Answer:
yes
Step-by-step explanation:
multiply 6 for the numerators and denominaters
Solution:
iv) Given equation: 
Since we are given two variables, we need to split the original expression into two quadratic expressions as
Factoring x first:
= 
= 
Factoring y now:
= 
= ![-1[ y(y+1) + 1(y+1)]](https://tex.z-dn.net/?f=-1%5B%20y%28y%2B1%29%20%2B%201%28y%2B1%29%5D)
= ![-1 [(y+1)(y+1)]](https://tex.z-dn.net/?f=-1%20%5B%28y%2B1%29%28y%2B1%29%5D)
= 
Therefore, the original expression becomes
Answer:
are they seperate equations?
Step-by-step explanation:
A variable can be used to represent an unknown amount. Then you write an equation using the variable and based on the information you are given, and you solve the equation to find the unknown amount.
Here is an example.
Mary has money in a bank account. She received $20 for her birthday. She deposited the $20 in her account. After the deposit, she was told the account had a balance of $85. How much money did she have in the account before depositing the $20?
We don't know the amount she had before depositing the $20, so we will call that amount x. x is a letter used as a variable to represent an unknown amount.
She had x dollars in her account before depositing the $20.
After she deposited the $20, she now has x plus 20, or x + 20.
We are told her account now has $85, so x + 20 must equal 85.
That allows us to write an equation which we can then solve to find the amount of money she had in the account before depositing the $20.
x + 20 = 85
Subtract 20 from both sides.
x + 20 - 20 = 85 - 20
x + 0 = 65
x = 65
Since the variable x represents the amount of money in the account before depositing the $20 birthday money, we can now answer the question.
Mary had $65.
