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enot [183]
3 years ago
9

Helppp please I’m desperate

Mathematics
1 answer:
IRINA_888 [86]3 years ago
7 0
The elimination method works by adding the two equations and eliminating one variable. Then you solve an equation in one variable. Finally, you use substitution or elimination again to find the other variable. Sometimes, by simply adding the equations, a variable is not eliminated. Then you need to multiply one or both equations by a factor to get a variable to be eliminated.

A)
2x - 4y = 8
x + 3y = -11

Adding the equations does not eliminate x or y.
Notice that in the first equation, every coefficient is even. We can divide both sides of the first equation by 2. Then the first term would be x. Instead, let's divide both sides of the first equation by -2. Then the x's will be eliminated.

-x + 2y = -4   <-- The first equation divided by -2
x + 3y = -11  <-- The original second equation. Now we add the equations.
----------------
      5y = -15

        y = -3

Now that we know y = -3, we substitute it into the first original equation and solve for x.

2x - 4y = 8

2x - 4(-3) = 8

2x + 12 = 8

2x = -4

x = -2

Answer: x = -2; y = -3

B)
We see that the x's will be eliminated by addition. Just add the equations.

-3x + 7y = 9
 3x  - 7y = 1
---------------
   0 + 0   = 10

0 = 10  <---- this is a false statement, so this system of equations has no solution.
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Hello, please consider the following.

We will multiply the numerator and denominator by

4+\sqrt{6x}

to get rid of the root in the denominator.

First of all, we cannot divide by 0, right? So, we need to make sure that the denominator is different from 0.

4-\sqrt{6x} =0\sqrt{6x}=4\\\\\text{Take the square}\\\\6x=4^2=16\\\\x=\dfrac{16}{6}=\dfrac{8}{3}

We need to take any x real number different from 8/3 then and simplify the expression.

Let's do it!

\begin{aligned}\dfrac{4}{4-\sqrt{6x}}&=\dfrac{4(4+\sqrt{6x})}{(4+\sqrt{6x})(4-\sqrt{6x})}\\\\&=\dfrac{4(4+\sqrt{6x})}{(4^2-\sqrt{6x}^2)}\\\\&=\dfrac{4(4+\sqrt{6x})}{(16-6x)}\\\\&=\dfrac{2(4+\sqrt{6x})}{(8-3x)}\\\\&\large \boxed{=\dfrac{8+2\sqrt{6x}}{8-3x}}\end{aligned}

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