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3 years ago

What value of x is in the solution set of 2x – 3 > 11 – 5x?

Mathematics

2 answers:

valentina_108 [34]3 years ago

8 0

Answer:

Option D is correct.

Value of x is in the solution set of given equation is., 4

Step-by-step explanation:

Given that: $2x -3 > 11-5x$ .....[1]

Find the value of x is in the solution set;

Add 5x to both sides of equation [1];

$2x -3 +5x > 11-5x+5x$

Simplify:

$7x -3> 11$

Add 3 to both sides we get;

$7x > 14$

Divide both sides by 7 we get;

x > 2.

Then, we have the solution set for x is: $(2, \infty)$

Therefore, the value of x is in the solution set is, 4

Alekssandra [29.7K]3 years ago

3 0

$2x - 3 \ \textgreater \ 11- 5x\\
7x\ \textgreater \ 14\\
x\ \textgreater \ 2$

4

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2 years ago

Read 2 more answers

10 + 3x < 4 or 2x + 5 > 11 in interval notation

Stella [2.4K]

Answer:

( − ∞ , ∞ ) hope i help

Step-by-step explanation:

First, solve each inequality. I'll solve the first one first.

7 ≥ 2 x − 5

12 ≥ 2 x

6 ≥ x

Therefore, x could be any number less than or equal to 6. In interval notation, this looks like:

( − ∞ , 6 ]

The parenthesis means that the lower end is not a solution, but every number above it is. (In this case, the lower end is infinity, so a parenthesis must be used, since infinity is not a real number and so it cannot be a solution.) The bracket means that the upper end is a solution. In this case, it indicates that not only could

be any number less than 6, but it could also be 6.

Let's try the second example:

3 x − 2 4 > 4

3 x − 2 > 16 3 x > 18 x > 6

Therefore, x could be any number greater than 6, but x couldn't be 6, since that would make the two sides of the inequality equal. In interval notation, this looks like:

( 6 , ∞ )

The parentheses mean that neither end of this range is included in the solution set. In this case, it indicates that neither 6 nor infinity are solutions, but every number in between 6 and infinity is a solution (that is, every real number greater than 6 is a solution).

Now, the problem used the word "OR", meaning that either of these equations could be true. That means that either x is on the interval ( − ∞ , 6 ] or the interval ( 6 , ∞ )

. In other words, x

is either less than or equal to 6, or it is greater than 6. When you combine these two statements, it becomes clear that

could be any real number, since no matter what number

is, it will fall in one of these intervals. The interval "all real numbers" is written like this:

( − ∞ , ∞ )

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3 years ago

Solve for x: 4 over x plus 4 over quantity x squared minus 9 equals 3 over quantity x minus 3. (2 points) Select one: a. x = -4

Kay [80]

Answer:

Step-by-step explanation:

$\dfrac{4}{x}+\dfrac{4}{x^2-9}=\dfrac{3}{x-3}$

Domain:

$x\neq0\ \wedge\ x^2-9\neq0\ \wedge\ x-3\neq0\\\\x\neq0\ \wedge\ x\neq\pm3$

solution:

$\dfrac{4}{x}+\dfrac{4}{x^2-3^2}=\dfrac{3}{x-3}$

use <em>(a - b)(a + b) = a² - b²</em>

$\dfrac{4}{x}+\dfrac{4}{(x-3)(x+3)}=\dfrac{3}{x-3}$

multiply both sides by (x - 3) ≠ 0

$\dfrac{4(x-3)}{x}+\dfrac{4(x-3)}{(x-3)(x+3)}=\dfrac{3(x-3)}{x-3}$

cancel (x - 3)

$\dfrac{4(x-3)}{x}+\dfrac{4}{x+3}=3$

subtract $\frac{4(x-3)}{x}$ from both sides

$\dfrac{4}{x+3}=3-\dfrac{4(x-3)}{x}\\\\\dfrac{4}{x+3}=\dfrac{3x}{x}-\dfrac{(4)(x)+(4)(-3)}{x}\\\\\dfrac{4}{x+3}=\dfrac{3x-\bigg(4x-12\bigg)}{x}\\\\\dfrac{4}{x+3}=\dfrac{3x-4x-(-12)}{x}\\\\\dfrac{4}{x+3}=\dfrac{-x+12}{x}$