Answer
Verified
429.9k+ views
Hint:We have to solve this algebraic equation \[8u = 3u + 35\] and find the value of ‘u’. So, shift the variable terms to one side and constants on the other. Thus, 3u and 8u should be shifted to one side to get subtracted and then keep only ‘u’ to one side and shift all the terms to another to calculate its
value.
Complete step by step solution:
We are given with the algebraic expression \[8u = 3u + 35\].
Now, this equation contains variables and constants which are separated by an arithmetic operatic that is addition. Always remember, only like terms are added or subtracted. That means, constant terms would be subtracted or added to constant terms only and variable terms would be added or subtracted from the variable only.
For example, 5+2=7, in this both the 5 and 2 are constant terms so they got added but addition of 5x+2 is not possible because 5x is a variable term and 2 is a constant, therefore addition or subtraction of unlike terms is not possible.
While multiplication of unlike terms is possible. For example, \[9 \times a = 9a\]or\[2x \times 8 = 16x\].
Similarly, division of unlike terms is also possible.
Now, we have to solve, \[8u = 3u + 35\]
In this equation, 8u and 3u are variable terms and 35 is constant. Therefore, 8u and 3u can be added or subtracted. So, shifting the 3u from RHS to LHS.
\[ \Rightarrow 8u = 3u + 35\]
\[ \Rightarrow 8u - 3u = 35\]
While shifting the terms from one side to another, signs of the terms would also change. 3u on RHS was positive (assume as\[8u = + 3u + 35\]) so on coming to LHS +3u would become -3u.
\[ \Rightarrow 5u = 35\]
Now this 5u term is actually\[5 \times u\], so to find the value of u, we need to shift 5 from LHS to RHS with change in sign, here it is in multiplication with u, whereas on RHS 5 would be in division.
\[
\Rightarrow 5 \times u = 35 \\
\Rightarrow u = \dfrac{{35}}{5} \\
\Rightarrow u = 7 \\
\]
Hence, the solution of this given algebraic equation is 7.
Note: Changing of sign is must while shifting the terms from one side to another.
Variable: variable are those terms whose values keep on changing from time to time. In the algebraic expressions or equations, the letters or alphabets basically represent the variables.
Constant: these are the terms whose values are always fixed and do not change. For example, numbers, integers, pi (22/7) etc.
value.
Complete step by step solution:
We are given with the algebraic expression \[8u = 3u + 35\].
Now, this equation contains variables and constants which are separated by an arithmetic operatic that is addition. Always remember, only like terms are added or subtracted. That means, constant terms would be subtracted or added to constant terms only and variable terms would be added or subtracted from the variable only.
For example, 5+2=7, in this both the 5 and 2 are constant terms so they got added but addition of 5x+2 is not possible because 5x is a variable term and 2 is a constant, therefore addition or subtraction of unlike terms is not possible.
While multiplication of unlike terms is possible. For example, \[9 \times a = 9a\]or\[2x \times 8 = 16x\].
Similarly, division of unlike terms is also possible.
Now, we have to solve, \[8u = 3u + 35\]
In this equation, 8u and 3u are variable terms and 35 is constant. Therefore, 8u and 3u can be added or subtracted. So, shifting the 3u from RHS to LHS.
\[ \Rightarrow 8u = 3u + 35\]
\[ \Rightarrow 8u - 3u = 35\]
While shifting the terms from one side to another, signs of the terms would also change. 3u on RHS was positive (assume as\[8u = + 3u + 35\]) so on coming to LHS +3u would become -3u.
\[ \Rightarrow 5u = 35\]
Now this 5u term is actually\[5 \times u\], so to find the value of u, we need to shift 5 from LHS to RHS with change in sign, here it is in multiplication with u, whereas on RHS 5 would be in division.
\[
\Rightarrow 5 \times u = 35 \\
\Rightarrow u = \dfrac{{35}}{5} \\
\Rightarrow u = 7 \\
\]
Hence, the solution of this given algebraic equation is 7.
Note: Changing of sign is must while shifting the terms from one side to another.
Variable: variable are those terms whose values keep on changing from time to time. In the algebraic expressions or equations, the letters or alphabets basically represent the variables.
Constant: these are the terms whose values are always fixed and do not change. For example, numbers, integers, pi (22/7) etc.
Recently Updated Pages
Fill in the blanks with suitable prepositions Break class 10 english CBSE
Fill in the blanks with suitable articles Tribune is class 10 english CBSE
Rearrange the following words and phrases to form a class 10 english CBSE
Select the opposite of the given word Permit aGive class 10 english CBSE
Fill in the blank with the most appropriate option class 10 english CBSE
Some places have oneline notices Which option is a class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
Which are the Top 10 Largest Countries of the World?
What is the definite integral of zero a constant b class 12 maths CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Define the term system surroundings open system closed class 11 chemistry CBSE
Full Form of IASDMIPSIFSIRSPOLICE class 7 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE