Answer
350.7k+ views
Hint: Rearrange the given equation such that the unknown terms arranged on one side and all other terms are arranged on other side and then compare both sides and with further simplification we can determine the value of unknown terms.
Given: \[\dfrac{{x - 1}}{2} - x + 13 = 5 - x\]
To find: the value of ‘x’
Complete step by step answer:
Step 1: Firstly we will determine the number of unknown terms in the given equation.
\[\dfrac{{x - 1}}{2} - x + 13 = 5 - x\]
So here we have only one unknown term so we require only one equation to determine the value of ‘x’
Step 2: multiply both sides of the equation with 2 such that all the denominator part get reduced and we get a simple linear equation
\[\dfrac{{x - 1}}{2} - x + 13 = 5 - x\]
\[2 \times (\dfrac{{x - 1}}{2} - x + 13) = (5 - x) \times 2\]
Opening the bracket and we get
\[(x - 1) - 2 \times x + 2 \times 13 = 5 \times 2 - x \times 2\]
Step 3: rearranging the terms such that the unknown term arranged on one side and all other terms are arranged on other side, that is
\[(x - 1) - 2 \times x + 2 \times 13 = 5 \times 2 - x \times 2\]
\[x - 2x - 1 + 26 = 10 - 2x\]
\[x - 2x + 2x = 10 + 1 - 26\]
On further simplification, we get
\[x - 2x + 2x = 10 + 1 - 26\]
\[x = - 15\]
Hence, on solving the given equation we determined the value of ‘x’ and it is equal to \[x = - 15\]
Note: We have different solution methods for different types of equation
We can use the substitution method
We can use the elimination method without multiplication.
We can use the elimination method with multiplication.
There might be the possibility of infinite solution or no solution.
Given: \[\dfrac{{x - 1}}{2} - x + 13 = 5 - x\]
To find: the value of ‘x’
Complete step by step answer:
Step 1: Firstly we will determine the number of unknown terms in the given equation.
\[\dfrac{{x - 1}}{2} - x + 13 = 5 - x\]
So here we have only one unknown term so we require only one equation to determine the value of ‘x’
Step 2: multiply both sides of the equation with 2 such that all the denominator part get reduced and we get a simple linear equation
\[\dfrac{{x - 1}}{2} - x + 13 = 5 - x\]
\[2 \times (\dfrac{{x - 1}}{2} - x + 13) = (5 - x) \times 2\]
Opening the bracket and we get
\[(x - 1) - 2 \times x + 2 \times 13 = 5 \times 2 - x \times 2\]
Step 3: rearranging the terms such that the unknown term arranged on one side and all other terms are arranged on other side, that is
\[(x - 1) - 2 \times x + 2 \times 13 = 5 \times 2 - x \times 2\]
\[x - 2x - 1 + 26 = 10 - 2x\]
\[x - 2x + 2x = 10 + 1 - 26\]
On further simplification, we get
\[x - 2x + 2x = 10 + 1 - 26\]
\[x = - 15\]
Hence, on solving the given equation we determined the value of ‘x’ and it is equal to \[x = - 15\]
Note: We have different solution methods for different types of equation
We can use the substitution method
We can use the elimination method without multiplication.
We can use the elimination method with multiplication.
There might be the possibility of infinite solution or no solution.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)