How do you solve \[\left| {\dfrac{2}{3}x - 9} \right| = 18\]?
Answer
Verified
444.9k+ views
Hint: We use the concept of modulus of a function and make two cases for the values of the right hand side of the equation. Remove the modulus and equate the left side to positive right side and negative right side simultaneously.
* Modulus of a function ‘x’ opens up as \[\left| x \right| = \left\{ {\begin{array}{*{20}{c}}
x&{;x \geqslant 0} \\
{ - x}&{;x < 0}
\end{array}} \right.\]
Complete step-by-step answer:
We have to solve the equation \[\left| {\dfrac{2}{3}x - 9} \right| = 18\]
We know that modulus opens up either positive or negative values.
We make two cases:
Case 1: \[\dfrac{2}{3}x - 9 = 18\]
Equate left hand side of the equation to right hand side of the equation
\[ \Rightarrow \dfrac{2}{3}x - 9 = 18\]
Shift constant values to right hand side of the equation
\[ \Rightarrow \dfrac{2}{3}x = 18 + 9\]
Add terms on right hand side
\[ \Rightarrow \dfrac{2}{3}x = 27\]
Cross multiply the values from left side of the equation to right side of the equation
\[ \Rightarrow x = 27 \times \dfrac{3}{2}\]
Calculate the product on right hand side of the equation
\[ \Rightarrow x = \dfrac{{81}}{2}\]
Divide the numerator from denominator
\[ \Rightarrow x = 40.5\] … (1)
Case 2: \[\dfrac{2}{3}x - 9 = - 18\]
Equate left hand side of the equation to right hand side of the equation
\[ \Rightarrow \dfrac{2}{3}x - 9 = - 18\]
Shift constant values to right hand side of the equation
\[ \Rightarrow \dfrac{2}{3}x = - 18 + 9\]
Add terms on right hand side
\[ \Rightarrow \dfrac{2}{3}x = - 9\]
Cross multiply the values from left side of the equation to right side of the equation
\[ \Rightarrow x = - 9 \times \dfrac{3}{2}\]
Calculate the product on right hand side of the equation
\[ \Rightarrow x = \dfrac{{ - 27}}{2}\]
Divide the numerator from denominator
\[ \Rightarrow x = - 13.5\] … (2)
So, from equations (1) and (2) we get the values for x as 40.5 and -13.5
\[\therefore \]Solution of the equation \[\left| {\dfrac{2}{3}x - 9} \right| = 18\] is \[x = 40.5\]and \[x = - 13.5\]
Note:
Many students make mistake of opening the left hand side of the equation using modulus i.e. they write positive and negative of left hand side equal to given right hand side, this gets complicated as we have fraction terms in left hand side and we will have to take LCM as well, so we choose to take positive and negative sign in right hand side of the equation as it contains only one value.
* Modulus of a function ‘x’ opens up as \[\left| x \right| = \left\{ {\begin{array}{*{20}{c}}
x&{;x \geqslant 0} \\
{ - x}&{;x < 0}
\end{array}} \right.\]
Complete step-by-step answer:
We have to solve the equation \[\left| {\dfrac{2}{3}x - 9} \right| = 18\]
We know that modulus opens up either positive or negative values.
We make two cases:
Case 1: \[\dfrac{2}{3}x - 9 = 18\]
Equate left hand side of the equation to right hand side of the equation
\[ \Rightarrow \dfrac{2}{3}x - 9 = 18\]
Shift constant values to right hand side of the equation
\[ \Rightarrow \dfrac{2}{3}x = 18 + 9\]
Add terms on right hand side
\[ \Rightarrow \dfrac{2}{3}x = 27\]
Cross multiply the values from left side of the equation to right side of the equation
\[ \Rightarrow x = 27 \times \dfrac{3}{2}\]
Calculate the product on right hand side of the equation
\[ \Rightarrow x = \dfrac{{81}}{2}\]
Divide the numerator from denominator
\[ \Rightarrow x = 40.5\] … (1)
Case 2: \[\dfrac{2}{3}x - 9 = - 18\]
Equate left hand side of the equation to right hand side of the equation
\[ \Rightarrow \dfrac{2}{3}x - 9 = - 18\]
Shift constant values to right hand side of the equation
\[ \Rightarrow \dfrac{2}{3}x = - 18 + 9\]
Add terms on right hand side
\[ \Rightarrow \dfrac{2}{3}x = - 9\]
Cross multiply the values from left side of the equation to right side of the equation
\[ \Rightarrow x = - 9 \times \dfrac{3}{2}\]
Calculate the product on right hand side of the equation
\[ \Rightarrow x = \dfrac{{ - 27}}{2}\]
Divide the numerator from denominator
\[ \Rightarrow x = - 13.5\] … (2)
So, from equations (1) and (2) we get the values for x as 40.5 and -13.5
\[\therefore \]Solution of the equation \[\left| {\dfrac{2}{3}x - 9} \right| = 18\] is \[x = 40.5\]and \[x = - 13.5\]
Note:
Many students make mistake of opening the left hand side of the equation using modulus i.e. they write positive and negative of left hand side equal to given right hand side, this gets complicated as we have fraction terms in left hand side and we will have to take LCM as well, so we choose to take positive and negative sign in right hand side of the equation as it contains only one value.
Recently Updated Pages
Master Class 10 General Knowledge: Engaging Questions & Answers for Success
Master Class 10 Computer Science: Engaging Questions & Answers for Success
Master Class 10 Science: Engaging Questions & Answers for Success
Master Class 10 Social Science: Engaging Questions & Answers for Success
Master Class 10 Maths: Engaging Questions & Answers for Success
Master Class 10 English: Engaging Questions & Answers for Success
Trending doubts
Assertion The planet Neptune appears blue in colour class 10 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
The term disaster is derived from language AGreek BArabic class 10 social science CBSE
Imagine that you have the opportunity to interview class 10 english CBSE
10 examples of evaporation in daily life with explanations
Differentiate between natural and artificial ecosy class 10 biology CBSE