
How do you solve the absolute value $\dfrac{1}{2}\left| {3c + 5} \right| = 6c + 4?$
Answer
551.7k+ views
Hint: First of all we will frame the two equations to get the absolute values from the equation given in the modulus and then will find the value for the unknown term “c”.
Complete step-by-step solution:
Take the given expression: $\dfrac{1}{2}\left| {3c + 5} \right| = 6c + 4$
Do cross multiplication where the denominator of one side is multiplied to the numerator on the opposite side.
$\left| {3c + 5} \right| = 2(6c + 4)$
Simplify the above equation-
$\left| {3c + 5} \right| = 12c + 8$
Now remove mode and get the two equations-
$ - (3c + 5) = 12c + 8$
$(3c + 5) = 12c + 8$
Now solve equations (A) and (B) one by one-
$ - (3c + 5) = 12c + 8$
Open the brackets, remember when there is a negative sign outside the bracket then the sign of the term inside the bracket changes.
$ - 3c - 5 = 12c + 8$
Make like terms together. Move constants on one side and the term with variables on the other side. . When you move any term from one side to another then the sign of the term also changes. Positive term becomes negative and negative term becomes positive.
$ - 5 - 8 = 12c + 3c$
Simplify the above equation
$ - 13 = 15c$
The above equation can be re-written as –
$15c = - 13$
Term multiplicative on one side if moved to the opposite side then it goes to the denominator.
$c = - \dfrac{{13}}{{15}}$ …. (C)
$3c + 5 = 12c + 8$
Make like terms together. Move constants on one side and the term with variables on the other side. . When you move any term from one side to another then the sign of the term also changes. Positive term becomes negative and negative term becomes positive.
$ + 5 - 8 = 12c - 3c$
Simplify the above equation
$ - 3 = 9c$
Make “C” the subject
$ \Rightarrow c = - \dfrac{3}{9}$
Common factors from the numerator and the denominator cancel each other.
$ \Rightarrow c = - \dfrac{1}{3}$ … (D)
Hence, the required solution is $c = - \dfrac{{13}}{{15}}$or $c = - \dfrac{1}{3}$
Note: Be careful about the sign convention while simplification. When you simplify between the two negative terms or two positive terms you have to do addition and give sign of bigger number while when you simplify between one negative term or the positive term you have to do subtraction and give sign of the bigger number to the resultant value.
Complete step-by-step solution:
Take the given expression: $\dfrac{1}{2}\left| {3c + 5} \right| = 6c + 4$
Do cross multiplication where the denominator of one side is multiplied to the numerator on the opposite side.
$\left| {3c + 5} \right| = 2(6c + 4)$
Simplify the above equation-
$\left| {3c + 5} \right| = 12c + 8$
Now remove mode and get the two equations-
$ - (3c + 5) = 12c + 8$
$(3c + 5) = 12c + 8$
Now solve equations (A) and (B) one by one-
$ - (3c + 5) = 12c + 8$
Open the brackets, remember when there is a negative sign outside the bracket then the sign of the term inside the bracket changes.
$ - 3c - 5 = 12c + 8$
Make like terms together. Move constants on one side and the term with variables on the other side. . When you move any term from one side to another then the sign of the term also changes. Positive term becomes negative and negative term becomes positive.
$ - 5 - 8 = 12c + 3c$
Simplify the above equation
$ - 13 = 15c$
The above equation can be re-written as –
$15c = - 13$
Term multiplicative on one side if moved to the opposite side then it goes to the denominator.
$c = - \dfrac{{13}}{{15}}$ …. (C)
$3c + 5 = 12c + 8$
Make like terms together. Move constants on one side and the term with variables on the other side. . When you move any term from one side to another then the sign of the term also changes. Positive term becomes negative and negative term becomes positive.
$ + 5 - 8 = 12c - 3c$
Simplify the above equation
$ - 3 = 9c$
Make “C” the subject
$ \Rightarrow c = - \dfrac{3}{9}$
Common factors from the numerator and the denominator cancel each other.
$ \Rightarrow c = - \dfrac{1}{3}$ … (D)
Hence, the required solution is $c = - \dfrac{{13}}{{15}}$or $c = - \dfrac{1}{3}$
Note: Be careful about the sign convention while simplification. When you simplify between the two negative terms or two positive terms you have to do addition and give sign of bigger number while when you simplify between one negative term or the positive term you have to do subtraction and give sign of the bigger number to the resultant value.
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