How do you solve #(x)/(2x+7)=(x-5)/(x-1)#?

Question

Abigail Allen · Accepted Answer

#x=-5# or #x=7#

Recall that you multiply to the numerator when you perform a fractional multiplication.

First, let's simplify the LHS by multiplying #(x-5)/(x-1)# with #2x+7#

#(x-5*2x+7)/(x-1)#

#(2x^2-10x+7x-35)/(x-1)#

Thus, we have

#x=(2x^2-3x-35)/(x-1)#

Then let's simplify the RHS by multiplying #x# with #x-1#.

#x/1=(2x^2-3x-35)/(x-1)# since #x=x/1#,

Then

#x*x-1=2x^2-3x-35#

Then #x^2-x=2x^2-3x-35#

#0=x^2-2x-35#

Cross multiplying is what we have just done.

We can factor this quadratic equation to get the solution we need.

#(x+5)(x-7) =0#

Set each factor equal to #0#, since if either #(x+5)# or #(x-7)# was 0, the equation would hold, because #0# multiplied by any number will always be 0.

#x+5=0# #x=-5#

#x-7=0# #x=7#

These are our two responses.

Julia Adams · Answer

undefined To solve the equation (x)/(2x+7)=(x-5)/(x-1), we can cross-multiply to eliminate the fractions. This gives us x(x-1) = (x-5)(2x+7). Expanding both sides of the equation, we get x^2 - x = 2x^2 - 3x - 35. Rearranging the terms, we have x^2 - 2x^2 + x - 3x + 35 = 0. Combining like terms, we obtain -x^2 - 2x + 35 = 0. To solve this quadratic equation, we can either factor it or use the quadratic formula. Factoring, we find (x-7)(x+5) = 0. Therefore, x = 7 or x = -5 are the solutions to the equation.