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Solve following equations.
1/(x+1) = 1- (5/2x - 4)
Solution
Just like along with the linear equations the primary thing which we're going to need to do here is to clear the denominators out through multiplying by the LCD. Remember that we will also have to note value(s) of x which will give division by zero so that we can ensure that these aren't involved in the solution.
1/ (x+1) = 1 - (5/2x - 4)
The LCD for this problem is ( x + 1)(2x - 4) and we will have to avoid x =-1 and x = 2 to ensure we don't get division by zero. Following is the work for this equation.
( x + 1) ( 2x - 4) (1/(x+1))= (x + 1) ( 2 x - 4) (1 - 5/2x - 4)
2x - 4 = ( x + 1) ( 2 x - 4) - 5 ( x + 1)
2x - 4 =2 x2 - 2 x - 4 - 5x - 5
0 = 2x2 - 9x - 5
0 = ( 2x + 1) ( x - 5)
Thus, it seem like the two solutions to this equation are following,
x =- 1/2 and x = 5
Notice that neither of these are the values of x which we needed to avoid nor so both are solutions.
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Two sides of an isosceles triangle are 7and 3. The perimeter of the triangle is?
three ordered pair
Logarithm form In this definition y = log b x is called the logarithm form Exponential form In this definition b y = x is called the exponential form.
4(11-5)=
can you help me?
We'll begin this section by defining just what a root or zero of a polynomial is. We say that x = r is a root or zero of a polynomial, P ( x) , if P ( r )= 0 . In other terms x=
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x squred y+ x
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