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x/3=4/x
Let's begin by highlighting the all of the different factors in the denominators. This will help us find the least common denominator (LCD).
LHS * (3) (x)=RHS* (3) (x)
Multiply
Cancel out common factors
Simplify quotient
Finally, we will check if either x= sqrt(12) or x= - sqrt(12) is an extraneous solution. To do this, we need to substitute sqrt(12) and - sqrt(12) for x in the original equation. Let's start with x= sqrt(12).
x= sqrt(12)
a/b=a * sqrt(12)/b * sqrt(12)
sqrt(a^2)=a
a/b=.a /4./.b /4.
Now, we will check our second solution, x= - sqrt(12).
x= - sqrt(12)
a/b=a * - sqrt(12)/b * - sqrt(12)
(- a)^2=a^2
sqrt(a^2)=a
a/b=.a /4./.b /4.
Neither of our solutions is an extraneous solution. Therefore, the solutions to the given equation are x=sqrt(12) and x=- sqrt(12).
x/x-1=4/x
Let's begin by highlighting the all of the different factors in the denominators. This will help us find the LCD.
x/x-1=4/x
LHS * (x-1) (x)=RHS* (x-1) (x)
Multiply
Cancel out common factors
Simplify quotient
Note that we have a quadratic equation now. Let's identify the values of a, b, and c. x^2-4x+4=0 ⇔ 1x^2 +(-4)x+4=0 We have that a= 1, b=-4, and c=4. Let's substitute these values into the Quadratic Formula and solve for x.
Substitute values
- (- a)=a
Calculate power and product
Subtract term
Calculate root
Identity Property of Addition
Calculate quotient
Finally, we will check if x=2 is an extraneous solution. To do this, we need to substitute 2 for x in the original equation.
x=2 is not an extraneous solution, so it solves our equation.
1/x+1/3x=6
Let's begin by highlighting the all of the different factors in the denominators. This will help us find the LCD.
LHS * ( 3x)=RHS* ( 3x)
Multiply
Cancel out common factors
Simplify quotient
Add terms
.LHS /18.=.RHS /18.
a/b=.a /2./.b /2.
Rearrange equation
Finally, we will check if x= 29 is an extraneous solution. To do this, we need to substitute 29 for x in the original equation.
x= 2/9
a*b/c= a* b/c
a/b/c= a * c/b
a/b=a * 3/b * 3
Add fractions
Calculate quotient
x = 29 is not an extraneous solution, so it solves our equation.
1/x+1/x+1=3
Let's begin by highlighting the all of the different factors in the denominators. This will help us find the LCD.
1/x+1/x+1=3
LHS * (x) (x+1)=RHS* (x) (x+1)
Multiply
Cancel out common factors
Simplify quotient
Note that we have a quadratic equation now. Let's identify the values of a, b, and c. -3x^2-x+1=0 ⇔ - 3x^2 +(-1)x+1=0 We have that a= - 3, b=-1, and c=1. Let's substitute these values into the Quadratic Formula and solve for x.
Substitute values
- (- a)=a
Calculate power and product
Add terms
Let's calculate both solutions by using the positive and negative signs.
| x=1 ± sqrt(13)/- 6 | |
|---|---|
| x=1 + sqrt(13)/- 6 | x=1 - sqrt(13)/- 6 |
| x=-(1 + sqrt(13))/6 | x=-(1 - sqrt(13))/6 |
| x=-1 - sqrt(13)/6 | x=-1 + sqrt(13)/6 |
Finally, we will check if either x= -1 - sqrt(13)6 or x= -1 + sqrt(13)6 is an extraneous solution. To do this, we need to substitute -1 - sqrt(13)6 and -1 + sqrt(13)6 for x in the original equation. Let's start with x= -1 - sqrt(13)6.
x= -1 - sqrt(13)6
a/b=a * 6/b * 6
Add fractions
Add terms
a/b/c= a * c/b
a/b=a * (- sqrt(13)+5)/b * (- sqrt(13)+5)
a/b=a * (-1 - sqrt(13))/b * (-1 - sqrt(13))
Add fractions
Distribute 6
Add and subtract terms
Distribute (-1 - sqrt(13))
Distribute - sqrt(13)
Distribute 5
Add and subtract terms
Factor out 3
a/b=.a /(8-4sqrt(13))./.b /(8-4sqrt(13)).
Calculate quotient
Now, we will check our second solution, x= -1 + sqrt(13)6.
x= -1 + sqrt(13)6
a/b=a * 6/b * 6
Add fractions
Add terms
a/b/c= a * c/b
a/b=a * ( sqrt(13)+5)/b * ( sqrt(13)+5)
a/b=a * (-1 + sqrt(13))/b * (-1 + sqrt(13))
Add fractions
Distribute 6
Add and subtract terms
Distribute (-1 - sqrt(13))
Distribute sqrt(13)
Distribute 5
Add and subtract terms
Factor out 3
a/b=.a /(8+4sqrt(13))./.b /(8+4sqrt(13)).
Calculate quotient
Neither of our solutions is an extraneous solution. Therefore, the solutions to the given equation are x= -1 - sqrt(13)6 and x= -1 + sqrt(13)6.