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2 years ago

10

Fractional linear equations 10= 3 over 5 (x+5)

1 answer:

vitfil [10]2 years ago

5 0

At first glance I believe you meant 10 = (3/5)(x+5). It's important that you share the instructions for each problem you post; I assume they state: "solve the following fractional linear equation for x."

Here are two approaches. Take your pick.
1) Distribute the multiplier (3/5) over (x+5). Resulting equation is

10=(3/5)x + 3. Subtr. 3 from both sides: 7=(3/5)x. Mult both sides by (5/3). Result: (35/3)=x.

2) Mult. the original equation by 5 to remove fractions: 50=3(x+5). Multiply out 3(x+5), to obtain 50=3x +15. Subtr. 15 from both sides. Result:
35=3x. Divide both sides by 3.

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Oksanka [162]

The answer is Option B $(x + 6)(x^2 + 8)$

Step-by-step explanation:

Step 1: Group the given cubic polynomial into two sections.

So polynomial can be grouped as

$(x^3 + 6x^2) + (8x + 48)$

Step 2:Find what's the common in each section.

In section $(x^3 + 6x^2)$ the come term is $x^2$

In section (8x + 48) the come term is 8

Step 3:Factor the commonalities out of the two terms.

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Factoring out 8 from the second section(8x + 48) , we will get 8(x + 6).

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3 years ago

Which classification describes the following system of equations? inconsistent and dependent consistent and dependent consistent

sveta [45]

The system of equations is consistent and independent and the solutions are x = 4, y = 5, and z = -1 option third is correct.

<h3>What is a linear equation?</h3>

It is defined as the relation between two variables, if we plot the graph of the linear equation we will get a straight line.

If in the linear equation, one variable is present, then the equation is known as the linear equation in one variable.

The question is incomplete.

The complete question is in the picture, please refer to the attached picture.

We have three linear equations in three variables:

x - y - z = 0 ..(1)

2x - y + 2z = 1 ..(2)

x - y + z = -2 ..(3)

From the equation (1):

$\rm x=y+z$

Plug this value in the equation (2) and (3):

$\rm 2\left(y+z\right)-y+2z=1\\\\ y+z-y+z=-2$

After solving we get:

z = -1

y = 5

Plug the above two values in equation (1) we get:

x = 4

Thus, the system of equations is consistent and independent and the solutions are x = 4, y = 5, and z = -1 option third is correct.

Learn more about the linear equation here:

brainly.com/question/11897796

#SPJ1

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1 year ago