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erma4kov [3.2K]
3 years ago
15

How to solve equations with variable in denominator?

Mathematics
1 answer:
OlgaM077 [116]3 years ago
4 0
In general you can multiply through by the LCD

For example solve :-

2/x  +  1 / (1+ x) =  11/12

The LCD is 12x(x + 1).

Now we multiply each term by 12x(x + 1) 

2*12(x+1) + 12x = 11x(x + 1)

and this is solved in the usual way.
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velikii [3]

Answer:

there is no intercept

Step-by-step explanation:

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3 years ago
For each system of equations, drag the true statement about its solution set to the box under the system?
natta225 [31]

Answer:

y = 4x + 2

y = 2(2x - 1)

Zero solutions.

4x + 2 can never be equal to 4x - 2

y = 3x - 4

y = 2x + 2

One solution

3x - 4 = 2x + 2 has one solution

Step-by-step explanation:

* Lets explain how to solve the problem

- The system of equation has zero number of solution if the coefficients

 of x and y are the same and the numerical terms are different

- The system of equation has infinity many solutions if the

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- The system of equation has one solution if at least one of the

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* Lets solve the problem

∵ y = 4x + 2 ⇒ (1)

∵ y = 2(2x - 1) ⇒ (2)

- Lets simplify equation (2) by multiplying the bracket by 2

∴ y = 4x - 2

- The two equations have same coefficient of y and x and different

  numerical terms

∴ They have zero equation

y = 4x + 2

y = 2(2x - 1)

Zero solutions.

4x + 2 can never be equal to 4x - 2

∵ y = 3x - 4 ⇒ (1)

∵ y = 2x + 2 ⇒ (2)

- The coefficients of x and y are different, then there is one solution

- Equate equations (1) and (2)

∴ 3x - 4 = 2x + 2

- Subtract 2x from both sides

∴ x - 4 = 2

- Add 4 to both sides

∴ x = 6

- Substitute the value of x in equation (1) or (2) to find y

∴ y = 2(6) + 2

∴ y = 12 + 2 = 14

∴ y = 14

∴ The solution is (6 , 14)

y = 3x - 4

y = 2x + 2

One solution

3x - 4 = 2x + 2 has one solution

3 0
3 years ago
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If he puts in $25 every month and there are 12 months in one year, just multiply 25 by 12 to find that Maurice will save $300 a year.
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3 years ago
Read 2 more answers
What two numbers multiply to 15 and add to 6
Katen [24]
Since there are no number that multiply to 155 and add to 6 you must now use the Quadratic Formula.

x=(-b(+/-)√b^{2}-4ac)
                    2a
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Answer:

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