(7/4x)+(3/8x)=48/12

Solution for (7/4x)+(3/8x)=48/12 equation:

(7/4x)+(3/8x)=48/12

We move all terms to the left:

(7/4x)+(3/8x)-(48/12)=0


Domain of the equation: 4x)!=0

x!=0/1

x!=0

x∈R



Domain of the equation: 8x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+7/4x)+(+3/8x)-4=0

We get rid of parentheses

7/4x+3/8x-4=0

We calculate fractions


56x/32x^2+12x/32x^2-4=0

We multiply all the terms by the denominator


56x+12x-4*32x^2=0

We add all the numbers together, and all the variables

68x-4*32x^2=0

Wy multiply elements

-128x^2+68x=0

a = -128; b = 68; c = 0;
Δ = b2-4ac
Δ = 682-4·(-128)·0
Δ = 4624
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4624}=68$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(68)-68}{2*-128}=\frac{-136}{-256}
=17/32
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(68)+68}{2*-128}=\frac{0}{-256}
=0
$