3/5(20y-5)=12-2/7(14-42y)

Solution for 3/5(20y-5)=12-2/7(14-42y) equation:

3/5(20y-5)=12-2/7(14-42y)

We move all terms to the left:

3/5(20y-5)-(12-2/7(14-42y))=0


Domain of the equation: 5(20y-5)!=0

y∈R



Domain of the equation: 7(14-42y))!=0

y∈R


We add all the numbers together, and all the variables

3/5(20y-5)-(12-2/7(-42y+14))=0

We calculate fractions


(21y(-)/(5(20y-5)*7(-42y+14)))+(-(-10y2)/(5(20y-5)*7(-42y+14)))=0


We calculate terms in parentheses: +(21y(-)/(5(20y-5)*7(-42y+14))), so:

21y(-)/(5(20y-5)*7(-42y+14))

We add all the numbers together, and all the variables

21y0/(5(20y-5)*7(-42y+14))

We multiply all the terms by the denominator


21y0

We add all the numbers together, and all the variables

21y

Back to the equation:

+(21y)



We calculate terms in parentheses: +(-(-10y2)/(5(20y-5)*7(-42y+14))), so:

-(-10y2)/(5(20y-5)*7(-42y+14))

We add all the numbers together, and all the variables

-(-10y^2)/(5(20y-5)*7(-42y+14))

We multiply all the terms by the denominator


-(-10y^2)

We get rid of parentheses

10y^2

Back to the equation:

+(10y^2)


determiningTheFunctionDomain
10y^2+21y=0

a = 10; b = 21; c = 0;
Δ = b2-4ac
Δ = 212-4·10·0
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-21}{2*10}=\frac{-42}{20}
=-2+1/10
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+21}{2*10}=\frac{0}{20}
=0
$