
Add the following algebraic expression using both horizontal and vertical methods.
Did you get the same answer with both methods?
$2{x^2} - 6x + 3; - 3{x^2} - x + 4;1 + 2x - 3{x^2}$
Answer
216.3k+ views
Hint: The horizontal method and the vertical method of adding the expression are two different ways to add the expression but the resultant answer is the same in both the cases.
Complete step-by-step answer:
Given algebraic expressions are $2{x^2} - 6x + 3; - 3{x^2} - x + 4;1 + 2x - 3{x^2}$
Assume the given algebraic expression as the functions:
$f\left( x \right) = 2{x^2} - 6x + 3$ ,
$g\left( x \right) = - 3{x^2} - x + 4$
$h\left( x \right) = 1 + 2x - 3{x^2}$
First, we add these expressions using the horizontal method. In the horizontal method of adding the expression, we first need to write all the expression in a horizontal line and then make groups of like terms, after that add all the like terms together.
$S\left( x \right) = f\left( x \right) + g\left( x \right) + h\left( x \right)$
Substitute $2{x^2} - 6x + 3$as the value of$f\left( x \right)$, $ - 3{x^2} - x + 4$as the value of$g\left( x \right)$and $1 + 2x - 3{x^2}$as the value of $h\left( x \right)$in the above expression:
$S\left( x \right) = \left( {2{x^2} - 6x + 3} \right) + \left( { - 3{x^2} - x + 4} \right) + \left( {1 + 2x - 3{x^2}} \right)$
Open the braces and express the terms:
$S\left( x \right) = 2{x^2} - 6x + 3 - 3{x^2} - x + 4 + 1 + 2x - 3{x^2}$
Now, collect all the like terms together and add them.
$S\left( x \right) = 2{x^2} - 3{x^2} - 3{x^2} - 6x - x + 2x + 3 + 4 + 1$
$S\left( x \right) = 4{x^2} - 5x + 8$
The sum of the given expression using the horizontal method is$4{x^2} - 5x + 8$.
Now, add the functions in the vertical method. In this method of adding expression, we first need to arrange the expressions in a vertical line in such a manner that the like terms and their signs are one below the other, and then add them.

The sum of the given expression using the vertical method is$4{x^2} - 5x + 8$.
When we added the given expression using the horizontal method, then we have the solution:
$4{x^2} - 5x + 8$
And when we added the given expression using the vertical method, then we have the solution:
$4{x^2} - 5x + 8$
It can be seen that both solutions are the same. The conclusion is that we get the same answer using horizontal and vertical methods both.
Note: The like terms are the terms of the expression that have the same variables raised to the same power. In the given expressions, $2{x^2} - 6x + 3; - 3{x^2} - x + 4;1 + 2x - 3{x^2}$, the like terms are $\left( {2{x^2}, - 3{x^2}, - 3{x^2}} \right),\left( { - 6x, - x,2x} \right){\text{ and }}\left( {3,4,1} \right)$.
Complete step-by-step answer:
Given algebraic expressions are $2{x^2} - 6x + 3; - 3{x^2} - x + 4;1 + 2x - 3{x^2}$
Assume the given algebraic expression as the functions:
$f\left( x \right) = 2{x^2} - 6x + 3$ ,
$g\left( x \right) = - 3{x^2} - x + 4$
$h\left( x \right) = 1 + 2x - 3{x^2}$
First, we add these expressions using the horizontal method. In the horizontal method of adding the expression, we first need to write all the expression in a horizontal line and then make groups of like terms, after that add all the like terms together.
$S\left( x \right) = f\left( x \right) + g\left( x \right) + h\left( x \right)$
Substitute $2{x^2} - 6x + 3$as the value of$f\left( x \right)$, $ - 3{x^2} - x + 4$as the value of$g\left( x \right)$and $1 + 2x - 3{x^2}$as the value of $h\left( x \right)$in the above expression:
$S\left( x \right) = \left( {2{x^2} - 6x + 3} \right) + \left( { - 3{x^2} - x + 4} \right) + \left( {1 + 2x - 3{x^2}} \right)$
Open the braces and express the terms:
$S\left( x \right) = 2{x^2} - 6x + 3 - 3{x^2} - x + 4 + 1 + 2x - 3{x^2}$
Now, collect all the like terms together and add them.
$S\left( x \right) = 2{x^2} - 3{x^2} - 3{x^2} - 6x - x + 2x + 3 + 4 + 1$
$S\left( x \right) = 4{x^2} - 5x + 8$
The sum of the given expression using the horizontal method is$4{x^2} - 5x + 8$.
Now, add the functions in the vertical method. In this method of adding expression, we first need to arrange the expressions in a vertical line in such a manner that the like terms and their signs are one below the other, and then add them.

The sum of the given expression using the vertical method is$4{x^2} - 5x + 8$.
When we added the given expression using the horizontal method, then we have the solution:
$4{x^2} - 5x + 8$
And when we added the given expression using the vertical method, then we have the solution:
$4{x^2} - 5x + 8$
It can be seen that both solutions are the same. The conclusion is that we get the same answer using horizontal and vertical methods both.
Note: The like terms are the terms of the expression that have the same variables raised to the same power. In the given expressions, $2{x^2} - 6x + 3; - 3{x^2} - x + 4;1 + 2x - 3{x^2}$, the like terms are $\left( {2{x^2}, - 3{x^2}, - 3{x^2}} \right),\left( { - 6x, - x,2x} \right){\text{ and }}\left( {3,4,1} \right)$.
Recently Updated Pages
JEE Main 2024 (January 24 Shift 1) Question Paper with Solutions [PDF]

Progressive Wave: Meaning, Types & Examples Explained

Temperature Dependence of Resistivity Explained

JEE Main 2024 (January 25 Shift 1) Physics Question Paper with Solutions [PDF]

Difference Between Vectors and Scalars: JEE Main 2026

Salt Hydrolysis IIT JEE | Aсіdіtу and Alkаlіnіtу of Sаlt Sоlutіоns JEE Chemistry

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

JEE Main 2026 Chapter-Wise Syllabus for Physics, Chemistry and Maths – Download PDF

JEE Main Correction Window 2026 Session 1 Dates Announced - Edit Form Details, Dates and Link

JEE Main Previous Year Question Paper with Answer Keys and Solutions

Understanding Newton’s Laws of Motion

JEE Main Cut Off 2026 - Expected Qualifying Marks and Percentile Category Wise

Other Pages
NCERT Solutions For Class 9 Maths Chapter 9 Circles

NCERT Solutions for Class 9 Maths Chapter 11 Surface Area and Volume 2025-26

NCERT Solutions For Class 9 Maths Chapter 11 Surface Areas And Volumes

Fuel Cost Calculator – Estimate Your Journey Expenses Easily

NCERT Solutions For Class 9 Maths Chapter 12 Statistics

NCERT Solutions For Class 9 Maths Chapter 10 Heron's Formula


