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Post a LessonAnswered on 23 Feb Learn CBSE/Class 6/Maths/algebra/Introduction to formation of Variables
Sadika
Let's denote: Length of the rectangle = 2332 of its breadth Breadth of the rectangle = xx cm
Then, we can express the length of the rectangle in terms of its breadth: Length=23×BreadthLength=32×Breadth Length=23×xLength=32×x Length=2x3Length=32x
Now, we can calculate the area of the rectangle: Area=Length×BreadthArea=Length×Breadth Area=(2x3)×xArea=(32x)×x Area=2x23Area=32x2
So, the area of the rectangle is 2x2332x2 square centimeters.
Answered on 23 Feb Learn CBSE/Class 6/Maths/algebra/Introduction to formation of Variables
Sadika
If there are xx rows of chairs and each row contains x2x2 chairs, then the total number of chairs can be found by multiplying the number of rows by the number of chairs in each row.
Given: Number of rows (xx) Number of chairs in each row (x2x2)
Total number of chairs = Number of rows * Number of chairs in each row Total number of chairs = x×x2x×x2
So, the total number of chairs is x3x3.
Answered on 23 Feb Learn CBSE/Class 6/Maths/algebra/Introduction to formation of Variables
Sadika
If there are x2x2 rows of chairs and each row contains 2×2x22×2x2 chairs, then the total number of chairs can be found by multiplying the number of rows by the number of chairs in each row.
Given: Number of rows (x2x2) Number of chairs in each row (2×2x2=4x22×2x2=4x2)
Total number of chairs = Number of rows * Number of chairs in each row Total number of chairs = x2×4x2x2×4x2
So, the total number of chairs in the room is 4x44x4.
Answered on 23 Feb Learn CBSE/Class 6/Maths/algebra/Conversion of Statement to Algebraic Expression
Sadika
Sure, here are two equations for which 22 is the solution:
When x=2x=2: 2−4=−22−4=−2 −2=−2−2=−2
When y=2y=2: 2×24=142×2=1 44=144=1 1=11=1
In both equations, 22 is the solution.
Answered on 23 Feb Learn CBSE/Class 6/Maths/algebra/Conversion of Statement to Algebraic Expression
Sadika
Here's an example of an equation whose solution is not a whole number:
2x+1=52x+1=5
To solve for xx, we need to isolate it: 2x=5−12x=5−1 2x=42x=4 x=42x=24 x=2x=2
So, the solution to this equation is x=2x=2, which is a whole number.
Let me provide another example:
3x−2=73x−2=7
To solve for xx, we need to isolate it: 3x=7+23x=7+2 3x=93x=9 x=93x=39 x=3x=3
In this equation, the solution is also a whole number.
Let's try another example:
2x+3=72x+3=7
To solve for xx, we need to isolate it: 2x=7−32x=7−3 2x=42x=4 x=42x=24 x=2x=2
Once again, the solution is a whole number.
It seems that I provided examples where the solution is a whole number. Let me adjust:
2x+1.5=52x+1.5=5
To solve for xx: 2x=5−1.52x=5−1.5 2x=3.52x=3.5 x=3.52x=23.5 x=1.75x=1.75
In this equation, the solution x=1.75x=1.75 is not a whole number.
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