5/4x+7=7/8x+19

Solution for 5/4x+7=7/8x+19 equation:

5/4x+7=7/8x+19

We move all terms to the left:

5/4x+7-(7/8x+19)=0


Domain of the equation: 4x!=0

x!=0/4

x!=0

x∈R



Domain of the equation: 8x+19)!=0

x∈R


We get rid of parentheses

5/4x-7/8x-19+7=0

We calculate fractions


40x/32x^2+(-28x)/32x^2-19+7=0

We add all the numbers together, and all the variables

40x/32x^2+(-28x)/32x^2-12=0

We multiply all the terms by the denominator


40x+(-28x)-12*32x^2=0

Wy multiply elements

-384x^2+40x+(-28x)=0

We get rid of parentheses

-384x^2+40x-28x=0

We add all the numbers together, and all the variables

-384x^2+12x=0

a = -384; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·(-384)·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12}{2*-384}=\frac{-24}{-768}
=1/32
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*-384}=\frac{0}{-768}
=0
$