1/2(2y+8)-6=-1/5(10y-35)

Solution for 1/2(2y+8)-6=-1/5(10y-35) equation:

1/2(2y+8)-6=-1/5(10y-35)

We move all terms to the left:

1/2(2y+8)-6-(-1/5(10y-35))=0


Domain of the equation: 2(2y+8)!=0

y∈R



Domain of the equation: 5(10y-35))!=0

y∈R


We calculate fractions


(5y1/(2(2y+8)*5(10y-35)))+(-(-2y2)/(2(2y+8)*5(10y-35)))-6=0


We calculate terms in parentheses: +(5y1/(2(2y+8)*5(10y-35))), so:

5y1/(2(2y+8)*5(10y-35))

We multiply all the terms by the denominator


5y1

We add all the numbers together, and all the variables

5y

Back to the equation:

+(5y)



We calculate terms in parentheses: +(-(-2y2)/(2(2y+8)*5(10y-35))), so:

-(-2y2)/(2(2y+8)*5(10y-35))

We add all the numbers together, and all the variables

-(-2y^2)/(2(2y+8)*5(10y-35))

We multiply all the terms by the denominator


-(-2y^2)

We get rid of parentheses

2y^2

Back to the equation:

+(2y^2)


determiningTheFunctionDomain
2y^2+5y-6=0

a = 2; b = 5; c = -6;
Δ = b2-4ac
Δ = 52-4·2·(-6)
Δ = 73
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{73}}{2*2}=\frac{-5-\sqrt{73}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{73}}{2*2}=\frac{-5+\sqrt{73}}{4}
$