x=3β2d,5,β2(1+d),5,β27β2d,5,β2(2+d),5,2(yβd),5
Step-by-step explanation:Solving for x. Want to solve for y or solve for d instead?
1 Simplify Β 0-20β2 Β to Β -2β2.
3,-2,-27,-2-2,02y=1,5x+2d3,β2,β27,β2β2,02y=1,5x+2d
2 Simplify Β -2-2β2β2 Β to Β -4β4.
3,-2,-27,-4,02y=1,5x+2d3,β2,β27,β4,02y=1,5x+2d
3 Subtract 2d2d from both sides.
3-2d,-2-2d,-27-2d,-4-2d,02y-2d=1,5x3β2d,β2β2d,β27β2d,β4β2d,02yβ2d=1,5x
4 Divide both sides by 1,51,5.
\frac{3-2d}{1},5,\frac{-2-2d}{1},5,\frac{-27-2d}{1},5,\frac{-4-2d}{1},5,\frac{02y-2d}{1},5=x
1
3β2d
,5,
1
β2β2d
,5,
1
β27β2d
,5,
1
β4β2d
,5,
1
02yβ2d
,5=x
5 Factor out the common term 22.
\frac{3-2d}{1},5,\frac{-2(1+d)}{1},5,\frac{-27-2d}{1},5,\frac{-4-2d}{1},5,\frac{02y-2d}{1},5=x
1
3β2d
,5,
1
β2(1+d)
,5,
1
β27β2d
,5,
1
β4β2d
,5,
1
02yβ2d
,5=x
6 Factor out the common term 22.
\frac{3-2d}{1},5,\frac{-2(1+d)}{1},5,\frac{-27-2d}{1},5,\frac{-2(2+d)}{1},5,\frac{02y-2d}{1},5=x
1
3β2d
,5,
1
β2(1+d)
,5,
1
β27β2d
,5,
1
β2(2+d)
,5,
1
02yβ2d
,5=x
7 Factor out the common term 22.
\frac{3-2d}{1},5,\frac{-2(1+d)}{1},5,\frac{-27-2d}{1},5,\frac{-2(2+d)}{1},5,\frac{2(y-d)}{1},5=x
1
3β2d
,5,
1
β2(1+d)
,5,
1
β27β2d
,5,
1
β2(2+d)
,5,
1
2(yβd)
,5=x
8 Simplify Β \frac{3-2d}{1}
1
3β2d
Β to Β (3-2d)(3β2d).
3-2d,5,\frac{-2(1+d)}{1},5,\frac{-27-2d}{1},5,\frac{-2(2+d)}{1},5,\frac{2(y-d)}{1},5=x3β2d,5,
1
β2(1+d)
,5,
1
β27β2d
,5,
1
β2(2+d)
,5,
1
2(yβd)
,5=x
9 Simplify Β \frac{-2(1+d)}{1}
1
β2(1+d)
Β to Β (-2(1+d))(β2(1+d)).
3-2d,5,-2(1+d),5,\frac{-27-2d}{1},5,\frac{-2(2+d)}{1},5,\frac{2(y-d)}{1},5=x3β2d,5,β2(1+d),5,
1
β27β2d
,5,
1
β2(2+d)
,5,
1
2(yβd)
,5=x
10 Simplify Β \frac{-27-2d}{1}
1
β27β2d
Β to Β (-27-2d)(β27β2d).
3-2d,5,-2(1+d),5,-27-2d,5,\frac{-2(2+d)}{1},5,\frac{2(y-d)}{1},5=x3β2d,5,β2(1+d),5,β27β2d,5,
1
β2(2+d)
,5,
1
2(yβd)
,5=x
11 Simplify Β \frac{-2(2+d)}{1}
1
β2(2+d)
Β to Β (-2(2+d))(β2(2+d)).
3-2d,5,-2(1+d),5,-27-2d,5,-2(2+d),5,\frac{2(y-d)}{1},5=x3β2d,5,β2(1+d),5,β27β2d,5,β2(2+d),5,
1
2(yβd)
,5=x
12 Simplify Β \frac{2(y-d)}{1}
1
2(yβd)
Β to Β (2(y-d))(2(yβd)).
3-2d,5,-2(1+d),5,-27-2d,5,-2(2+d),5,2(y-d),5=x3β2d,5,β2(1+d),5,β27β2d,5,β2(2+d),5,2(yβd),5=x
13 Switch sides.
x=3-2d,5,-2(1+d),5,-27-2d,5,-2(2+d),5,2(y-d),5x=3β2d,5,β2(1+d),5,β27β2d,5,β2(2+d),5,2(yβd),5
Done