Given:
QR is parallel to RS.
To find:
The value of x.
Solution:
In the given figure, we have,
[tex]PR=12[/tex]
[tex]RT=2x-2
[/tex]
[tex]QS=21[/tex]
[tex]ST=3x-1[/tex]
Using basic proportionality theorem (BPT), we get
[tex]\dfrac{PR}{RT}=\dfrac{QS}{ST}[/tex]
[tex]\dfrac{12}{2x-2}=\dfrac{21}{3x-1}[/tex]
On cross multiplication, we get
[tex]12(3x-1)=21(2x-2)[/tex]
[tex]36x-12=42x-42[/tex]
[tex]36x-42x=12-42[/tex]
[tex]-6x=-30[/tex]
Divide both sides by -6.
[tex]x=\dfrac{-30}{-6}[/tex]
[tex]x=5[/tex]
Therefore, the value of x is 5.
