u – v + 1)

Example 2.
Factorize 8a³ + 27b³ – 90ab + 125?
Solution:
Given Expression = 8a³ + 27b³ – 90ab + 125
= (2a)³ + (3b)³ + (5)³ – 3 ∙ 2a ∙ 3b ∙ 5
= (2a + 3b + 5)(4x² + 9y² – 6ab – 15b – 10a + 25)

Example 3.
Factorize the Expression 8x³ + y³+ 27z³ –18xyz?
Solution:
Given Expression = 8x³ + y³+ 27z³ –18xyz
=(2x)³ +(y)³ +(3z)³-3.2x.y.3z[As per Identity a³ + b³ + c³ – 3abc =  (a + b + c)(a² + b² + c² – bc – ca – ab)]
=(2x+y+3z)((2x)²+(y)²+(3z)²-y.3z-3z.2x-2x.y)
=(2x+y+3z)(4x²+y²+9z²-3yz-6xz-2xy)

Example 4.
Factorize the Expression 125x³ + 64y³ + 216z³– 360xyz?
Solution:
Given Expression = 125x³ + 64y³ + 216z³– 360xyz
=(5x)³+(4y)³+(6z)³-3.5x.4y.6z[As per Identity a³ + b³ + c³ – 3abc =  (a + b + c)(a² + b² + c² – bc – ca – ab)]
=(5x+4y+6z)((5x)²+(4y)²+(6z)²-4y.6z-6z.5x-5x.4y)
=(5x+4y+6z)(25x²+16y²+36z²-24yz-30xz-20xy)

Example 5.
Factorize the Expression 27a³+b³+8c³-18abc?
Solution:
Given Expression = 27a³+b³+8c³-18abc
= (3a)³+(b)³+(2c)³-3.3a.b.2c[As per Identity a³ + b³ + c³ – 3abc =  (a + b + c)(a² + b² + c² – bc – ca – ab)]
=(3a+b+2c)((3a)²+(b)²+(2c)²-b.2c-2c.3a-3a.b)
=(3a+b+2c)(9a²+b²+4c²-2bc-6ac-3ab)

Example 6.
Factorize the Expression 125u³ + v³ –15uv + 1?
Solution:
Given Expression =125u³ + v³ –15uv + 1
= (5u)³+(v)³+(1)³-3.5u.v.1[As per Identity a³ + b³ + c³ – 3abc =  (a + b + c)(a² + b² + c² – bc – ca – ab)]
=(5u+v+1)((5u)²+(v)²+(1)²-v.1-1.5u-5u.v)
=(5u+v+1)(25u²+v²+1-v-5u-5uv)

In mathematics, an Index is a power or exponent which is raised to a variable or number. Want to master solving large number index calculations in algebra? Then, go with these laws of indices article and memorize all the rules of indices with examples. In algebra, the index is dealing with regard to numbers. Let’s dive into further modules and learn about the laws/rules of the indices with formulas and worked-out examples.

Do Refer:

• Exponents and Indices
• Number Formed by Any Power
• Expansion of Powers of Binomials and Trinomials
• Application Problems on Expansion of Powers of Binomials and Trinomials
• Worksheet on Powers of Literal Numbers

Definition of Index

A variable or a number may have an index which is a value that raises the power of a number or variable. Powers or exponents are also called Indices. In short, indices show the number of times a given number has to be multiplied. Also, it is called to be a compressed method of writing large numbers and calculations. The representation of the index is in the form:

aⁿ = a x a x a x a x……..x a (n times)

Where a is the base and n is the index.

Laws of Indices – Rules & its Explanation

We have so many primary rules or laws of indices that are important to grasp before we start knowing about indices. These index laws are utilized while solving algebraic expressions and performing the algebraic operations on indices.

