
What is the solution set for $\left| {3x - 1} \right| = x + 5$ ?
Answer
523.8k+ views
Hint: Here we are going to find the solution of the given absolute value of the expression. Generally, the absolute value of a number is a non-negative number.
Complete step by step solution:
Given $\left| {3x - 1} \right| = x + 5$
The first thing that you need to notice here is that the expression on the right side of the equation must be positive because it represents the absolute value in the expression $3x - 1$.
So, any solution that does satisfy the condition, that is ,
$x + 5 \geqslant 0$
$ \to x \geqslant - 5$
will be an extraneous solution.
Now, we need to take into account two possibilities for this equation, that is ,
First case is $3x - 1 > 0$ and another one is $3x - 1 < 0$,
which means that for the first case,
$\left| {3x - 1} \right| = 3x - 1$ and the equation becomes,
$3x - 1 = x + 5$
Isolate the $x$ term and find the values of $x$we get,
$3x - x = 5 + 1$
On simplifying it we get,
$2x = 6$
Therefore we get, $x = \dfrac{6}{2} = 3$
$x = 3$
Now, for the second case $3x - 1 < 0$ which means that,
$\left| {3x - 1} \right| = - \left( {3x - 1} \right)$ and the equation becomes,
$ - \left( {3x - 1} \right) = x + 5$
$ - 3x + 1 = x + 5$
Isolate the $x$ term and find the values of $x$we get,
$3x + x = 1 - 5$
On simplifying it we get,
$4x = - 4$
Therefore we get, $x = \dfrac{{ - 4}}{4} = - 1$
$x = - 1$
Since both values satisfy the condition $x \geqslant - 5$ , they are both valid solutions to the equation.
Note: Absolute value means to remove any negative sign in front of a number, and to think of all numbers as positive or zero. $\left| x \right|$ which says that absolute value of $x$ equals , $\left| x \right| = x$ when $x$ is greater than zero, and $\left| x \right| = 0$ when $x$ is equals to zero, $\left| x \right| = - x$ when $x$ is less than zero this flips the number back to positive. So, when a number is positive or zero we leave it alone, when it is negative we change it to positive using$ - x$.
Complete step by step solution:
Given $\left| {3x - 1} \right| = x + 5$
The first thing that you need to notice here is that the expression on the right side of the equation must be positive because it represents the absolute value in the expression $3x - 1$.
So, any solution that does satisfy the condition, that is ,
$x + 5 \geqslant 0$
$ \to x \geqslant - 5$
will be an extraneous solution.
Now, we need to take into account two possibilities for this equation, that is ,
First case is $3x - 1 > 0$ and another one is $3x - 1 < 0$,
which means that for the first case,
$\left| {3x - 1} \right| = 3x - 1$ and the equation becomes,
$3x - 1 = x + 5$
Isolate the $x$ term and find the values of $x$we get,
$3x - x = 5 + 1$
On simplifying it we get,
$2x = 6$
Therefore we get, $x = \dfrac{6}{2} = 3$
$x = 3$
Now, for the second case $3x - 1 < 0$ which means that,
$\left| {3x - 1} \right| = - \left( {3x - 1} \right)$ and the equation becomes,
$ - \left( {3x - 1} \right) = x + 5$
$ - 3x + 1 = x + 5$
Isolate the $x$ term and find the values of $x$we get,
$3x + x = 1 - 5$
On simplifying it we get,
$4x = - 4$
Therefore we get, $x = \dfrac{{ - 4}}{4} = - 1$
$x = - 1$
Since both values satisfy the condition $x \geqslant - 5$ , they are both valid solutions to the equation.
Note: Absolute value means to remove any negative sign in front of a number, and to think of all numbers as positive or zero. $\left| x \right|$ which says that absolute value of $x$ equals , $\left| x \right| = x$ when $x$ is greater than zero, and $\left| x \right| = 0$ when $x$ is equals to zero, $\left| x \right| = - x$ when $x$ is less than zero this flips the number back to positive. So, when a number is positive or zero we leave it alone, when it is negative we change it to positive using$ - x$.
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