Question

Solve the following system of equation graphically:
$$\begin{gathered} 2x+3y+5=0 \\ 3x-2y-12=0 \end{gathered}$$

Answer 1

From the first equation, write y in terms of x
$y=-\left(\frac{5+2x}{3}\right).....\left(i\right)$
Substitute different values of x in (i) to get different values of y
$$\begin{gathered} \mathrm{For}x=-1,y=-\frac{5-2}{3}=-1 \\ \mathrm{For}x=2,y=-\frac{5+4}{3}=-3 \\ \mathrm{For}x=5,y=-\frac{5+10}{3}=-5 \end{gathered}$$
Thus, the table for the first equation ( 2x + 3y + 5 = 0 ) is

x	−1	2	5
y	−1	−3	−5

Now, plot the points A(−1,−1), B(2,−3) and C(5,−5) on a graph paper and join
them to get the graph of 2x + 3y + 5 = 0.
From the second equation, write y in terms of x
$y=\frac{3x-12}{2}.....\left(ii\right)$
Now, substitute different values of x in (ii) to get different values of y
$$\begin{gathered} \mathrm{For}x=0,y=\frac{0-12}{2}=-6 \\ \mathrm{For}x=2,y=\frac{6-12}{2}=-3 \\ \mathrm{For}x=4,y=\frac{12-12}{2}=0 \end{gathered}$$
So, the table for the second equation ( 3x − 2y − 12 = 0 ) is

x	0	2	4
y	−6	−3	0

Now, plot the points D(0,−6), E(2,−3) and F(4,0) on the same graph paper and join
D, E and F to get the graph of 3x − 2y − 12 = 0.



From the graph it is clear that, the given lines intersect at (2,−3).
Hence, the solution of the given system of equation is (2,−3).