SOLUTION: I have two problems that I just go blank on when I see these types of equations. I dont know were to start or how to do them. In my textbook there isnt anything like them other tha

Question 154668: I have two problems that I just go blank on when I see these types of equations. I dont know were to start or how to do them. In my textbook there isnt anything like them other than when y1=y2. There are no examples of the given conditions. The answers from the book are #1 is 8 and #2 is 5. I tried everything to come up with those answers, but it is eluding me on how to do the following rational equations. Please help. Thanks.
Find all values of x satisfying the given conditions.
y1= 10(2x-1),y2= 2x+1, and y1 is 14 more than 8 times y2.
y1= 1/x, y2= 1/2x, y3= 1/x-1, and the sum of 3 times y1 and 4 times y2 is the product of 4 and y2.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Find all values of x satisfying the given conditions.
:
y1 = 10(2x-1), y2 = 2x+1, and y1 is 14 more than 8 times y2.
:
Use some logic here, it says:
y1 = 8(y2) + 14
:
From the 1st two given equations:
we can substitute 10(2x-1) for y1 and 2x+1 for y2, in the above; therefore:
:
10(2x-1) = 8(2x+1) + 14
Multiply the terms inside the brackets, and solve for x
20x - 10 = 16x + 8 + 14
20x - 10 = 16x + 22
Some basic algebra:
20x - 16x = 22 + 10
4x = 32
x =
x = 8
:
:
y1= 1/x, y2= 1/2x, y3= 1/x-1, and the sum of 3 times y1 and 4 times y2 is the product of 4 and y3(corrected this typo).
:
Write an equation for the statement:
"the sum of 3 times y1 and 4 times y2 is the product of 4 and y3"
3(y1) + 4(y2) = 4(y3)
:
Substitute: (1/x) for y1; (1/2x) for y2; and (1/(x-1)) for y3

which is:

2 will cancel into 4 so we have:

we can add like fractions

Cross multiply and find x:
4x = 5(x-1)
4x = 5x - 5
4x -5x = -5
-x = -5
x = 5
:
:
did this make sense to you now?
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