solve for b
c=a+2b/ab-1
I know this is easy but I'm having a mind block right now studying for the final. I know the answer is
b=a+c/ac-2 but how?
c=a+2b/ab-1
I know this is easy but I'm having a mind block right now studying for the final. I know the answer is
b=a+c/ac-2 but how?
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http://www.flickr.com/photos/dwread/6976…
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Since b is on the right-hand side of the equation switch the sides so it is on the left-hand side of the equation.
a+2b/ab-1=c
Cancel the common factor of b in 2b/ab.
a+2/a-1=c
Find the LCD (least common denominator) of a+2/a-1=c.
LCD:a
Multiply each term in the equation by a in order to remove all the denominators from the equation.
a*a+2*a-1=c*a
Simplify the left-hand side of the equation by canceling the common factors.
a^20a+2=ac
Since ac contains the variable to solve for, move it to the left-hand side of the equation by subtracting ac from both sides.
a^2-a+2-ac=0
Move all terms not containing a to the right-hand side of the equation.
a^2-a-ac+2=0
Use the quadratic formula to find the solutions. In this case, the values are a=1,b=-1, and c=-1c+2.
Substitute in the values of a=1, b=-1, and c=-1c+2.
a=(-(-1)[plus or minus sign]√(-1)^2-4(1)(-1c+2) )/(2(1))
Notice: The first and last parentheses before the "/" and after, are used to show that they are on top and bottom of a fraction. You do not have to worry about the parentheses before the "-(-1)", after the "2)", and before and after the "2(1)".
Simplify the section inside the radical.
a=( 1[plus or minus sign]√(4c-7) )/( 2 )
The final answer is the combination of both solutions.
a=( 1+√( 4c-7 )/( 2 ), ( 1-√( 4c-7 )/( 2 )
Check that all solutions found are valid and are part of the domain by substituting them into the original equation.
b=( 1+√( 4c-7 )/( 2 ), ( 1-√( 4c-7 )/( 2 )
Sorry. Not quite sure how you got b=a+c/ac-2, and I am sure that I did it correctly. I am very good at math :D.
Oh and in addition before I submit... I love how you play WoW. I play as well!:P Just saying.
a+2b/ab-1=c
Cancel the common factor of b in 2b/ab.
a+2/a-1=c
Find the LCD (least common denominator) of a+2/a-1=c.
LCD:a
Multiply each term in the equation by a in order to remove all the denominators from the equation.
a*a+2*a-1=c*a
Simplify the left-hand side of the equation by canceling the common factors.
a^20a+2=ac
Since ac contains the variable to solve for, move it to the left-hand side of the equation by subtracting ac from both sides.
a^2-a+2-ac=0
Move all terms not containing a to the right-hand side of the equation.
a^2-a-ac+2=0
Use the quadratic formula to find the solutions. In this case, the values are a=1,b=-1, and c=-1c+2.
Substitute in the values of a=1, b=-1, and c=-1c+2.
a=(-(-1)[plus or minus sign]√(-1)^2-4(1)(-1c+2) )/(2(1))
Notice: The first and last parentheses before the "/" and after, are used to show that they are on top and bottom of a fraction. You do not have to worry about the parentheses before the "-(-1)", after the "2)", and before and after the "2(1)".
Simplify the section inside the radical.
a=( 1[plus or minus sign]√(4c-7) )/( 2 )
The final answer is the combination of both solutions.
a=( 1+√( 4c-7 )/( 2 ), ( 1-√( 4c-7 )/( 2 )
Check that all solutions found are valid and are part of the domain by substituting them into the original equation.
b=( 1+√( 4c-7 )/( 2 ), ( 1-√( 4c-7 )/( 2 )
Sorry. Not quite sure how you got b=a+c/ac-2, and I am sure that I did it correctly. I am very good at math :D.
Oh and in addition before I submit... I love how you play WoW. I play as well!:P Just saying.
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2b/ab = 2/a, leaving no b
Did you mean (a+2b) / (ab-1) or ( (a+2b) / (ab) ) - 1 or maybe a + (2b / (ab - 1)), or something else?
Did you mean (a+2b) / (ab-1) or ( (a+2b) / (ab) ) - 1 or maybe a + (2b / (ab - 1)), or something else?
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cab-c = a+2b
b(ac-2)=a+c
b = (a+c)/(ac-2)
b(ac-2)=a+c
b = (a+c)/(ac-2)