
How do you solve \[\left| {8x + - 2} \right| = 10\]?
Answer
545.4k+ views
Hint: Here in this question, we have to solve the given absolute equation to the x variable. The given equation is the algebraic equation with one variable x, this can be solve by add or subtract the necessary term from each side of the equation to isolate the term with the variable x, then multiply or divide each side of the equation by the appropriate value, while keeping the equation balanced then solve the resultant balance equation for the x value.
Complete step by step solution:
The absolute value function takes any negative or positive term and transforms it to its positive form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent.
Consider the given equation
\[\left| {8x + - 2} \right| = 10\]
First, we have to remove modulus we get the two solution, i.e.,
\[\left( {8x + - 2} \right) = 10\]------(1)
\[ - \left( {8x + - 2} \right) = 10\]----(2)
Solution of equation (1) is
Consider
\[ \Rightarrow \,\,\,\left( {8x + - 2} \right) = 10\]
Or
\[ \Rightarrow \,\,\,8x + - 2 = 10\]
Add 2 on both the side, then
\[ \Rightarrow \,\,\,8x + - 2 + 2 = 10 + 2\]
On simplification, we get
\[ \Rightarrow \,\,\,8x + 0 = 12\]
\[ \Rightarrow \,\,\,8x = 12\]
To isolate x, divide 8 by both side
\[ \Rightarrow \,\,\,\dfrac{8}{8}x = \dfrac{{12}}{8}\]
\[ \Rightarrow \,\,\,x = \dfrac{{12}}{8}\]
Divide both numerator and denominator of RHS by 4, then
\[ \Rightarrow \,\,\,x = \dfrac{3}{2}\]
Solution of equation (2) is
Consider
\[ \Rightarrow \,\,\, - \left( {8x + - 2} \right) = 10\]
Multiply both side by -1
\[ \Rightarrow \,\,\,8x + - 2 = - 10\]
Add 2 on both the side, then
\[ \Rightarrow \,\,\,8x + - 2 + 2 = - 10 + 2\]
On simplification, we get
\[ \Rightarrow \,\,\,8x + 0 = - 8\]
\[ \Rightarrow \,\,\,8x = - 8\]
To isolate x, divide 8 by both side
\[ \Rightarrow \,\,\,\dfrac{8}{8}x = \dfrac{{ - 8}}{8}\]
\[ \Rightarrow \,\,\,x = - 1\]
Hence, the required solutions are \[x = \dfrac{3}{2}\] and \[x = - 1\].
So, the correct answer is “\[x = \dfrac{3}{2}\] and \[x = - 1\]”.
Note: The given equation contains the mod symbol, as we know that the mod can take both positive and negative. so we consider both positive and negative and with the help of simple arithmetic operations we are going to solve the equation for the variable “x”. We can also verify the equation.
Complete step by step solution:
The absolute value function takes any negative or positive term and transforms it to its positive form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent.
Consider the given equation
\[\left| {8x + - 2} \right| = 10\]
First, we have to remove modulus we get the two solution, i.e.,
\[\left( {8x + - 2} \right) = 10\]------(1)
\[ - \left( {8x + - 2} \right) = 10\]----(2)
Solution of equation (1) is
Consider
\[ \Rightarrow \,\,\,\left( {8x + - 2} \right) = 10\]
Or
\[ \Rightarrow \,\,\,8x + - 2 = 10\]
Add 2 on both the side, then
\[ \Rightarrow \,\,\,8x + - 2 + 2 = 10 + 2\]
On simplification, we get
\[ \Rightarrow \,\,\,8x + 0 = 12\]
\[ \Rightarrow \,\,\,8x = 12\]
To isolate x, divide 8 by both side
\[ \Rightarrow \,\,\,\dfrac{8}{8}x = \dfrac{{12}}{8}\]
\[ \Rightarrow \,\,\,x = \dfrac{{12}}{8}\]
Divide both numerator and denominator of RHS by 4, then
\[ \Rightarrow \,\,\,x = \dfrac{3}{2}\]
Solution of equation (2) is
Consider
\[ \Rightarrow \,\,\, - \left( {8x + - 2} \right) = 10\]
Multiply both side by -1
\[ \Rightarrow \,\,\,8x + - 2 = - 10\]
Add 2 on both the side, then
\[ \Rightarrow \,\,\,8x + - 2 + 2 = - 10 + 2\]
On simplification, we get
\[ \Rightarrow \,\,\,8x + 0 = - 8\]
\[ \Rightarrow \,\,\,8x = - 8\]
To isolate x, divide 8 by both side
\[ \Rightarrow \,\,\,\dfrac{8}{8}x = \dfrac{{ - 8}}{8}\]
\[ \Rightarrow \,\,\,x = - 1\]
Hence, the required solutions are \[x = \dfrac{3}{2}\] and \[x = - 1\].
So, the correct answer is “\[x = \dfrac{3}{2}\] and \[x = - 1\]”.
Note: The given equation contains the mod symbol, as we know that the mod can take both positive and negative. so we consider both positive and negative and with the help of simple arithmetic operations we are going to solve the equation for the variable “x”. We can also verify the equation.
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