Question

Solve the following system of equation graphically:
$$\begin{gathered} 2x-3y+13=0 \\ 3x-2y+12=0 \end{gathered}$$

Answer 1

From the first equation, write y in terms of x
$y=\frac{2x+13}{3}.....\left(i\right)$
Substitute different values of x in (i) to get different values of y
$$\begin{gathered} \mathrm{For}x=-5,y=\frac{-10+13}{3}=1 \\ \mathrm{For}x=1,y=\frac{2+13}{3}=5 \\ \mathrm{For}x=4,y=\frac{8+13}{3}=7 \end{gathered}$$
Thus, the table for the first equation ( 2x − 3y + 13 = 0 ) is

x	−5	1	4
y	1	5	7

Now, plot the points A(−5,1), B(1,5) and C(4,7) on a graph paper and join
A, B and C to get the graph of 2x − 3y + 13 = 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
$$\begin{gathered} \mathrm{For}x=-4,y=\frac{-12+12}{2}=0 \\ \mathrm{For}x=-2,y=\frac{-6+12}{2}=3 \\ \mathrm{For}x=0,y=\frac{0+12}{2}=6 \end{gathered}$$
So, the table for the second equation ( 3x − 2y + 12 = 0 ) is

x	−4	−2	0
y	0	3	6

Now, plot the points D(−4,0), E(−2,3) and F(0,6) 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).