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m_a_m_a [10]
2 years ago
15

What is the solution set of the equation 3X=-6/1-X for x # 1?

Mathematics
1 answer:
Irina-Kira [14]2 years ago
3 0

The solution set of the equation 3X=-6/1-X for x # 1 is {-2,1)

3x = -6/(1-x)

3x(1-x) = -6

3x – 3x2 = -6

Taking 3 as common from both sides

x –x2 = -2

x2 – x +2 =0

Using factorisation method we will get two factors

Factoring quadratics is a method of expressing the quadratic equation ax2 + bx + c = 0 as a product of its linear factors as (x - k)(x - h), where h, k are the roots of the quadratic equation ax2 + bx + c = 0. This method is also is called the method of factorization of quadratic equations. Factorization of quadratic equations can be done using different methods such as splitting the middle term, using the quadratic formula, completing the squares, etc.

( x – 2)(x+1)=0

So, x = 2 , -1

Now, when x ≠ 1

The solution of the equation will vary from {-2,1) .

Learn more about the factorisation here brainly.com/question/13496719

#SPJ9

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Answer:

1/8

Step-by-step explanation:

-5/8+3/4 =( -20+24)/32 = 4/32

4/32= 1/8

4 0
3 years ago
A system of linear equations is graphed.
marysya [2.9K]

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Step-by-step explanation:

It cant be A or C since the intersection does not touch the y axis and it isn't D since it is closer to -4 than -4 1/2

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3 years ago
Simplify 42 ⋅ 48. Can someone please help me?
defon
Since simplify means multiply you would do
42 × 48 = 2,016

Your answer is 2,016

Hope I helped :)
3 0
4 years ago
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You have to pick a marble twice. Explain what you must do to make the picking of the marbles independent of one another.
Serggg [28]

Answer:

You make it independent by putting it back.

Step-by-step explanation:

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8 0
3 years ago
(3xy/3x^2-12)*(x^2+3x+2/xy+y)
TiliK225 [7]

Value of expression (3xy/3x^2-12)*(x^2+3x+2/xy+y) is  \frac{x}{x-2} .

<u>Step-by-step explanation:</u>

Here we need to evaluate expression : (3xy/3x^2-12)*(x^2+3x+2/xy+y)  or ,

(3xy/3x^2-12)*(x^2+3x+2/xy+y)

Let's simplify this

⇒ (\frac{3xy}{3x^2-12})(\frac{x^2+3x+2}{xy+y})

Factorizing the terms we get:

⇒ (\frac{3xy}{3(x^2-4)})(\frac{x^2+2x+x+2}{y(x+1)})              { a^2-b^2=(a+b)(a-b)      }

⇒ (\frac{xy}{(x-2)(x+2)})(\frac{x(x+2)+1(x+2)}{y(x+1)})

⇒ (\frac{xy}{(x-2)(x+2)})(\frac{(x+1)(x+2)}{y(x+1)})

Cancelling similar terms we get:

⇒ \frac{x}{x-2}

Therefore , Value of expression (3xy/3x^2-12)*(x^2+3x+2/xy+y) is  \frac{x}{x-2} .

3 0
3 years ago
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