Question

(x-5)(x+1)=3x-5

Answer 1

$(x-5)(x+1)=3x-5$

$x^{2} -5x+x-5=3x-5$

$x^{2} -5x+x-5-3x+5=0$

$x^{2} -7x=0$

$x(x-7)=0$

$x=0$   или  $x-7=0$

                      $x=7$

Ответ:   $0;$  $7$

Answer 2

To solve the equation $(x-5)(x+1) = 3x-5$, let's simplify and solve step by step:

Expanding the left side: $(x-5)(x+1) = x(x+1) - 5(x+1) = x^2 + x - 5x - 5 = x^2 - 4x - 5$

So now the equation becomes: $x^2 - 4x - 5 = 3x - 5$

Rearranging the equation: $x^2 - 4x - 3x - 5 + 5 = 0$ $x^2 - 7x = 0$

Factoring out an 'x': $x(x - 7) = 0$

Now, either $x = 0$ or $x - 7 = 0$.

If $x = 0$, then the equation holds true: $(0-5)(0+1) = -5 \neq 3(0)-5 = -5$ This is not a solution.

If $x - 7 = 0$, then $x = 7$, and we can check if it satisfies the equation: $(7-5)(7+1) = 2(8) = 16 = 3(7)-5 = 16$ This is a valid solution.

Therefore, the equation $(x-5)(x+1) = 3x-5$ has a single solution: $x = 7$.