Find the points of intersection of the lines

Solution:

Suppose the given two lines intersect at a point $P\left(x_{1}, y_{1}\right) .$ Then, $\left(x_{1}, y_{1}\right)$ 

satisfies each of the given equations.

$\therefore 4 x+3 y=5$

or $4 x+3 y-5=0 \ldots$ (i)

and $x=2 y-7$

or $x-2 y+7=0$...(ii)

Now, we find the point of intersection of eq. (i) and (ii)

Multiply the eq. (ii) by 4, we get

$4 x-8 y+28=0$ ...............(iii)

On subtracting eq. (iii) from (i), we get

$4 x-8 y+28-4 x-3 y+5=0$

$\Rightarrow-11 y+33=0$

$\Rightarrow-11 y=-33$

$\Rightarrow y=\frac{33}{11}=3$

Putting the value of $y$ in eq. (i), we get

$4 x+3(3)-5=0$

$\Rightarrow 4 x+9-5=0$

$\Rightarrow 4 x+4=0$

$\Rightarrow 4 x=-4$

$\Rightarrow x=-1$

Hence, the point of intersection $\mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$ is $(-1,3)$