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How do you solve for u in \[8u = 3u + 35\] ?

Last updated date: 01st Mar 2024
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IVSAT 2024
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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

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.
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