If you were to make both x's equivalent, you'd have to multiply the first equation by 2 to get rid of the coefficient, 1/2. However, you'd have to multiply 2 onto -8 as well, therefore the equation would turn into x - 16 = 18. That is not equivalent to x - 8 = 18.
Answer:
it so blurry
Step-by-step explanation:
The answer is x=2 i hope this helps !!
To answer this question we will have to understand the nature of
.
can be rewritten as:
. Now,
is an irrational number and thus any of the four operations of addition, multiplication, subtraction or division placed anywhere between any of these terms will always result in an irrational number.
The only way irrational numbers can be made rational is by dividing or multiplying the given irrational number by the same irrational number.
Answer: ![\frac{\sqrt[4]{10xy^3}}{2y}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B4%5D%7B10xy%5E3%7D%7D%7B2y%7D)
where y is positive.
The 2y in the denominator is not inside the fourth root
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Work Shown:
![\sqrt[4]{\frac{5x}{8y}}\\\\\\\sqrt[4]{\frac{5x*2y^3}{8y*2y^3}}\ \ \text{.... multiply top and bottom by } 2y^3\\\\\\\sqrt[4]{\frac{10xy^3}{16y^4}}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{16y^4}} \ \ \text{ ... break up the fourth root}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{(2y)^4}} \ \ \text{ ... rewrite } 16y^4 \text{ as } (2y)^4\\\\\\\frac{\sqrt[4]{10xy^3}}{2y} \ \ \text{... where y is positive}\\\\\\](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cfrac%7B5x%7D%7B8y%7D%7D%5C%5C%5C%5C%5C%5C%5Csqrt%5B4%5D%7B%5Cfrac%7B5x%2A2y%5E3%7D%7B8y%2A2y%5E3%7D%7D%5C%20%5C%20%5Ctext%7B....%20multiply%20top%20and%20bottom%20by%20%7D%202y%5E3%5C%5C%5C%5C%5C%5C%5Csqrt%5B4%5D%7B%5Cfrac%7B10xy%5E3%7D%7B16y%5E4%7D%7D%5C%5C%5C%5C%5C%5C%5Cfrac%7B%5Csqrt%5B4%5D%7B10xy%5E3%7D%7D%7B%5Csqrt%5B4%5D%7B16y%5E4%7D%7D%20%5C%20%5C%20%5Ctext%7B%20...%20break%20up%20the%20fourth%20root%7D%5C%5C%5C%5C%5C%5C%5Cfrac%7B%5Csqrt%5B4%5D%7B10xy%5E3%7D%7D%7B%5Csqrt%5B4%5D%7B%282y%29%5E4%7D%7D%20%5C%20%5C%20%5Ctext%7B%20...%20rewrite%20%7D%2016y%5E4%20%5Ctext%7B%20as%20%7D%20%282y%29%5E4%5C%5C%5C%5C%5C%5C%5Cfrac%7B%5Csqrt%5B4%5D%7B10xy%5E3%7D%7D%7B2y%7D%20%5C%20%5C%20%5Ctext%7B...%20where%20y%20is%20positive%7D%5C%5C%5C%5C%5C%5C)
The idea is to get something of the form
in the denominator. In this case, 
To be able to reach the
, your teacher gave the hint to multiply top and bottom by
For more examples, search out "rationalizing the denominator".
Keep in mind that
only works if y isn't negative.
If y could be negative, then we'd have to say
. The absolute value bars ensure the result is never negative.
Furthermore, to avoid dividing by zero, we can't have y = 0. So all of this works as long as y > 0.