Question 624779
Here's how I did it:

<font style = arial>{{{4x(x-5)  -  5x(x-4) = -1}}}          This is given

{{{4x^2-20x - 5x^2+20x=-1}}}            Distribute 4x and -5x to the contents of the parenthesis next to them. DPMA?
{{{-x^2=-1}}}    Add like terms: {{{4x^2+-5x^2}}} and {{{-20x+20x}}}
{{{-x^2+1=0}}}   Transpose -1 to the other side of the equation, which makes it positive
{{{x^2-1=0}}}    In solving equations like this, oyu are allowed to multiply a number to both sides of the equation, so long as you do it for both sides of the equation. Here, multiply -1 to both sides turning {{{-x^2}}} to positive {{{x^2}}} and {{{+1}}} to {{{-1}}}

If you are familiar with the special product called DOTS <I>(Difference of two squares)</I> this will make solving the equation easier. Given {{{x^2-1=0}}} you can factor it to {{{(x+1)}}} and {{{(x-1)}}}. Equate both of these to 0 and you get:

{{{x+1=0}}} -->{{{x=-1}}}
and
{{{x-1=0}}} --> {{{x=1}}}

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I can't find a way to reply to your comment so, I just edited this.
when {{{x=1}}};

{{{4*(1)*((1)-5)  -  5*(1)*((1)-4) = -1}}} 
{{{4*(-4)-5*(-3)=-1}}}
{{{-16-(-15)=-16+15=-1}}}

You can try the same with {{{x=-1}}}

Hope this does it for your problem.

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