1/2(2y+4)-7=-1/8(16y-32)

Solution for 1/2(2y+4)-7=-1/8(16y-32) equation:

1/2(2y+4)-7=-1/8(16y-32)

We move all terms to the left:

1/2(2y+4)-7-(-1/8(16y-32))=0


Domain of the equation: 2(2y+4)!=0

y∈R



Domain of the equation: 8(16y-32))!=0

y∈R


We calculate fractions


(8y1/(2(2y+4)*8(16y-32)))+(-(-2y2)/(2(2y+4)*8(16y-32)))-7=0


We calculate terms in parentheses: +(8y1/(2(2y+4)*8(16y-32))), so:

8y1/(2(2y+4)*8(16y-32))

We multiply all the terms by the denominator


8y1

We add all the numbers together, and all the variables

8y

Back to the equation:

+(8y)



We calculate terms in parentheses: +(-(-2y2)/(2(2y+4)*8(16y-32))), so:

-(-2y2)/(2(2y+4)*8(16y-32))

We add all the numbers together, and all the variables

-(-2y^2)/(2(2y+4)*8(16y-32))

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+8y-7=0

a = 2; b = 8; c = -7;
Δ = b2-4ac
Δ = 82-4·2·(-7)
Δ = 120
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{30}}{2*2}=\frac{-8-2\sqrt{30}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{30}}{2*2}=\frac{-8+2\sqrt{30}}{4}
$