x = 11 (or) x = –13
Solution:
Theorem:
If three or more parallel lines are cut by a transversal then they divide the transversals proportionally.
Given lines l, m, n are parallel lines cut by a two transversal lines.
Therefore, they are in proportion by the above theorem.
[tex]$\Rightarrow\frac{x+5}{10}=\frac{12.8}{x-3}[/tex]
Do cross multiplication.
[tex]$\Rightarrow(x+5)\times(x-3)=12.8\times10[/tex]
[tex]$\Rightarrow x^2+5x-3x-15=128[/tex]
[tex]$\Rightarrow x^2+2x-15=128[/tex]
Arrange all terms in one side.
[tex]$\Rightarrow x^2+2x-15-128=0[/tex]
[tex]$\Rightarrow x^2+2x-143=0[/tex]
[tex]$\Rightarrow x^2-11x+13x-143=0[/tex]
Take common terms outside
[tex]$\Rightarrow x(x-11)+13(x-11)=0[/tex]
[tex]$\Rightarrow (x-11)(x+13)=0[/tex]
[tex]$\Rightarrow (x-11)=0\ \text{(or)}\ (x+13)=0[/tex]
⇒ x = 11 (or) x = –13
Hence x = 11 (or) x = –13.
