Answer: [tex]x=\frac{5}{2}, x=\frac{1}{3}[/tex]
Explanation:
The equation is:
[tex](2x-5)(3x-1)=0[/tex]
The term on the left consists of a product of two different factors: therefore, this product can be zero if either the first term (2x-5) or the second term (3x-1) is equal to zero.
This means that we can solve separately for the two terms:
[tex]2x-5=0\\3x-1=0[/tex]
Solving the first equation:
[tex]2x-5=0\\2x=5\\x=\frac{5}{2}[/tex]
Solving the second equation:
[tex]3x-1=0\\3x=1\\x=\frac{1}{3}[/tex]
Answer:
[tex]\textbf{The solution of the given equation are : }\frac{5}{2}\textbf{ and }\frac{1}{3}[/tex]
Step-by-step explanation:
The equation is given to be : (2x – 5)(3x – 1) = 0
We need to find the values of x such that the value of the equation is 0 and those corresponding values of x will be our required solution of the given equation.
Now, the product of two factors is given to be 0
⇒ Either of the two factors is equal to 0
So, first taking 2x - 5 = 0 and finding the value of x
⇒ 2x = 5
[tex]\implies\bf x=\frac{5}{2}[/tex]
Now, taking the second factor equal to 0 and finding the other value of x
⇒ 3x - 1 = 0
⇒ 3x = 1
[tex]\implies\bf x = \frac{1}{3}[/tex]
So, we get :
[tex]\bf x =\frac{5}{2}\textbf{ and }x=\frac{1}{3}[/tex]
Hence, these two values of x are the required solution of the given equation
[tex]\textbf{Hence, the solution of the given equation are : }\frac{5}{2}\textbf{ and }\frac{1}{3}[/tex]
