We want to solve the given .
s/3s+2+s+3/2s-4=-2 s/3s^2-4s-4
We will start by the denominators to find the (LCD).
Note that the first denominator is already factored. Let's factor the second one.
We can now rewrite the given expression.
s/3s+2+s+3/2s-4=-2 s/3s^2-4s-4
⇕
s/3s+2+s+3/2 (s-2)=-2 s/(s-2) (3s+2)
We will solve the equation by multiplying each side by the LCD to clear denominators.
s/3s+2+s+3/2(s-2)=-2 s/(s-2)(3s+2)
(s/3s+2+s+3/2(s-2)) * 2 (3s+2) (s-2)=-2 s/(s-2)(3s+2) * 2 (3s+2) (s-2)
s/3s+2 * 2(3s+2)(s-2)+s+3/2(s-2)* 2(3s+2)(s-2)=-2 s/(s-2)(3s+2)* 2(3s+2)(s-2)
s(2)(3s+2)(s-2)/3s+2 +(s+3)(2)(3s+2)(s-2)/2(s-2)=-2 s(2)(3s+2)(s-2)/(s-2)(3s+2)
s(2)(3s+2)(s-2)/3s+2 +(s+3)(2)(3s+2)(s-2)/2(s-2)=-2 s(2)(3s+2)(s-2)/(s-2)(3s+2)
s(2)(s-2)+(s+3)(3s+2)=-2 s(2)
2s(s-2)+(s+3)(3s+2)=-4 s
2s(s-2)+s(3s+2)+3(3s+2)=-4 s
2s^2-4s+s(3s+2)+3(3s+2)=-4 s
2s^2-4s+3s^2+2s+3(3s+2)=-4 s
2s^2-4s+3s^2+2s+9s+6=-4 s
2s^2+3s^2+2s+9s+6=0
5s^2+11s+6=0
We obtained a . Let's identify the values of a, b, and c.
5s^2+ 11s+ 6=0
We see that a = 5, b = 11, and c = 6. Next, we will substitute these values into the .
s=- b±sqrt(b^2-4ac)/2a
s=- 11±sqrt(11^2-4( 5)( 6))/2( 5)
â–¼
Solve using the quadratic formula
s=-11±sqrt(121-4(5)(6))/2(5)
s=-11±sqrt(121-120)/10
s=-11±sqrt(1)/10
s=-11± 1/10
We will find the values for s by using the positive and the negative signs.
| s=-11± 1/10
|
| s=-11+ 1/10
|
s=-11- 1/10
|
| s=-10/10
|
s=-12/10
|
| s=-1
|
s=-6/5
|
We found that s=-1 and s=- 65 are possible solutions. Now we have to check them!
Checking the Solutions
We check our solutions to see whether any of them are extraneous. To do so, we will substitute s=-1 and s=- 65 into the given equation. Let's start with s=-1.
s/3s+2+s+3/2s-4=-2 s/3s^2-4s-4
-1/3( -1)+2+-1+3/2( -1)-4? =-2 ( -1)/3( -1)^2-4( -1)-4
-1/3(-1)+2+-1+3/2(-1)-4? =-2 (-1)/3(1)-4(-1)-4
-1/3(-1)+2+-1+3/2(-1)-4? =-2 (-1)/3-4(-1)-4
-1/-3+2+-1+3/-2-4? =-2 (-1)/3-4(-1)-4
-1/-3+2+-1+3/-2-4? =2/3+4-4
-1/-1+2/-6? =2/3
1/1+2/-6? =2/3
1/1+(-2/6)? =2/3
1/1-2/6? =2/3
1/1-1/3? =2/3
3/3-1/3? =2/3
2/3=2/3 ✓
Since we obtained a true statement, s=-1 is a solution of the equation. Let's now substitute s=- 65 into the equation.
s/3s+2+s+3/2s-4=-2 s/3s^2-4s-4
- 65/3( - 65)+2+- 65+3/2( - 65)-4? =-2 ( - 65)/3( - 65)^2-4( - 65)-4
- 65/3(- 65)+2+- 65+3/2(- 65)-4? =-2 (- 65)/3( 65)^2-4(- 65)-4
- 65/3(- 65)+2+- 65+3/2(- 65)-4? =-2 (- 65)/3( 3625)-4(- 65)-4
- 65/-3( 65)+2+- 65+3/-2( 65)-4? =-2(- 65)/3( 3625)-4(- 65)-4
- 65/-3( 65)+2+- 65+3/-2( 65)-4? =2 ( 65)/3( 3625)+4( 65)-4
- 65/- 185+2+- 65+3/- 125-4? =125/10825+ 245-4
- 65/- 185+ 105+- 65+ 155/- 125- 205? =125/10825+ 245- 205
- 65/- 185+ 105+- 65+ 155/- 125- 205? =125/10825+ 12025- 10025
-65/-185+ 105+-65+ 155/-125- 205? =125/10825+ 12025- 10025
-65/-85+95/-325? =125/12825
- 65/- 85+95/- 325? =125/12825
65/85+95/- 325? =125/12825
Recall that dividing by a fraction is the same as multiplying by its reciprocal.
65/85+95/- 325? =125/12825
⇕
6/5(5/8)+9/5(-5/32)? = 12/5(25/128)
Let's simplify to finally see whether the equation is satisfied when s=- 65.
6/5(5/8)+9/5(-5/32)? = 12/5(25/128)
6/5(5/8)-9/5(5/32)? = 12/5(25/128)
30/40-45/160? = 300/640
3/4-45/160? = 300/640
3/4-9/32? = 300/640
3/4-9/32? = 15/32
24/32-9/32? = 15/32
15/32=15/32 ✓
Neither of our solutions is an extraneous solution. Therefore, the solutions to the given equation are s=-1 and s=- 65.
