Already have an account? Get multiple benefits of using own account!
Login in your account..!
Remember me
Don't have an account? Create your account in less than a minutes,
Forgot password? how can I recover my password now!
Enter right registered email to receive password!
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.
5x-3y-11=0 and 3x+10y+17=0 Solve for all variables in each system of equations
28,14...is the sequence arithmetic or geometric?
5x+2x-17=53
SUPPOSE Y IS DIRECTLY PROPORTIONAL TO X AND THAT Y = 35 WHEN X = 5 FIND THE CONSTANT OF PROPORTIONALITY K K=
v(5)
In this section we will discussed at solving exponential equations There are two way for solving exponential equations. One way is fairly simple, however requires a very specia
can you explain to me how to solve x^3+x+5
I need to do a coursework for math but do not know what to do?
I dont really understand it can you help me
4=x+5 3x+4=-11
Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!
whatsapp: +91-977-207-8620
Phone: +91-977-207-8620
Email: [email protected]
All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd