Solving for x in $\frac{6x}{150}$ - $\frac{5x}{150}$ = 1

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In summary, the conversation discusses finding the value of x in the equation $\frac{x}{25}$ - $\frac{x}{30}$ = 1 by finding the least common multiple of 25 and 30 and using it to simplify the equation. It is determined that the value of x is 150.
  • #1
mathlearn
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$\frac{x}{25}$ - $\frac{x}{30}$ = 1 (Happy)

Okay, I see that they can come to a common denominator of,

$\frac{6x}{150}$ - $\frac{5x}{150}$ = 1 , Now to find x; (Happy)
 
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  • #2
I would begin by finding the LCM of 25 and 30. To do that, consider the prime factorizations:

\(\displaystyle 25=5^2\)

\(\displaystyle 30=2\cdot3\cdot5\)

Therefore:

\(\displaystyle \text{lcm}(25,30)=2\cdot3\cdot5^2=150\)

So, if we multiply the equation by 150, we obtain:

\(\displaystyle 6x-5x=150\)

\(\displaystyle x=150\)

Does that make sense?
 
  • #3
After getting into common denominator and subtracting,

$\frac{x}{150}$ =1

$\frac{x}{150}*150 =1*150 $

$ x = 150$ , Correct ? (Thinking)
 

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  • #4
mathlearn said:
$\frac{x}{25}$ - $\frac{x}{30}$ = 1 (Happy)

Okay, I see that they can come to a common denominator of,

$\frac{6x}{150}$ - $\frac{5x}{150}$ = 1 , Now to find x; (Happy)

Good work! Now, 6x/150 - 5x/150 = x/150 = 1, so x = 150. :)
 

FAQ: Solving for x in $\frac{6x}{150}$ - $\frac{5x}{150}$ = 1

What is the equation for solving for x in $\frac{6x}{150}$ - $\frac{5x}{150}$ = 1?

The equation for solving for x in $\frac{6x}{150}$ - $\frac{5x}{150}$ = 1 is:
$\frac{6x}{150}$ - $\frac{5x}{150}$ = 1
To solve for x, we need to isolate it on one side of the equation by using algebraic operations.

How do I solve for x in $\frac{6x}{150}$ - $\frac{5x}{150}$ = 1?

To solve for x in this equation, we need to combine the fractions on the left side by finding a common denominator. In this case, the common denominator is 150. So we can rewrite the equation as:
$\frac{6x - 5x}{150}$ = 1
Then we can simplify by combining like terms on the numerator:
$\frac{x}{150}$ = 1
To isolate x, we can multiply both sides by 150:
x = 150

Are there any restrictions or limitations for the value of x in this equation?

Yes, there are limitations for the value of x in this equation. Since we cannot have a denominator of 0, x cannot be equal to 0. Additionally, since x is in the denominator, it cannot be equal to any value that would make the fraction equal to 0. Therefore, x cannot equal 0 and x cannot equal any multiple of 150.

Can this equation have multiple solutions for x?

Yes, this equation can have multiple solutions for x. Since 150 is a common denominator, any value of x that is a multiple of 150 will satisfy the equation. For example, x = 150, x = 300, x = -150, etc.

How would I check if my solution for x is correct?

To check if your solution for x is correct, you can substitute the value of x back into the original equation and see if it satisfies the equation. For example, if you solved for x and got x = 150, you can plug it back into the equation:
$\frac{6(150)}{150}$ - $\frac{5(150)}{150}$ = 1
If the left side equals 1, then your solution for x is correct.

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