1/7y+3/4y=50/49

Solution for 1/7y+3/4y=50/49 equation:

1/7y+3/4y=50/49

We move all terms to the left:

1/7y+3/4y-(50/49)=0


Domain of the equation: 7y!=0

y!=0/7

y!=0

y∈R



Domain of the equation: 4y!=0

y!=0/4

y!=0

y∈R


We add all the numbers together, and all the variables

1/7y+3/4y-(+50/49)=0

We get rid of parentheses

1/7y+3/4y-50/49=0

We calculate fractions


(-5600y^2)/5488y^2+784y/5488y^2+4116y/5488y^2=0

We multiply all the terms by the denominator


(-5600y^2)+784y+4116y=0

We add all the numbers together, and all the variables

(-5600y^2)+4900y=0

We get rid of parentheses

-5600y^2+4900y=0

a = -5600; b = 4900; c = 0;
Δ = b2-4ac
Δ = 49002-4·(-5600)·0
Δ = 24010000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{24010000}=4900$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4900)-4900}{2*-5600}=\frac{-9800}{-11200}
=7/8
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4900)+4900}{2*-5600}=\frac{0}{-11200}
=0
$