
How do you solve $\left| {8 - 2x} \right| = 4$?
Answer
554.1k+ views
Hint: Here we solve this absolute value equation by clear the absolute values bars by splitting the equation into its two cases. Here, one is the positive case and the remaining for the negative case. On doing some simplification we get the required answer.
Complete step-by-step solution:
Here absolute value equation is $\left| {8 - 2x} \right| = 4$,
Rearrange this absolute value equation and clear the absolute value bars by splitting the equation into its two cases, one for the positive case and the other for the negative case.
The absolute value term is$\left| {8 - 2x} \right|$ ,
For the negative case we will use $ - \left( {8 - 2x} \right)$ and for the positive case we will use $\left( {8 - 2x} \right)$ ,
Solve the negative case$ - \left( {8 - 2x} \right)$,
Given equation becomes$ - \left( {8 - 2x} \right) = 4$,
Multiply the $' - 1'$into both sides we get,
$ \Rightarrow 8 - 2x = - 4$
Our targeting variable is $x$ , rearrange the balance term we get,
\[ \Rightarrow - 2x = - 4 - 8\]
On add the term and we get,
$ \Rightarrow - 2x = - 12$
Divide $' - 2'$ both sides we get,
$ \Rightarrow \dfrac{{ - 2x}}{{ - 2}} = \dfrac{{ - 12}}{{ - 2}}$
Let us cancel the term and we get,
$ \Rightarrow x = 6$
This is the solution of negative case
Now for the positive case,
Given equation becomes $\left( {8 - 2x} \right) = 4$
$ \Rightarrow 8 - 2x = 4$
Our targeting variable is $x$, rearrange the balance term we get,
\[ \Rightarrow - 2x = 4 - 8\]
On subtract the term and we get,
$ \Rightarrow - 2x = - 4$
Divide $' - 2'$ both sides we get,
$ \Rightarrow \dfrac{{ - 2x}}{{ - 2}} = \dfrac{{ - 4}}{{ - 2}}$
On cancel the term and we get,
$ \Rightarrow x = 2$
This is the solution to the positive case.
Now the solution of the absolute value equation is $x = 6$ , $x = 2$
Note: The absolute value of $x$ is written as $\left| x \right|$. It was the following properties:
If $x \geqslant 0$, then $\left| x \right| = x$.
If $x < 0$ , then $\left| x \right| = - x$.
For real numbers $A$ and $B$, an equation of the form $\left| A \right| = B$, with $B \geqslant 0$ , will have solutions when $A = B$ or $A = - B$. If $B < 0$ , the equation $\left| A \right| = B$has no solution.
An absolute value equation in the form $\left| {ax + b} \right| = c$ has the following property:
If $c < 0$ , $\left| {ax + b} \right| = c$ has no solution.
If $c = 0$ , $\left| {ax + b} \right| = c$ has one solution.
If $c > 0$ , $\left| {ax + b} \right| = c$ has two solutions.
Complete step-by-step solution:
Here absolute value equation is $\left| {8 - 2x} \right| = 4$,
Rearrange this absolute value equation and clear the absolute value bars by splitting the equation into its two cases, one for the positive case and the other for the negative case.
The absolute value term is$\left| {8 - 2x} \right|$ ,
For the negative case we will use $ - \left( {8 - 2x} \right)$ and for the positive case we will use $\left( {8 - 2x} \right)$ ,
Solve the negative case$ - \left( {8 - 2x} \right)$,
Given equation becomes$ - \left( {8 - 2x} \right) = 4$,
Multiply the $' - 1'$into both sides we get,
$ \Rightarrow 8 - 2x = - 4$
Our targeting variable is $x$ , rearrange the balance term we get,
\[ \Rightarrow - 2x = - 4 - 8\]
On add the term and we get,
$ \Rightarrow - 2x = - 12$
Divide $' - 2'$ both sides we get,
$ \Rightarrow \dfrac{{ - 2x}}{{ - 2}} = \dfrac{{ - 12}}{{ - 2}}$
Let us cancel the term and we get,
$ \Rightarrow x = 6$
This is the solution of negative case
Now for the positive case,
Given equation becomes $\left( {8 - 2x} \right) = 4$
$ \Rightarrow 8 - 2x = 4$
Our targeting variable is $x$, rearrange the balance term we get,
\[ \Rightarrow - 2x = 4 - 8\]
On subtract the term and we get,
$ \Rightarrow - 2x = - 4$
Divide $' - 2'$ both sides we get,
$ \Rightarrow \dfrac{{ - 2x}}{{ - 2}} = \dfrac{{ - 4}}{{ - 2}}$
On cancel the term and we get,
$ \Rightarrow x = 2$
This is the solution to the positive case.
Now the solution of the absolute value equation is $x = 6$ , $x = 2$
Note: The absolute value of $x$ is written as $\left| x \right|$. It was the following properties:
If $x \geqslant 0$, then $\left| x \right| = x$.
If $x < 0$ , then $\left| x \right| = - x$.
For real numbers $A$ and $B$, an equation of the form $\left| A \right| = B$, with $B \geqslant 0$ , will have solutions when $A = B$ or $A = - B$. If $B < 0$ , the equation $\left| A \right| = B$has no solution.
An absolute value equation in the form $\left| {ax + b} \right| = c$ has the following property:
If $c < 0$ , $\left| {ax + b} \right| = c$ has no solution.
If $c = 0$ , $\left| {ax + b} \right| = c$ has one solution.
If $c > 0$ , $\left| {ax + b} \right| = c$ has two solutions.
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