Solve :
14x+3−1=5x+1\dfrac{14}{x + 3} - 1 = \dfrac{5}{x + 1}x+314−1=x+15
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Given,
⇒14x+3−1=5x+1⇒14−(x+3)x+3=5x+1⇒14−x−3x+3=5x+1⇒11−xx+3=5x+1⇒(11−x)(x+1)=5(x+3)⇒11x+11−x2−x=5x+15⇒−x2+10x+11=5x+15⇒−x2+10x−5x+11−15=0⇒−x2+5x−4=0⇒x2−5x+4=0⇒x2−4x−x+4=0⇒x(x−4)−1(x−4)=0⇒(x−1)(x−4)=0⇒x=1 or x=4.\Rightarrow \dfrac{14}{x + 3} - 1 = \dfrac{5}{x + 1} \\[1em] \Rightarrow \dfrac{14 - (x + 3)}{x + 3} = \dfrac{5}{x + 1} \\[1em] \Rightarrow \dfrac{14 - x - 3}{x + 3} = \dfrac{5}{x + 1} \\[1em] \Rightarrow \dfrac{11 - x}{x + 3} = \dfrac{5}{x + 1} \\[1em] \Rightarrow (11 - x)(x + 1) = 5(x + 3) \\[1em] \Rightarrow 11x + 11 - x^2 - x = 5x + 15 \\[1em] \Rightarrow -x^2 + 10x + 11 = 5x + 15 \\[1em] \Rightarrow -x^2 + 10x - 5x + 11 - 15 = 0 \\[1em] \Rightarrow -x^2 + 5x - 4 = 0 \\[1em] \Rightarrow x^2 - 5x + 4 = 0 \\[1em] \Rightarrow x^2 − 4x − x + 4 = 0 \\[1em] \Rightarrow x(x − 4) −1(x − 4) = 0 \\[1em] \Rightarrow (x − 1)(x − 4) = 0 \\[1em] \Rightarrow x = 1 \text{ or } x = 4.⇒x+314−1=x+15⇒x+314−(x+3)=x+15⇒x+314−x−3=x+15⇒x+311−x=x+15⇒(11−x)(x+1)=5(x+3)⇒11x+11−x2−x=5x+15⇒−x2+10x+11=5x+15⇒−x2+10x−5x+11−15=0⇒−x2+5x−4=0⇒x2−5x+4=0⇒x2−4x−x+4=0⇒x(x−4)−1(x−4)=0⇒(x−1)(x−4)=0⇒x=1 or x=4.
Hence, x = 1, 4.
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If x ≠ 0 and a ≠ 0, solve :
xa−a+bx=b(a+b)ax\dfrac{x}{a} - \dfrac{a + b}{x} = \dfrac{b(a + b)}{ax}ax−xa+b=axb(a+b)
(1200x+2)(x−10)−1200=60.\Big(\dfrac{1200}{x} + 2\Big)(x - 10) - 1200 = 60.(x1200+2)(x−10)−1200=60.
2x2 + ax - a2 = 0
2x+9+x=13\sqrt{2x + 9} + x = 132x+9+x=13
