How do you solve \[7(2x + 5) = 4x - 9 - x\;?\]
Answer
Verified
439.8k+ views
Hint: We will multiply and subtract all the variables and constant terms separately and then we will calculate the value of \[x\]. On doing some calculation we get the required answer.
Formula used: If an equation contains a single variable in the following format, we can calculate the value of the variable in the following way.
Let's say, \[A(mx + n) = c\] is the given equation and we have to calculate the value of \[x\].
So, we can interpret the equation in following way:
\[ \Rightarrow Amx + An = c\].
Now, we can take the constant terms in R.H.S:
\[ \Rightarrow Amx = c - An\].
So, we can get the value of \[x\] by dividing the constant term by the coefficient of \[x\].
\[ \Rightarrow x = \dfrac{{c - An}}{{Am}}\].
Complete Step by Step Solution:
The given equation is \[7(2x + 5) = 4x - 9 - x\;.\]
Now, by subtracting the variables on the R.H.S, we get:
\[ \Rightarrow 7(2x + 5) = 3x - 9\].
Now, by doing the multiplication on L.H.S, we get:
\[ \Rightarrow 14x + 35 = 3x - 9.\]
Now, take the constant term on R.H.S and the variable term on L.H.S, we get:
\[ \Rightarrow 14x - 3x = - 35 - 9.\]
Now, perform the required operations on R.H.S and L.H.S, we get the following equation:
\[ \Rightarrow 11x = - 44.\]
Now, divide the constant term on L.H.S by the coefficient of\[x\] on R.H.S, we get:
\[ \Rightarrow x = \dfrac{{ - 44}}{{\;\;11}}.\]
Now, perform the division on R.H.S, we get:
\[ \Rightarrow x = - 4.\]
Therefore, the value of \[x\] or the solution of the equation is \[ - 4\].
Note: Alternate way to solution:
The given equation is\[7(2x + 5) = 4x - 9 - x\;.\]
We can divide both the sides of the equation by \[7\].
SO, by performing it, we get:
\[ \Rightarrow 2x + 5 = \dfrac{{4x - 9 - x}}{7}\].
Now, by solving on the R.H.S, we get:
\[ \Rightarrow 2x + 5 = \dfrac{{3x - 9}}{7}\].
Now, further splitting up the terms in R.H.S, we get:
\[ \Rightarrow 2x + 5 = \dfrac{{3x}}{7} - \dfrac{9}{7}\].
Now, taking the variable terms on L.H.S and constant terms on R.H.S, we get:
\[ \Rightarrow 2x - \dfrac{{3x}}{7} = - 5 - \dfrac{9}{7}\].
Now, by doing the subtractions on both the sides, we get:
\[ \Rightarrow \dfrac{{14x - 3x}}{7} = \dfrac{{ - 35 - 9}}{7}\].
Now, simplify it further:
\[ \Rightarrow \dfrac{{11x}}{7} = \dfrac{{ - 44}}{7}\].
Now, cancel out the same terms on the denominators, we get:
\[ \Rightarrow 11x = - 44.\]
By doing the division, we get:
\[ \Rightarrow x = \dfrac{{ - 44}}{{\;\;11}}\]
\[ \Rightarrow x = - 4.\]
\[\therefore \]The solution of the equation is \[x = - 4.\]
Formula used: If an equation contains a single variable in the following format, we can calculate the value of the variable in the following way.
Let's say, \[A(mx + n) = c\] is the given equation and we have to calculate the value of \[x\].
So, we can interpret the equation in following way:
\[ \Rightarrow Amx + An = c\].
Now, we can take the constant terms in R.H.S:
\[ \Rightarrow Amx = c - An\].
So, we can get the value of \[x\] by dividing the constant term by the coefficient of \[x\].
\[ \Rightarrow x = \dfrac{{c - An}}{{Am}}\].
Complete Step by Step Solution:
The given equation is \[7(2x + 5) = 4x - 9 - x\;.\]
Now, by subtracting the variables on the R.H.S, we get:
\[ \Rightarrow 7(2x + 5) = 3x - 9\].
Now, by doing the multiplication on L.H.S, we get:
\[ \Rightarrow 14x + 35 = 3x - 9.\]
Now, take the constant term on R.H.S and the variable term on L.H.S, we get:
\[ \Rightarrow 14x - 3x = - 35 - 9.\]
Now, perform the required operations on R.H.S and L.H.S, we get the following equation:
\[ \Rightarrow 11x = - 44.\]
Now, divide the constant term on L.H.S by the coefficient of\[x\] on R.H.S, we get:
\[ \Rightarrow x = \dfrac{{ - 44}}{{\;\;11}}.\]
Now, perform the division on R.H.S, we get:
\[ \Rightarrow x = - 4.\]
Therefore, the value of \[x\] or the solution of the equation is \[ - 4\].
Note: Alternate way to solution:
The given equation is\[7(2x + 5) = 4x - 9 - x\;.\]
We can divide both the sides of the equation by \[7\].
SO, by performing it, we get:
\[ \Rightarrow 2x + 5 = \dfrac{{4x - 9 - x}}{7}\].
Now, by solving on the R.H.S, we get:
\[ \Rightarrow 2x + 5 = \dfrac{{3x - 9}}{7}\].
Now, further splitting up the terms in R.H.S, we get:
\[ \Rightarrow 2x + 5 = \dfrac{{3x}}{7} - \dfrac{9}{7}\].
Now, taking the variable terms on L.H.S and constant terms on R.H.S, we get:
\[ \Rightarrow 2x - \dfrac{{3x}}{7} = - 5 - \dfrac{9}{7}\].
Now, by doing the subtractions on both the sides, we get:
\[ \Rightarrow \dfrac{{14x - 3x}}{7} = \dfrac{{ - 35 - 9}}{7}\].
Now, simplify it further:
\[ \Rightarrow \dfrac{{11x}}{7} = \dfrac{{ - 44}}{7}\].
Now, cancel out the same terms on the denominators, we get:
\[ \Rightarrow 11x = - 44.\]
By doing the division, we get:
\[ \Rightarrow x = \dfrac{{ - 44}}{{\;\;11}}\]
\[ \Rightarrow x = - 4.\]
\[\therefore \]The solution of the equation is \[x = - 4.\]
Recently Updated Pages
Identify how many lines of symmetry drawn are there class 8 maths CBSE
State true or false If two lines intersect and if one class 8 maths CBSE
Tina had 20m 5cm long cloth She cuts 4m 50cm lengt-class-8-maths-CBSE
Which sentence is punctuated correctly A Always ask class 8 english CBSE
Will Mr Black be at home Saturday evening Yes hell class 8 english CBSE
An electrician sells a room heater for Rs 3220 gaining class 8 maths CBSE
Trending doubts
How is the Lok Sabha more powerful than the Rajya class 8 social science CBSE
Write a letter to your friend telling himher how you class 8 english CBSE
Write the following in HinduArabic numerals XXIX class 8 maths CBSE
Differentiate between the farms in India and the U class 8 social science CBSE
The strategy of Divide and rule was adopted by A Lord class 8 social science CBSE
When will we use have had and had had in the sente class 8 english CBSE