Rule 1: In case a constant or variable has index ‘0’, then the result will be equal to one, irrespective of any base value.

a⁰ = 1

Rule 2: If the negative value is represented as an index, then it should be shown as the reciprocal of the positive index of the same variable.

a^-p = 1/a^p

Rule 3: Whenever we have two variables with the same base and to multiply them, we have to add their powers or indices raised to the respective base.

a^p.a^q = a^p+q

Rule 4: If you want to divide the two variables with the same base, we have to subtract the power of the denominator from the numerator power and raise it to the base.

a^p/a^q = a^p-q

Rule 5: When a variable with some index is again raised with another index, then we have to multiply both indices together raised to the power of the same base.

(a^p)^q = a^pq

Rule 6: When we have two variables with the same indices, but different bases are multiplied together. First, we have to multiply the bases and raise the same index to it.

a^p.b^p = (ab)^p

Rule 7: If two variables with various bases, but the same indices are divided, we have to divide the bases and raise the same index to it.

a^p/b^p = (a/b)^p

Rule 8: An index in the form of a fraction can be denoted as the radical form.

a^p/q = ^q√a^p

List of Laws of Indices Formula

The list of important index laws is given here in the shareable image format, kindly make use of it for quick reference before exams and memorize the laws frequently.

Indices Examples | Questions on Laws of Indices

Example 1:
Find the numerical value for each of the following (not containing exponents):
(i) 8⁰
(ii)5^-10
(iii)27^2/3

Solution:
(i)8⁰ = 1 [by using the law of indices].
(ii)5^-10 = 1/5¹⁰ = 1/5x5x5x5x5x5x5x5x5x5 = 1/9765625.
(iii)27^2/3 = 3√27² = 3² = 9.

Example 2:
Write the following expressions more concisely by using an index.
(i) b x b x b x b (ii) (x/y) x (x/y)  (iii) ac x ac x ac x ac x ac

Solution:
(i) b x b x b x b = b⁴
(ii) (x/y) x (x/y) = (x/y)²
(iii) ac x ac x ac x ac x ac = (ac)⁵

Example 3:
Multiply x²y⁴z³ and xy³z^-1

Solution:
Given that x²y⁴z³ and xy³z^-1

= x².x³ .y⁴.y³.z³.z^-1

= x²⁺³.y⁴⁺³.z^3-1

= x⁵.y⁷.z²

FAQs on Index Laws

1. What are the laws of indices with examples?

Laws of Indices are the rules for simplifying expressions including powers of the same base number. For instance, (−2)³ = −8 and (−2)⁴ = 16, so (−x)⁷ = −x⁷ and (−x)⁸ = -x⁸.

2. How many index laws are there mainly?

Majorly, there are three laws of indices. They are as such:

1. a^m x aⁿ= a^m+n
2. \(\frac { a^m }{ aⁿ } \) = a^m-n
3. (a^m)ⁿ = a ^mn

3. How do you explain an index?

A number that raises to power is called an index number. The power, also known as the index, states you know how many times you have to multiply the number by itself. As an example, 5 means that you have to multiply 2 by itself five times = 2×2×2×2×2 = 32.

If you want to learn the process of Factorization of Expressions of the Form a^3 – b^3 then this page is going to be extremely helpful as it covers how to factorize the difference of cubes. Refer to the further modules to know about the problem-solving approach when given an expression of the form a³ – b³. We have given step-by-step solutions for all the factorization example questions provided so that you can clearly understand the topic as well as resolve any doubts about the topic if any.

Formula for a³ – b³ = (a – b)(a² + ab + b²)

See More: Factorization of Expressions of the Form a^3 + b^3

Factorization of Expressions of the form a³ – b³

Example 1.
Factorize 8x³ – 27?
Solution:
Given Expression = 8x³ – 27
We can write the expression as (2x)³-(3)³
= (2x – 3){(2x)² + 2x ∙ 3 + 3²}
=(2x-3)(4x²+6x+9)

Example 2.
Factorize the Expression 1-64y³?
Solution:
Given Expression = 1-64y³
We can write the given expression as (1)³-(4y)³
=(1-4y)(1²+1.4y+(4y)²)
=(1-4y)(1=4y+16y²)

