#$&*
Mth 158
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How would you do the problem
-4/(2x+3) + 1/(x-1) = 1/(2x=3)(x-1)
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The denominator on the right-hand side requires parentheses, so the equation would be correctly written as
-4/(2x+3) + 1/(x-1) = 1/( (2x+3)(x-1) )
The least common denominator is
(2x+3)(x-1).
Multiplying both sides by the least common denominator you get
(2x+3)(x-1).* (-4/(2x+3)) + (2x+3)(x-1).* 1/(x-1) = (2x+3)(x-1).* 1/( (2x+3)(x-1) )
The 2x+3 in the numerator of the first term is divided by the 2x + 3 in the denominator of that term, leaving -4(x-1).
The x-1 in the numerator of the second term is divided by the x-1 in the denominator, leaving just 2x+3.
The (2x+3)(x-1). in the numerator of the right-hand side is divided by the (2x+3)(x-1). in the denominator, leaving just 1 on the right-hand side.
The equation thus becomes
-4(x-1) + (2x+3) = 1.
This equation simplifies to
-2x = 6
so that the solution is x = 3.
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#$&*
Mth 158
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I'm confused on how to solve the equation by completing the square.
x^2 - 6x =13
I know you have to get thirteen over to the right (which is already done) but where do you go from there?
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The perfect square must start out
x^2 - 6 x.
You've already got that on one side of the equation, which is a very good idea.
You need to find a number to add to this expression to make it a perfect square. Then after adding that number to both sides you'll have a perfect square on the left.
The rule for doing this isn't difficult to understand. If the coefficient of x^2 is 1, as it is here, we just take half the coefficient of x and square it. That gives us the number we need.
The coefficient of x is -6. Half of that is -3. Squaring -3 we get 9.
So we add 9 to both sides of the equation to get
x^2 - 6 x + 9 = 13 + 9.
We factor the left-hand side and simplify the right-hand side to get
(x-3)^2 = 22.
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To finish the problem we need to solve the equation
(x-3)^2 = 22.
If (x-3)^2 = 22, then
(x-3) = +- sqrt(22)
so that
x = 3 +- sqrt(22).
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