(2/3)x+1=16-x

Solution for (2/3)x+1=16-x equation:

(2/3)x+1=16-x

We move all terms to the left:

(2/3)x+1-(16-x)=0


Domain of the equation: 3)x!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+2/3)x-(-1x+16)+1=0

We multiply parentheses

2x^2-(-1x+16)+1=0

We get rid of parentheses

2x^2+1x-16+1=0

We add all the numbers together, and all the variables

2x^2+x-15=0

a = 2; b = 1; c = -15;
Δ = b2-4ac
Δ = 12-4·2·(-15)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-11}{2*2}=\frac{-12}{4}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+11}{2*2}=\frac{10}{4}
=2+1/2
$