Answer:
x = -5
Step-by-step explanation:
Angles "50" and "x+55" are alternate interior angles which means that they are the same.
So 50 = x + 55
That leaves you with x = -5
Answer:
![\frac{4x}{2x+y} +\frac{2y}{2x+y}=2](https://tex.z-dn.net/?f=%5Cfrac%7B4x%7D%7B2x%2By%7D%20%2B%5Cfrac%7B2y%7D%7B2x%2By%7D%3D2)
Step-by-step explanation:
Given:
The expression to simplify is given as:
![\frac{4x}{2x+y} +\frac{2y}{2x+y}](https://tex.z-dn.net/?f=%5Cfrac%7B4x%7D%7B2x%2By%7D%20%2B%5Cfrac%7B2y%7D%7B2x%2By%7D)
Since, the denominator is same, we add the numerators and divide it by the same denominator. This gives,
![\frac{4x+2y}{2x+y}](https://tex.z-dn.net/?f=%5Cfrac%7B4x%2B2y%7D%7B2x%2By%7D)
Now, we simplify further by factoring out the common terms from the numerator and denominator if possible.
We observe that, 2 is a common factor to both
. So, we factor out 2 from the numerator. This gives,
![\frac{2(2x+y)}{2x+y}](https://tex.z-dn.net/?f=%5Cfrac%7B2%282x%2By%29%7D%7B2x%2By%7D)
Now, the term
is common in both the numerator and denominator. Hence, ![\frac{2x+y}{2x+y}=1](https://tex.z-dn.net/?f=%5Cfrac%7B2x%2By%7D%7B2x%2By%7D%3D1)
So, the simplified form is:
![=2\times \frac{2x+y}{2x+y}\\\\=2\times 1\\\\=2](https://tex.z-dn.net/?f=%3D2%5Ctimes%20%5Cfrac%7B2x%2By%7D%7B2x%2By%7D%5C%5C%5C%5C%3D2%5Ctimes%201%5C%5C%5C%5C%3D2)
Answer:
![= { \tt{ \frac{2}{5} \div \frac{3}{7} }} \\ \\ = { \tt{ \frac{2}{5} \times \frac{7}{3} }} \\ \\ = { \underline{ \tt{ \: \: \frac{14}{15} \: \: }}}](https://tex.z-dn.net/?f=%20%3D%20%7B%20%5Ctt%7B%20%5Cfrac%7B2%7D%7B5%7D%20%20%5Cdiv%20%20%5Cfrac%7B3%7D%7B7%7D%20%7D%7D%20%5C%5C%20%20%5C%5C%20%20%3D%20%7B%20%5Ctt%7B%20%5Cfrac%7B2%7D%7B5%7D%20%5Ctimes%20%20%5Cfrac%7B7%7D%7B3%7D%20%20%7D%7D%20%5C%5C%20%20%5C%5C%20%20%3D%20%7B%20%5Cunderline%7B%20%5Ctt%7B%20%5C%3A%20%20%5C%3A%20%20%5Cfrac%7B14%7D%7B15%7D%20%20%5C%3A%20%20%5C%3A%20%7D%7D%7D)
<u>Answer</u><u>:</u><u> </u><u>B</u>
Hey, I really want to help you but I am going to need a clearer idea of the problem. You need more signs in there (addition, subtraction, multiplication, and/or division)
9514 1404 393
Answer:
C
Step-by-step explanation:
When you cross multiply by denominator expressions that share a zero, you are effectively making that zero a solution of the resulting equation. That zero cannot be part of the domain of the original equation, and so must be an extraneous solution.
This condition is seen in choice C.
![\dfrac{8}{x^2-9}=\dfrac{5}{2x-6}\qquad\text{original equation. $x\ne3$}\\\\8(2x-6)=5(x^2-9)\qquad\text{result of cross multiplying}\\\\16(x -3)=5(x+3)(x-3)\qquad\text{factored form}](https://tex.z-dn.net/?f=%5Cdfrac%7B8%7D%7Bx%5E2-9%7D%3D%5Cdfrac%7B5%7D%7B2x-6%7D%5Cqquad%5Ctext%7Boriginal%20equation.%20%24x%5Cne3%24%7D%5C%5C%5C%5C8%282x-6%29%3D5%28x%5E2-9%29%5Cqquad%5Ctext%7Bresult%20of%20cross%20multiplying%7D%5C%5C%5C%5C16%28x%20-3%29%3D5%28x%2B3%29%28x-3%29%5Cqquad%5Ctext%7Bfactored%20form%7D)
The solutions to this equation are x=1/5 and x=3. Of these, x=3 is extraneous.
__
A better solution method is multiplying by the least common denominator:
![2(8)=5(x+3)\qquad\text{result of multiplying by $2(x^2-9)$}](https://tex.z-dn.net/?f=2%288%29%3D5%28x%2B3%29%5Cqquad%5Ctext%7Bresult%20of%20multiplying%20by%20%242%28x%5E2-9%29%24%7D)