Example 3.
Factorize the Expression 216x⁶ – y⁶?
Solution:
Given Expression = (6x²)³ – (y²)³
= ( 6x²– y²){(6x²)² + 4x² ∙ y² + (y²)²}
= ( 6x²– y²)(36x⁴ + 4x²y² + y⁴)
= (6x + y)(6x – y)(36x⁴ + 4x²y² + y⁴)

Example 4.
Factorize  343x³ – 1/x³
Solution:
Given Expression = 343x³ – 1/x³
=(7x)³-(1/x)³
=(7x-1/x)((7x)²+7x.1/x+(1/x)²)
=(7x-1/x)(49x²-7+1/x²)

Example 5.
Factorize the Expression  512u³ – 64v³
Solution:
Given Expression = 512u³ – 64v³
= (8u)³– (4v)³
=(8u-4v)((8u)²+8u.4v+(4v)²)
=(8u-4v)(64u²+32uv+16v²)

Example 6.
Factorize the Expression y⁶ – z⁶
Solution:
Given Expression = y⁶ – z⁶
=(y²)³ – (z²)³
= (y² – z²){(y²)² + y² ∙ z ² + (z²)²}
= (y² – z²)(y⁴ + y² ∙ z ² + z⁴)

 

Learn how to factorize Expressions of the Form a³ + b³ by referring to the entire article. Here we will use the concept of algebraic identity to factor the given expressions. Factorizing is the process of writing an expression into a product of two expressions which results in the given expression.

Know how to factorize the expressions in the form or can be put in the form a³ + b³ with the step-by-step explanations provided. Check out the Problems on Factorization of Expressions of the form a³ + b³ and learn the problem-solving approach used.

Also, Check:

• Problems on Factorization of Expressions of the Form x^2 +(a + b)x +ab
• Factorization of Expressions of the Form x^2 + (a + b)x + ab

Problems on Factorization of Expressions a^3 + b^3

Example 1.
Factorize x³ + 64y³?
Solution:
Given Expression = x³ + 64y³
=(x)³+(4y)³
=(x+4y){(x)² – (x)(4y) + (4y)²}
=(x+4y)(x²-4xy+16y²)

Example 2.
Factorize a⁶ + b⁶?
Solution:
Given Expression = a⁶ + b⁶
=(a²)³ + (b²)³
= (a² + b²){(a²)² – a² ∙ b² + (b²)²}
= (a² + b²)(a⁴ – a²b² + b⁴)

Example 3.
Factorize 1 + 216z³?
Solution:
Given Expression = 1 + 216z³
=1³ + (6z)³
= (1 + 6z{1² – 1 ∙ 6z + (6z)² }
=(1 + 6z)(1 – 6z +36z²)

Example 4.
Factorize 27x³ + 1/x³?
Solution:
Given Expression = 27x³ + 1/x³
=(3x)³+(1/x)³
=(3x+1/x)((3x)²-3x.1/x+(1/x)²)
=(3x+1/x)(9x²-3+1/x²)

Example 5.
Factorize a⁶ + 1?
Solution:
Given Expression = a⁶ + 1
=(a²)³+(1)³
=(a²+1)((a²)²-a².1+(1)²)
=(a²+1)(a⁴-a².1+1)

Example 6.
Factorize the Expression 125u³ + 64v³?
Solution:
Given Expression = 125u³ + 64v³
= (5u)³+(4v)³
=(5u+4v)((5u)²-5u.4v+(4v)²)
=(5u+4v)(25u²-20uv+16v²)

 

In this article, students will completely understand the concept of the Power of a Number. We all know how to calculate the product of the number but if we have to find the product of the same number for n times then we can represent it as powers or exponents.

Learn the definitions of power, exponent, and many more along with basic rules for powers of the number. Also, you will find the solved examples on number powers or exponents calculations from this page so practice well and get a good grip on the topic.

