If it's not what You are looking for type in the equation solver your own equation and let us solve it.
5/8x+3/4=4+3/4x
We move all terms to the left:
5/8x+3/4-(4+3/4x)=0
Domain of the equation: 8x!=0
x!=0/8
x!=0
x∈R
Domain of the equation: 4x)!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
5/8x-(3/4x+4)+3/4=0
We get rid of parentheses
5/8x-3/4x-4+3/4=0
We calculate fractions
320x/512x^2+(-24x)/512x^2+24x/512x^2-4=0
We multiply all the terms by the denominator
320x+(-24x)+24x-4*512x^2=0
We add all the numbers together, and all the variables
344x+(-24x)-4*512x^2=0
Wy multiply elements
-2048x^2+344x+(-24x)=0
We get rid of parentheses
-2048x^2+344x-24x=0
We add all the numbers together, and all the variables
-2048x^2+320x=0
a = -2048; b = 320; c = 0;
Δ = b2-4ac
Δ = 3202-4·(-2048)·0
Δ = 102400
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{102400}=320$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(320)-320}{2*-2048}=\frac{-640}{-4096} =5/32 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(320)+320}{2*-2048}=\frac{0}{-4096} =0 $
| 27=3+x2 | | 6z-z=-3+3z-2-1 | | x=3+x2=27 | | m=4(2)^3 | | -4(x+1)+3(3x-3)=4(x-5)+12 | | 4^(x^2-27)=16^(3x) | | 4(n-5)-3(3-2n)=5n | | 4y+8-y=10+2y+8= | | 32=-16t^2+60t+2 | | -10.2q+1.2=-76.2-1.6q | | 16/10x=12 | | (x^2+5)=3 | | 6+(x^2-x+10)=10 | | 3. 3(2x−2)=2(3x+9) | | (2x-4)=(3x-2) | | (3x-7)=(2x-10) | | 5x+4x+5=19x-10 | | -5.2a=1.3 | | |2x|+9=19 | | x=20+(1/2x) | | -4=3(u+7)-8u | | 2x-97=90 | | 220=120+x-100 | | -32=4(x-5)-7 | | d=|110-30| | | 24+14x=17+15x | | 2/3/4x=6/9 | | 3/7=1/5/x/2 | | 500+102x+1.5x^2=350x+2.5x^2 | | (12x5)/4-8+12=60 | | t=21/1/3 | | -2u+6(u+8)=20 |