Question

The point of intersection of lines $3x+6y=9$ and $7x–15y=2$ is

• A. $(\frac{-49}{29},\frac{14}{39})$
• B. $(3,\frac{-19}{29})$
• C. $(\frac{-13}{29},\frac{14}{29})$
• D. $(\frac{49}{29},\frac{19}{29})$

Answer 1

The correct option is D $(\frac{49}{29},\frac{19}{29})$

$3x+6y=9$
$\Rightarrow 3(x+2y)=9$
$\Rightarrow x+2y=3$ …..(i)
and, $7x–15y=2$ …..(ii)
Multiplying (i) by 7, we get
$7(x+2y)=7\times 3$
$\Rightarrow 7x+14y=21$ …..(iii)
Subtracting (ii) from (iii), we get
$(7x+14y)–(7x–15y)=21–2$
$\Rightarrow 29y=19$
$\Rightarrow y=\frac{19}{29}$
Substituting $y=\frac{19}{29}$ in (i), we get
$x+2\left(\frac{19}{29}\right)=3$
$\Rightarrow x=3-\frac{38}{29}$
$\Rightarrow x=\frac{87-38}{29}=\frac{49}{29}$
Thus, the point of intersection of the lines is $(\frac{49}{29},\frac{19}{29})$
Hence, the correct answer is option (d).