Also Check:

• Exponents and Indices
• Number Formed by Any Power
• Expansion of Powers of Binomials and Trinomials
• Application Problems on Expansion of Powers of Binomials and Trinomials
• Worksheet on Powers of Literal Numbers

Powers and Exponents – Definition

We have an idea to find the value for the expression 2 x 2 x 2. But this expression can be defined in a shorter way with the help of powers and exponents.

ie., 2 x 2 x 2= 2³

An expression that denotes repeated multiplication of the same factor is known as a power.

The number 2 is called the base, and the number 3 is known as the exponent.

How to Find the Power of a Number without a Calculator?

The power of a number is also known as Exponents. Calculation exponents are very simple. Just follow the steps on how to solve exponents for both positive and negative numbers from here and practice some solved examples too for a better understanding of the concept.

Steps for Positive Exponents

• Find the base and power it is raised to.
• Write the base of the same number of times as exponent.
• Place a multiplication symbol between each base and multiply the same.

Steps to Find Exponent for Negative Numbers

• Define the base and power it is raised to.
• Address the reciprocal of the base and turn the sign of the exponent to positive.
• Write the reciprocal of the base the same as the number of times of exponent.
• Put a multiplication symbol between each and multiply to get the result.

Basic Rules for Power of Numbers

1. Multiplication Rule:

If we have the same base for two or more powers then we can multiply the powers. Now, we will see the rule ie., x^a ⋅ x^b=x^a+b, where we multiply two powers and add their exponents.

For Example, 2³. 2⁵ = 2³⁺⁵ = 2⁸

2. Division Rule:

In case the given power of numbers has the same base then we can divide the powers. When we divide powers we will do subtraction with the exponents. The rule is

\(\frac { x^a }{ x^b}\) = x^a-b

For Example, \(\frac { 6⁴ }{ 6²}\) = 6^4-2 = 6²

3. Power of Power Rule:

(x^a)^b=x^ab

4. Power of a Product Rule: 

(xy)^a= x^ay^a

Representation of Powers of Number from 1 to 12

Base number	2nd power	3rd power	4th power	5th power
1	1	1	1	1
2	4	8	16	32
3	9	27	81	243
4	16	64	256	1,024
5	25	125	625	3,125
6	36	216	1,296	7,776
7	49	343	2,401	16,807
8	64	512	4,096	32,768
9	81	729	6,561	59,049
10	100	1,000	10,000	100,000
11	121	1,331	14,641	161,051
12	144	1,728	20,736	248,832

Examples of Power of a Number

Example 1.
Find the value of 3⁵.
Solution:
3⁵ = 3 × 3 × 3 × 3 × 3 = 189.

Example 2.
Find the value of 2^-4.
Solution:
2^-4 = Reciprocal of 2⁴ = 1/2⁴ = 1/2×2×2×2 = 1/16.

Example 3.
Find the value of (1024)^1/4.
Solution:
(1024)^1/4
= (8⁴)^1/4
= 8^{4 x 1/4}
= 8

FAQs on Number Powers in Mathematics

1. Can I find Power of a Number online? 

Yes, absolutely you can find the power of a number online by using various valid & reliable free math calculators such as OnlineCalculator.Guru, and more.

2. How to write the Power of a Number in Word?

3. What is the Negative power of the number?

A negative power and exponent is represented as the multiplicative inverse of the base, raised to the power which is opposite to the given power. In short, we address the reciprocal of the number and then calculate it like positive exponents. For instance, (1/5)^-3 can be written as (5/1)³ ie., 5³.

4. What is the Fraction power of a number?

The numerator is the power and the denominator is the root in terms of fractinal exponent. For instance, in the variable x ^{\(\frac { a }{ b } \)}, where x, a, and b are real numbers then a is the power and b is the root ie., x^{\(\frac { a }{ b } \)} = ^b√x^a

5. What is the value of power zero of a number?

One is the value of power zero of any non-zero numbers. Whereas Zero is the value for zero to any positive exponent.

Previously we have learned the Introduction to Simultaneous Linear Equations. Kids should know that in linear equations in one variable, only one variable is introduced whose value has to solve using simple mathematical operations like +,-,/, and *.

