Solve 2y/y-4+2y-5/y-3=25/3
Answers
Answer:Δ−−√=2296−−−−√=4∗574−−−−−−√=4√∗574−−−√=2574−−−√
y1=−b−Δ√2a=−(−44)−2574√2∗6=44−2574√12
y2=−b+Δ√2a=−(−44)+2574√2∗6=44+2574√12
Step-by-step explanation:
We move all terms to the left:
2y/y-4+2y-5/y-3-(25/3)=0
Domain of the equation: y!=0
y∈R
We add all the numbers together, and all the variables
2y/y+2y-5/y-4-3-(+25/3)=0
We add all the numbers together, and all the variables
2y+2y/y-5/y-7-(+25/3)=0
We get rid of parentheses
2y+2y/y-5/y-7-25/3=0
We calculate fractions
2y+(2y-15)/3y+(-25y)/3y-7=0
We multiply all the terms by the denominator
2y*3y+(2y-15)+(-25y)-7*3y=0
Wy multiply elements
6y^2+(2y-15)+(-25y)-21y=0
We get rid of parentheses
6y^2+2y-25y-21y-15=0
We add all the numbers together, and all the variables
6y^2-44y-15=0
a = 6; b = -44; c = -15;
Δ = b2-4ac
Δ = -442-4·6·(-15)
Δ = 2296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
y1=−b−Δ√2ay2=−b+Δ√2a