All categories

4 years ago

Is 2 x+6-x=1/3(3x+18) a one solution ,no solution ,or an infinite numbers of solutions

1 answer:

5 0

To determine this, the first thing we should do is simplify the expression. As usual, bring like terms to either side of the "=" and then simplify.
2x + 6 - x = 1/3 (3x + 18)
x + 6 = 1/3 (3x + 18)
x + 6 = 1/3 (3x) + 1/3(18)
x + 6 = x + 6

We can already see that both sides have the same expression "x+6". Now, if you try to simplify any further, you will get 0=0. That means that for any value of x, the expression will simplify to 0=0. Because 0=0 is a true statement, for every value of x, the expression will be true. Therefore, there are an infinite number of solutions.

Hope this helped!! :D
Note: If you are still confused, let me know. I realize that I might have been a bit confusing in my explanation

You might be interested in

Graph the solution to the inequality<br> |y+5|> 2

kirill [66]

Answer:

y>-3 y>-7

Step-by-step explanation:

-(y+5)>2

y<-7

7 0

4 years ago

The solutions to a certain quadratic equation are x = -4 and x = 3. Write the equation in standard form below.

elena-14-01-66 [18.8K]

$y=a(x-x_1)(x-x_2)\\\\x_1=-4;\ x_2=3\ therefore\\\\y=a(x+4)(x-3)\\\\y=a(x^2+x-12)\\\\\boxed{y=ax^2+ax-12a}\ where\ a\in\mathbb{R}-\{0\}$

5 0

3 years ago

Read 2 more answers

Select the best answer for the question.<br> 10. Divide 715 by 5.

MAXImum [283]

Answer:143

Step-by-step explanation:5 goes into 715 143 times

7 0

3 years ago

In the diagram, PQRS TUVW$ . Find the value of x

Feliz [49]

Answer:

The answer is x = 8.

Step-by-step explanation:

6 0

3 years ago

A number is selected, at random, from the set {1,2,3,4,5,6,7,8}.

Olegator [25]

Applying the formula, you have:

A = the number is prime

B = the number is odd

I assume that with "random" you imply that all numbers can be chosen with the same probability. So, we have

$P(A) = \dfrac{4}{8} = \dfrac{1}{2}$

because 4 out of 8 numbers are prime: 2, 3, 5 and 7.

Similarly, we have

$P(B) = \dfrac{4}{8} = \dfrac{1}{2}$

because 4 out of 8 numbers are odd: 1, 3, 5 and 7.

Finally,

$P(A \land B) = \dfrac{3}{8}$

because 3 out of 8 numbers are prime and odd: 3, 5 and 7.

So, applying the formula, we have

$P(\text{prime } | \text{ odd}) = \dfrac{P(\text{prime and odd})}{P(\text{odd})} = \dfrac{\frac{3}{8}}{\frac{1}{2}} = \dfrac{3}{8}\cdot 2 = \dfrac{3}{4}$

Note:

I think that it is important to have a clear understanding of what's happening from a conceptual point of you: conditional probability simply changes the space you're working with: you are not asking "what is the probability that a random number, taken from 1 to 8, is prime?"

Rather, you are adding a bit of information, because you are asking "what is the probability that a random number, taken from 1 to 8, is prime, knowing that it's odd?"

So, we're not working anymore with the space {1,2,3,4,5,6,7,8}, but rather with {1,3,5,7} (we already know that our number is odd).

Out of these 4 odd numbers, 3 are primes. This is why the probability of picking a prime number among the odd numbers in {1,2,3,4,5,6,7,8} is 3/4: they are literally 3 out of 4.

5 0

4 years ago