But the theory of the linear equations remains unchanged in the case of linear equations in two variables also. Explore this guide and find detailed information on the Solution of a Linear Equation in Two Variables definition, how it works, what are the methods used, word problems, solved questions and answers, etc.

Also Check:

• Solvability of Linear Simultaneous Equations
• Word Problems on Linear Equations
• Problems on Linear Equations in One Variable

Solution of a Linear Equation in Two Variables Definition

The solution of linear equations in two variables, ax+by = c, is a specific point in the graph so that when x-coordinate is multiplied by a and y-coordinate is multiplied by b, then the sum of these two values will result as c. Basically, there are infinitely many solutions for a linear equation in two variables.

Linear Equation in Two Variable Formula

If a, b, and c are real numbers and if they both are equal to 0, then ax + by = c is known as a linear equation in two variables.

Where x and y are represented as two variables.
The numbers a and b are called the coefficients of the equation ax+by = c.
The number c is called the constant of the equation ax + by = c.

Example: 3x + 2y = 5 and -4x + 5y = 10 are linear equations in two variables.

How to Find the Solution of a Linear Equation in Two Variables?

Generally, to solve the equations involving two variables there is a possibility to use two methods. They are:

• Method of Substitution
• Method of Elimination

1. Substitution Method:

In this method, we have to follow some basic steps to solve the linear equations in two variables. Actually, we know that linear equations with two variables need at least two equations. Here, we discover the value of any one variable from the given expressions and substitute the result in the second equation to calculate the value of the variable. Now, apply the value of a variable in another equation and find the value of the other variables.

Look at the solved 2-variable equations examples mentioned below for a better understanding of the concept using the substitution method.

2. Elimination Method:

The method of determining the variables from the equations including two unknown quantities by eliminating one of the variables and then answering the resulting equation to find the value of one variable and then applying it into any one of the equations to obtain the other variable is called a method of elimination. By multiplying both equations with a number that any of the coefficients may have a multiple common, in such a way the elimination process is completed.

For getting a good grip on this method, let’s take a look at the problems of solving linear equations using the elimination method.

Graphical Method for Solving Linear Equations in Two Variables

Following are the necessary steps that included in solving linear equations in two variables graphically:

1. First, we have to graph each equation to solve the system of equations graphically.
2. Next, learn how to graph an equation manually. Just covert it to the form y = mx+b by solving the equation for y.
3. Now, put the values of x as 0, 1, 2, and so on and calculate the corresponding values of y, or vice-versa.
4. Determine the point where both lines intersect.
5. Finally, the point of intersection is the solution of the provided system.

Example: Consider a system of two linear equations; a₁ + b₁ + c₁ = 0 and a₂ + b₂ + c₂ = 0.

Problems on Solution of Linear Equations in Two Variables via Substitution & Elimination Methods | 2 Variable Equations Examples

Example 1:
Find the value of variables that satisfies the following equation:
x + 2y = 10 ……………… (i)
2x + y = 20 ……………… (ii)
Solution:
By using the substitution method, solve the pair of linear equations, we have:
x + 2y = 10 ……………… (i)
2x + y = 20 ……………… (ii)
From equation (ii), we get
y = 21 – 2x
Substitute the value of y in equation (i):
x + 2y = 10
x + 2(20-2x) = 10
x + 40 – 4x = 10
3x + 40 = 10
3x = 10 – 40
3x = -30
x = -30/3
x = -3
Substituting the x =-3 in equation (ii):
2x + y = 20
2(-3) + y = 20
-6 + y = 20
y = 20+6
y = 26
Hence x = -3 and y = 26 are the solutions of given linear equations in two variables via the substitution method.

Example 2:
Find the values of x and y by solving the given linear equations in two variables using elimination method:
3x + 4y = 10 ………. (i)
x + y = 20 ……….. (ii)
Solution:
Given equations are:
3x + 4y = 10 ………. (i)
x + y = 20 ……….. (ii)
Now, start with Multiplying equation (ii) by 3, we get;
3x + 3y = 60 ……….(iii)
Subtract (i) from (iii), we get
4y – 3y = 10 – 60
y = -50
Subsitute the y value in (ii), we get
x – 50 = 20
x = 20 + 50
x = 70
Therefore, x = 70 and y = -50 are the solutions of a linear equation in two variables.

