# Solve for v 2/5-7/(v+6)=9/(5(v-6)) 25-7v+6=95(v-6)
Subtract 25 from both sides of the equation.
-7v+6=95(v-6)-25
Find the LCD of the terms in the equation.
Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.
v+6,5(v-6),5
The LCM is the smallest positive number that all of the numbers divide into evenly.
1. List the prime factors of each number.
2. Multiply each factor the greatest number of times it occurs in either number.
The number 1 is not a prime number because it only has one positive factor, which is itself.
Not prime
Since 5 has no factors besides 1 and 5.
5 is a prime number
The LCM of 1,5,5 is the result of multiplying all prime factors the greatest number of times they occur in either number.
5
The factor for v+6 is v+6 itself.
(v+6)=v+6
(v+6) occurs 1 time.
The factor for v-6 is v-6 itself.
(v-6)=v-6
(v-6) occurs 1 time.
The LCM of v+6,v-6 is the result of multiplying all factors the greatest number of times they occur in either term.
(v+6)(v-6)
The Least Common Multiple LCM of some numbers is the smallest number that the numbers are factors of.
5(v+6)(v-6)
5(v+6)(v-6)
Multiply each term by 5(v+6)(v-6) and simplify.
Multiply each term in -7v+6=95(v-6)-25 by 5(v+6)(v-6) in order to remove all the denominators from the equation.
-7v+6⋅(5(v+6)(v-6))=95(v-6)⋅(5(v+6)(v-6))-25⋅(5(v+6)(v-6))
Simplify -7v+6⋅(5(v+6)(v-6)).
Cancel the common factor of v+6.
Move the leading negative in -7v+6 into the numerator.
-7v+6⋅(5(v+6)(v-6))=95(v-6)⋅(5(v+6)(v-6))-25⋅(5(v+6)(v-6))
Factor v+6 out of 5(v+6)(v-6).
-7v+6⋅((v+6)(5(v-6)))=95(v-6)⋅(5(v+6)(v-6))-25⋅(5(v+6)(v-6))
Cancel the common factor.
-7v+6⋅((v+6)(5(v-6)))=95(v-6)⋅(5(v+6)(v-6))-25⋅(5(v+6)(v-6))
Rewrite the expression.
-7⋅(5(v-6))=95(v-6)⋅(5(v+6)(v-6))-25⋅(5(v+6)(v-6))
-7⋅(5(v-6))=95(v-6)⋅(5(v+6)(v-6))-25⋅(5(v+6)(v-6))
Multiply 5 by -7.
-35⋅(v-6)=95(v-6)⋅(5(v+6)(v-6))-25⋅(5(v+6)(v-6))
Apply the distributive property.
-35v-35⋅-6=95(v-6)⋅(5(v+6)(v-6))-25⋅(5(v+6)(v-6))
Multiply -35 by -6.
-35v+210=95(v-6)⋅(5(v+6)(v-6))-25⋅(5(v+6)(v-6))
-35v+210=95(v-6)⋅(5(v+6)(v-6))-25⋅(5(v+6)(v-6))
Simplify 95(v-6)⋅(5(v+6)(v-6))-25⋅(5(v+6)(v-6)).
Simplify each term.
Rewrite using the commutative property of multiplication.
-35v+210=595(v-6)((v+6)(v-6))-25⋅(5(v+6)(v-6))
Cancel the common factor of 5.
Cancel the common factor.
-35v+210=595(v-6)((v+6)(v-6))-25⋅(5(v+6)(v-6))
Rewrite the expression.
-35v+210=9v-6((v+6)(v-6))-25⋅(5(v+6)(v-6))
-35v+210=9v-6((v+6)(v-6))-25⋅(5(v+6)(v-6))
Cancel the common factor of v-6.
Factor v-6 out of (v+6)(v-6).
-35v+210=9v-6((v-6)(v+6))-25⋅(5(v+6)(v-6))
Cancel the common factor.
-35v+210=9v-6((v-6)(v+6))-25⋅(5(v+6)(v-6))
Rewrite the expression.
-35v+210=9(v+6)-25⋅(5(v+6)(v-6))
-35v+210=9(v+6)-25⋅(5(v+6)(v-6))
Apply the distributive property.
-35v+210=9v+9⋅6-25⋅(5(v+6)(v-6))
Multiply 9 by 6.
-35v+210=9v+54-25⋅(5(v+6)(v-6))
Cancel the common factor of 5.
Move the leading negative in -25 into the numerator.
-35v+210=9v+54+-25⋅(5(v+6)(v-6))
Factor 5 out of 5(v+6)(v-6).
-35v+210=9v+54+-25⋅(5((v+6)(v-6)))
Cancel the common factor.
-35v+210=9v+54+-25⋅(5((v+6)(v-6)))
Rewrite the expression.
-35v+210=9v+54-2⋅((v+6)(v-6))
-35v+210=9v+54-2⋅((v+6)(v-6))
Expand (v+6)(v-6) using the FOIL Method.
Apply the distributive property.
-35v+210=9v+54-2⋅(v(v-6)+6(v-6))
Apply the distributive property.
-35v+210=9v+54-2⋅(v⋅v+v⋅-6+6(v-6))
Apply the distributive property.
-35v+210=9v+54-2⋅(v⋅v+v⋅-6+6v+6⋅-6)
-35v+210=9v+54-2⋅(v⋅v+v⋅-6+6v+6⋅-6)
Combine the opposite terms in v⋅v+v⋅-6+6v+6⋅-6.
Reorder the factors in the terms v⋅-6 and 6v.
