Answer
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Hint: To solve such questions first consider both the condition of absolute value. Then add the number $9$ to both the left-hand side and right-hand side of the given equation. Then multiply both LHS and RHS with the number $3$ . Next divide both LHS and RHS with number $2$ . Finally canceling out the common term in numerator and denominator we get the required solution.
Complete step by step answer:
Given $\left| {\dfrac{2}{3}x - 9} \right| = 18$
It is asked to find the solution to the given equation.
Write the given equation as shown below, that is,
$\left| {\dfrac{{2x}}{3} - 9} \right| = 18$
First, consider $\dfrac{{2x}}{3} - 9 = 18$
Next add the number $9$ to both the left-hand side and right-hand side of the equation, that is,
$\dfrac{{2x}}{3} - 9 + 9 = 18 + 9$
Simplifying further we get,
$\dfrac{{2x}}{3} = 27$
Multiply both the sides of the above equation with the number $3$, that is
$\dfrac{{2x}}{3} \times 3 = 27 \times 3$
We get
$2x = 81$
Next divide both sides of the equation with the number $2$, that is,
$\dfrac{{2x}}{2} = \dfrac{{81}}{2}$
Canceling the common terms in numerator and denominator we get
$x = \dfrac{{81}}{2}$
Next, consider $\dfrac{{2x}}{3} - 9 = - 18$
Next add the number $9$ to both the left-hand side and right-hand side of the equation, that is,
$\dfrac{{2x}}{3} - 9 + 9 = - 18 + 9$
Simplifying further we get,
$\dfrac{{2x}}{3} = - 9$
Multiply both the sides of the above equation with the number $3$, that is
$\dfrac{{2x}}{3} \times 3 = - 9 \times 3$
We get
$2x = - 27$
Next divide both sides of the equation with the number $2$, that is,
$\dfrac{{2x}}{2} = \dfrac{{ - 27}}{2}$
Canceling the common terms in numerator and denominator we get
$x = \dfrac{{ - 27}}{2}$
Hence the solution of the given equation are $x = \dfrac{{81}}{2}$ and $x = \dfrac{{ - 27}}{2}$
Additional information:
An equation that can be written in the form $ax + by + c = 0$, where $a$, $b$ and $c$ are real numbers, and $a$ and $b$ are not both zero, is known as a linear equation in two variables $x$ and $y$. Every solution of the equation is a point on the line representing it.
Note: One might not forget to consider the modulus, due to which we have to consider two values for the right-hand side. While solving this type of question always start by reducing the left-hand side of the given equation and then solve. To check whether the obtained solution is correct or not, substitute the obtained solution in the given equation.
Complete step by step answer:
Given $\left| {\dfrac{2}{3}x - 9} \right| = 18$
It is asked to find the solution to the given equation.
Write the given equation as shown below, that is,
$\left| {\dfrac{{2x}}{3} - 9} \right| = 18$
First, consider $\dfrac{{2x}}{3} - 9 = 18$
Next add the number $9$ to both the left-hand side and right-hand side of the equation, that is,
$\dfrac{{2x}}{3} - 9 + 9 = 18 + 9$
Simplifying further we get,
$\dfrac{{2x}}{3} = 27$
Multiply both the sides of the above equation with the number $3$, that is
$\dfrac{{2x}}{3} \times 3 = 27 \times 3$
We get
$2x = 81$
Next divide both sides of the equation with the number $2$, that is,
$\dfrac{{2x}}{2} = \dfrac{{81}}{2}$
Canceling the common terms in numerator and denominator we get
$x = \dfrac{{81}}{2}$
Next, consider $\dfrac{{2x}}{3} - 9 = - 18$
Next add the number $9$ to both the left-hand side and right-hand side of the equation, that is,
$\dfrac{{2x}}{3} - 9 + 9 = - 18 + 9$
Simplifying further we get,
$\dfrac{{2x}}{3} = - 9$
Multiply both the sides of the above equation with the number $3$, that is
$\dfrac{{2x}}{3} \times 3 = - 9 \times 3$
We get
$2x = - 27$
Next divide both sides of the equation with the number $2$, that is,
$\dfrac{{2x}}{2} = \dfrac{{ - 27}}{2}$
Canceling the common terms in numerator and denominator we get
$x = \dfrac{{ - 27}}{2}$
Hence the solution of the given equation are $x = \dfrac{{81}}{2}$ and $x = \dfrac{{ - 27}}{2}$
Additional information:
An equation that can be written in the form $ax + by + c = 0$, where $a$, $b$ and $c$ are real numbers, and $a$ and $b$ are not both zero, is known as a linear equation in two variables $x$ and $y$. Every solution of the equation is a point on the line representing it.
Note: One might not forget to consider the modulus, due to which we have to consider two values for the right-hand side. While solving this type of question always start by reducing the left-hand side of the given equation and then solve. To check whether the obtained solution is correct or not, substitute the obtained solution in the given equation.
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