FAQs on Linear Equations in Two Variables Questions

1. What are the different methods used to solve Linear Equations in Two Variables?

There are five methods used for finding the solutions of a linear equation in two variables. They are as follows:

• Substitution Method
• Cross Multiplication Method
• Elimination Method
• Graphical Method
• Determinant Method

2. How to find a solution of a linear equation in two variables?

3. What is the solution of the linear equation?

A point at which the planes or lines representing the linear equations cross each other is called the solution of a linear equation.

4. How many solutions are there for linear equations in two variables?

The answer for linear equations in two variables has how many solutions question is stated here. They are infinitely many solutions.

Worksheet on Factorization of the Trinomial ax² + bx + c comes in handy while solving factorizing trinomial questions. Put in extra effort and learn from the best preparation resource i.e. Factorization of Trinomial ax² + bx + c Worksheet with Answers.

Solve all the problems available and get a good grip on the concept of Factorizing Trinomials ax² + bx + c and maximize your learning whenever the time is available. Improve your preparation level by regularly answering the Questions from the Factorization of the Trinomial ax² + bx + c Worksheet PDF.

Also, See:

• Problems on Factorization of Expressions of the Form x² +(a + b)x +ab
• Factorization of Expressions of the Form ax² + bx + c, a ≠ 1

Free Factoring Trinomial Worksheet PDF

Example 1.
Factor the Trinomial 2b² + 11b + 12?
Solution:
Given Expression = 2b² + 11b + 12
= 2b² + 8b+3b + 12
=2b(b+4)+3(b+4)
=(2b+3)(b+4)

Example 2.
Factorize x² + 8x – 105?
Solution:
Given Expression = x² + 8x – 105
= x² + 15x –7x-105
=x(x+15)-7(x+15)
=(x-7)(x+15)

Example 3.
Factorize 9c² – c – 8?
Solution:
Given Expression = 9c² – c – 8
=9c² – 9c+8c – 8
=9c(c-1)+8(c-1)
=(9c+8)(c-1)

Example 4.
Factorize Trinomial 7m² – 15m – 18?
Solution:
Given Expression = 7m² – 15m – 18
=7m² – 21m+6m – 18
=7m(m-3)+6(m-3)
=(7m+6)(m-3)

Example 5.
Factorize the Trinomial 18 + 11x + x²
Solution:
Given Expression = 18 + 11x + x²
=18+9x+2x+x²
=x(x+9)+2(x+9)
=(x+9)(x+2)

Example 6.
Factorize 5p²+ 19 p + 12
Solution:
Given Expression = 5p²+ 19 p + 12
=5p²+ 15p +4p+ 12
=5p(p+3)+4(p+3)
=(5p+4)(p+3)

Example 7.
Factorize 6r²− 17r + 12?
Solution:
Given Expression = 6r²− 17r + 12
=6r²− 17r + 12
=6r²− 8r-9r + 12
=2r(3r-4)-3(3r-4)
=(3r-4)(2r-3)

Example 8.
Factorize 3r²+ 40r + 100
Solution:
Given Expression = 3r²+ 40r + 100
=3r²+ 40r + 100
=3r²+ 30r +10r+ 100
=3r(r+10)+10(r+10)
=(3r+10)(r+10)

Want to master in techniques of Simultaneous Linear Equations? Then, this page plays a vital role in your practice session. To help you achieve all your academic goals, we have again come up with this guide on  Introduction to Simultaneous Linear Equations.

Excited to learn what it teaches you and how this concept assists you in solving complex problems. Just go with the further section and gain detailed knowledge about the solutions of linear equations, methods to solve the system of linear equations, examples on simultaneous equations, etc.

• Solution of a Linear Equation in Two Variables
• Method of Elimination
• Method of Substitution
• Method of Cross Multiplication

What is Simultaneous Linear Equation with Example?

