4/x+1 - 3/x+2 = 1 how do you solve this thanks!
A alfard New member Joined Oct 8, 2014 Messages 1 Oct 8, 2014 #1 4/x+1 - 3/x+2 = 1 how do you solve this thanks!
D Deleted member 4993 Guest Oct 8, 2014 #2 alfard said: 4/x+1 - 3/x+2 = 1 how do you solve this thanks! Click to expand... Is your expression: \(\displaystyle \dfrac{4}{x} \ + \ 1 - \dfrac{3}{x} \ + \ 2 \ = \ 1\) or \(\displaystyle \dfrac{4}{x \ + \ 1} \ - \dfrac{3}{x \ + \ 2} \ = \ 1\) In either case, simplify the left-hand-side using law of addition (or subtraction) of fractions (as if you have numbers instead of 'x' and 'y'). If you are still stuck, let us know, including your work.
alfard said: 4/x+1 - 3/x+2 = 1 how do you solve this thanks! Click to expand... Is your expression: \(\displaystyle \dfrac{4}{x} \ + \ 1 - \dfrac{3}{x} \ + \ 2 \ = \ 1\) or \(\displaystyle \dfrac{4}{x \ + \ 1} \ - \dfrac{3}{x \ + \ 2} \ = \ 1\) In either case, simplify the left-hand-side using law of addition (or subtraction) of fractions (as if you have numbers instead of 'x' and 'y'). If you are still stuck, let us know, including your work.
T TeacherSatya New member Joined Oct 19, 2014 Messages 10 Oct 19, 2014 #3 Solution 4/(x+1) - 3/(x+2) = 1 Alright here we go: So what we first do is cross multiply the left hand side: [4(x + 2) -3(x + 1)]/[(x + 1)(x + 2)]=1 [4x + 8 -3x - 3]/[x^2 + 3x + 2] = 1 4x + 8 -3x - 3 = x^2 + 3x + 2 x + 5 = x^2 + 3x + 2 x^2 + 3x + 2 - x - 5 = 0 x^2 + 2x - 3 = 0 >>>> (Now we factorize them) (x + 3)(x - 1) = 0 i) x + 3 = 0 x = -3 ii) x - 1 = 0 x = 1 So the answer of x = 1 or x = -3 Hope it helps Last edited by a moderator: Oct 19, 2014
Solution 4/(x+1) - 3/(x+2) = 1 Alright here we go: So what we first do is cross multiply the left hand side: [4(x + 2) -3(x + 1)]/[(x + 1)(x + 2)]=1 [4x + 8 -3x - 3]/[x^2 + 3x + 2] = 1 4x + 8 -3x - 3 = x^2 + 3x + 2 x + 5 = x^2 + 3x + 2 x^2 + 3x + 2 - x - 5 = 0 x^2 + 2x - 3 = 0 >>>> (Now we factorize them) (x + 3)(x - 1) = 0 i) x + 3 = 0 x = -3 ii) x - 1 = 0 x = 1 So the answer of x = 1 or x = -3 Hope it helps