-35v+210=9v+54-2⋅(v⋅v-6v+6v+6⋅-6)
-35v+210=9v+54-2⋅(v⋅v+0+6⋅-6)
-35v+210=9v+54-2⋅(v⋅v+6⋅-6)
-35v+210=9v+54-2⋅(v⋅v+6⋅-6)
Simplify each term.
Multiply v by v.
-35v+210=9v+54-2⋅(v2+6⋅-6)
Multiply 6 by -6.
-35v+210=9v+54-2⋅(v2-36)
-35v+210=9v+54-2⋅(v2-36)
Apply the distributive property.
-35v+210=9v+54-2v2-2⋅-36
Multiply -2 by -36.
-35v+210=9v+54-2v2+72
-35v+210=9v+54-2v2+72
-35v+210=9v-2v2+126
-35v+210=9v-2v2+126
-35v+210=9v-2v2+126
Solve the equation.
Since v is on the right side of the equation, switch the sides so it is on the left side of the equation.
9v-2v2+126=-35v+210
Set the equation equal to zero.
Move all the expressions to the left side of the equation.
Move 35v to the left side of the equation by adding it to both sides.
9v-2v2+126+35v=210
Move 210 to the left side of the equation by subtracting it from both sides.
9v-2v2+126+35v-210=0
9v-2v2+126+35v-210=0
Simplify 9v-2v2+126+35v-210.
44v-2v2+126-210=0
Subtract 210 from 126.
44v-2v2-84=0
44v-2v2-84=0
44v-2v2-84=0
Factor the left side of the equation.
Factor 2 out of 44v-2v2-84.
Factor 2 out of 44v.
2(22v)-2v2-84
Factor 2 out of -2v2.
2(22v)+2(-v2)-84
Factor 2 out of -84.
2(22v)+2(-v2)+2(-42)
Factor 2 out of 2(22v)+2(-v2).
2(22v-v2)+2(-42)
Factor 2 out of 2(22v-v2)+2(-42).
2(22v-v2-42)
2(22v-v2-42)
Reorder terms.
2(-v2+22v-42)
Replace the left side with the factored expression.
2(-v2+22v-42)=0
2(-v2+22v-42)=0
Divide each term by 2 and simplify.
Divide each term in 2(-v2+22v-42)=0 by 2.
2(-v2+22v-42)2=02
Cancel the common factor of 2.
Cancel the common factor.
2(-v2+22v-42)2=02
Divide -v2+22v-42 by 1.
-v2+22v-42=02
-v2+22v-42=02
Divide 0 by 2.
-v2+22v-42=0
-v2+22v-42=0
Factor -1 out of -v2+22v-42.
Factor -1 out of -v2.
-(v2)+22v-42=0
Factor -1 out of 22v.
-(v2)-(-22v)-42=0
Rewrite -42 as -1(42).
-(v2)-(-22v)-1⋅42=0
Factor -1 out of -(v2)-(-22v).
-(v2-22v)-1⋅42=0
Factor -1 out of -(v2-22v)-1(42).
-(v2-22v+42)=0
-(v2-22v+42)=0
Multiply each term in -(v2-22v+42)=0 by -1
Multiply each term in -(v2-22v+42)=0 by -1.
-(v2-22v+42)⋅-1=0⋅-1
Simplify -(v2-22v+42)⋅-1.
Apply the distributive property.
(-v2-(-22v)-1⋅42)⋅-1=0⋅-1
Simplify.
Multiply -22 by -1.
(-v2+22v-1⋅42)⋅-1=0⋅-1
Multiply -1 by 42.
(-v2+22v-42)⋅-1=0⋅-1
(-v2+22v-42)⋅-1=0⋅-1
Apply the distributive property.
-v2⋅-1+22v⋅-1-42⋅-1=0⋅-1
Simplify.
Multiply -v2⋅-1.
Multiply -1 by -1.
1v2+22v⋅-1-42⋅-1=0⋅-1
Multiply v2 by 1.
v2+22v⋅-1-42⋅-1=0⋅-1
v2+22v⋅-1-42⋅-1=0⋅-1
Multiply -1 by 22.
v2-22v-42⋅-1=0⋅-1
Multiply -42 by -1.
v2-22v+42=0⋅-1
v2-22v+42=0⋅-1
v2-22v+42=0⋅-1
Multiply 0 by -1.
v2-22v+42=0
v2-22v+42=0
Use the quadratic formula to find the solutions.
-b±b2-4(ac)2a
Substitute the values a=1, b=-22, and c=42 into the quadratic formula and solve for v.
22±(-22)2-4⋅(1⋅42)2⋅1
Simplify.
Simplify the numerator.
Raise -22 to the power of 2.
v=22±484-4⋅(1⋅42)2⋅1
Multiply 42 by 1.
v=22±484-4⋅422⋅1
Multiply -4 by 42.
v=22±484-1682⋅1
Subtract 168 from 484.
v=22±3162⋅1
Rewrite 316 as 22⋅79.
Factor 4 out of 316.
v=22±4(79)2⋅1
Rewrite 4 as 22.
v=22±22⋅792⋅1
v=22±22⋅792⋅1
Pull terms out from under the radical.
v=22±2792⋅1
v=22±2792⋅1
Multiply 2 by 1.
v=22±2792
Simplify 22±2792.
v=11±79
v=11±79
The final answer is the combination of both solutions.
v=11+79,11-79
v=11+79,11-79
The result can be shown in multiple forms.
Exact Form:
v=11+79,11-79
Decimal Form:
v=19.88819441…,2.11180558…
Solve for v 2/5-7/(v+6)=9/(5(v-6))

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