Two linear equations in two or three variables determined simultaneously to get a common solution is known as simultaneous linear equations. Simultaneous linear equations in two variables include two strange quantities to depict real-life problems. It aids in setting up a relationship between quantities, prices, speed, time, distance, etc. outcome in a better knowledge of the problems.

The result of the system of simultaneous linear equations is the ordered pair (x, y) that meets both the linear equations.

For Example, x + y = 20 and x – y = 10 are two linear equation (simultaneous equations). In case, we take x = 15 and y = 5, then the two equations are met, so we can say the solution of the system of given simultaneous linear equations is (10, 5).

Important steps to form & solve Simultaneous Linear Equations

Let’s consider any mathematical question to denote and understand the important steps for forming simultaneous equations:

In a book shop, the cost of 3 books exceeds the cost of 4 pencils by $6. Also, the total cost of 5 books and 7 pencils is $20.

Step 1: Detect the unknown variables; let’s take one of them as x and the other as y.
Here two unknown variables are:
Cost of each book = $x
Cost of each pencil = $y

Step 2: Now, find the relation between the unknown quantities.
Cost of 3 books = $3x
Cost of 4 pencils = $4y
Hence, the first condition is 3x-4y = 6

Step 3: Communicate the given conditions of the problem in terms of x and y
Now, go for the other condition ie., cost of 5 books = $5x
cost of 7 pencils = $7y
Therefore, the second condition is 5x+7y = 10

Simultaneous equations formed from the given conditions of the problem:
(i) 3x-4y = 6
(ii) 5x+7y = 10

How To Solve Simultaneous Linear Equations Using Substitution Method?

From this section, you can explore the detailed steps on how to solve a system of linear equations (Simultaneous equations) using the substitution method. They are as follows:

• Firstly, obtain a relation that divides one of the variables by changing the subject of a formula.
• Now, substitute the relation into the other expression(s) to diminish the no. of variables by 1.
• Just go with the same steps till we’re left with a single variable, and solve for it.
• Finally, at this step, we have to substitute the end result in the relations and state the complete solution.

Steps for Solving Simultaneous Linear Algebraic Equations via Elimination Method?

The process of the Elimination method is repeated until the values of n variables are found. Want to learn the simple steps to solve the system of linear equations using the elimination method? then look at the below lines:

• First, identify two equations that have the same variable.
• Now, perform the multiplication operation on each equation by a number to become their coefficients equal.
• Next, subtract the two equations
• You need to do the same process until you are left with a single variable, and solve for it.
• At last, substitute the resultant value into the original equations and find the values of unknown variables.

Why Simultaneous Linear Equations Usage?

• To memorize the process of framing Simultaneous Linear Equations from Mathematical Problems.
• By using the method of comparison and method of elimination, we can remember to solve the simultaneous equations.
• For obtaining the ability to find simultaneous equations via the method of substitution and method of cross-multiplication.
• To take the skill to solve mathematical problems framing simultaneous equations.
• To understand the condition for a pair of linear expressions to become simultaneous equations

Problems on Simultaneous Linear Equations

Example 1:
Show that x = 3 and y = 6 is the solution of the system of linear equation x + 2y = 15 and 2x + y =12.

Solution:
Given that equation 1 is x + 2y = 4
equation 2 is x + 3y = 8
Now, put x = 3 and y = 6 in equation 1 ie., x + 2y = 15
LHS = x + 2y = 3 + 2×6 = 3 + 12 = 15. which is equal to RHS.
Now, put x = 3 and y = 6 in LHS of 2nd equation ie., 2x + y =12
LHS = 2x + y = 2×3 + 6 = 6 + 6 = 12, which is equal to RHS.
Thus, x = 3 and y = 6 is the solution of the given system of equations.

