By, Cross multiplying, we get
[tex]⇢2x(x - 1) = (2x - 1)(x + 2)[/tex]
[tex]⇢2 {x}^{2} - 2x = 2x(x + 2) - 1( x+ 2)[/tex]
[tex]⇢2 {x}^{2} - 2x = 2 {x}^{2} + 4x - x - 2[/tex]
Here, [tex]2 {x}^{2} [/tex] from both sides will get cancelled
[tex] ⇢- 2x = 3x - 2[/tex]
[tex]⇢ - 2x - 3x = - 2[/tex]
[tex] ⇢- 5x = - 2[/tex]
[tex] ⇢x = \frac{ - 2}{ - 5} [/tex]
[tex] ⇢ \underline { \boxed{ x = \frac{2}{5} }} [/tex]
Answer:
[tex]x=\frac{2}{5}[/tex]
Step-by-step explanation:
[tex]\frac{2 x}{x+2}-\frac{x}{x-1}=1[/tex]
[tex]\begin{aligned} & \Leftrightarrow \frac{2 x(x-1)}{(x+2)(x-1)}-\frac{x(x+2)}{(x-1)(x+2)}=1 \\& \Leftrightarrow \frac{2 x^{2}-2 x}{(x+2)(x-1)}-\frac{\left(x^{2}+2 x\right)}{(x+2)(x-1)}=1 \\& \Leftrightarrow \frac{2 x^{2}-2 x-\left(x^{2}+2 x\right)}{(x+2)(x-1)}=1 \\& \Leftrightarrow \frac{2 x^{2}-2 x-x^{2}-2 x}{(x+2)(x-1)}=1 \\& \Leftrightarrow \frac{x^{2}-4 x}{(x+2)(x-1)}=1 \\\end{aligned}[/tex]
[tex]\begin{aligned}& \Leftrightarrow x^{2}-4 x=(x+2)(x-1) \\& \Leftrightarrow x^{2}-4 x=x^{2}-x+2 x-2 \\& \Leftrightarrow x^{2}-4 x=x^{2}+x-2 \\& \Leftrightarrow-4 x=x-2 \\& \Leftrightarrow 5 x=2\end{aligned}[/tex]
therefore,
[tex]x=\frac{2}{5}[/tex]
