Question

Find the coordinates of the intersection of the lines 7x+9y-48=0 and 7x+5y- 18=0

• A. ($\frac{-19}{14}$, $\frac{-4}{7}$)
• B. ($\frac{-39}{14}$, $\frac{15}{7}$)
• C. ($\frac{-39}{14}$, $\frac{15}{4}$)
• D. ($\frac{-39}{14}$, $\frac{15}{2}$)

Answer 1

The correct option is D ($\frac{-39}{14}$, $\frac{15}{2}$)

$7x+9y-48=0$ - - - - - (1)

$7x+5y-18=0$ - - - - - (2)

Equation (1) can be written as $7x+9y=48$ - - - - - (3)

Equation (2) can be written as $7x+5y=18$ - - - - - (4)

(3) - (4) $⟹$ $4y=30$ $⟹$ $y=\frac{15}{2}$

Substituting $y=\frac{15}{2}$ in equation (4),

$7x+5\left(\frac{15}{2}\right)=18$

$⟹7x+\frac{75}{2}=18$

$⟹7x=18-\frac{75}{2}$
$⟹7x=\frac{(36-75)}{2}$
$⟹7x=\frac{(-39)}{2}$

$⟹x=\frac{-39}{14}$

$\therefore$ The coordinates of the intersection of the lines $2x-3y+1=0$ and $2x+4y+5=0$ are ($\frac{-39}{14}$, $\frac{15}{2}$).