Example 2:
Consider the two following simultaneous linear equations:
x + y = 5 ………… (i)
2x – 3y = 15 ………… (ii)

Solution:
The given equations are:
x + y = 5 ………… (i)
2x – 3y = 15 ………… (ii)
From (i) we get y = 5-x
Now, substituting the y value in equation (ii) then we get;
2x – 3(5-x) = 15
2x -15 + 3x = 15
2x + 3x = 30
5x = 30
x = 5
Substitute the solved x value in equation (i), we get;
x + y = 5
5 + y = 5
y = 0
Hence, (5, 0) is the solution of the system of equation x + y = 5 and 2x – 3y = 15.

FAQs on Framing Simultaneous Equations

1. What are simultaneous equations?

The two or more algebraic equations that share variables e.g. x and y are called simultaneous equations.

2. What is the graph of two linear equations?

The graph of two linear equations is Two straight lines.

3. Where on the graph do we get the solution of a system of linear equations?

At the point of intersection of two lines, we got to know a pair of simultaneous equations is equivalent.

Factorizing Algebraic Expressions can be a difficult task for certain students who aren’t strong in the fundamentals of maths. Such students will find this article quite a handy resource as it explains different problems on the factorization of expressions of the form x² +(a + b)x +ab.

Go through the entire article and learn how to factorize algebraic expressions using the identity x² +(a + b)x +ab and arrive at the result easily. Practice the example questions on the factorization of expressions of the form x² +(a + b)x +ab regularly and attempt the exam with utmost confidence and score well.

See More:

• Factorization of Expressions of the Form x² + (a + b)x + ab
• Factorization of Expressions of the Form ax² + bx + c, a ≠ 1

Examples on Factorization of Algebraic Expressions using Identity x² +(a + b)x +ab

Example 1.
Factorize x² + 6x + 8?
Solution:
Given Expression = x² + 6x + 8
Here constant term = 8= 4*2=8 Coefficient of x = 6 i.e.(4+2)
= x² + 4x +2x+ 8
= x(x+4)+2(x+4)
=(x+4)(x+2)

Example 2.
Factorize x² + 21x + 108?
Solution:
Given Expression = x² + 21x + 108
Here constant term = 108= 12*9=108 Coefficient of x = 21 i.e.(12+9)
=x² + 12x +9x+ 108
=x(x+12)+9(x+12)
=(x+9)(x+12)

Example 3.
Factorize x²y² – 2xy – 63?
Solution:
Given Expression = x²y² – 2xy – 63
here Constant Term = -63=7*-9 =-63, Coefficient of xy=-2=-9+7
= x²y² – 9xy+7xy – 63
=xy(xy-9)+7(xy-9)
=(xy-9)(xy+7)

Example 4.
Factorize 6x² +5x +6?
Solution:
Given Expression = 6x² +5 x +6
Here Constant Term = 6*6 =36 =9*4, Coefficient of x= 5=9-4
=6x² +9x-4x +6
=3x(2x+3)-2(2x-3)
=(3x-2)(2x+3)

Example 5.
Factorize 5x²+25x+25?
Solution:
Given Expression = 5x²+25x+25
Here Constant Term = 5*25=125, Coefficient of x =25=20+5
= 5x²+20x+5x+25
= 5x(x+5)+5(x+5)
= (5x+5)(x+5)

Example 6.
Factorize 2x²-5x-3?
Solution:
Given Expression = 2x² -5x -3
Here Constant Term = 2*-3 =-6, Coefficient of x= -5=-6+1
=2x²-5x-3
=2x²-6x+x-3
=2x(x-3)+1(x-3)
=(2x+1)(x-3)

Example 7.
Factorize 3x²+13x+4?
Solution:
Given Expression = 3x²+13x+4
Here Constant Term =3*4 =12, Coefficient of x= 13=12+1
=3x²+13x+4
=3x²+12x+x+4
=3x(x+4)+1(x+4)
=(3x+1)(x+4)

Example 8.
Factorize r² -10r + 21?
Solution:
Given Expression = r² -10r + 21
Here Constant Term =1*21 =21, Coefficient of a=-10=-7-3
=r² -10r + 21
=r² -7r-3r + 21
=r(r-7)-3(r-7)
=(r-7)(r-3)