Answer:
y =
x + 
Step-by-step explanation:
Linear function is defined by y = ax+b
where a and b are constants and (x,y ) are variables given as point
using point (3,2) that is plugging x =3 and y =2 ,we
2 = 3a+b ............. equation(1)
likewise using point (-2,13) and plugging x =-2 and y =13 ,we get
13 = -2a+b .......... equation(2)
Solving the equation (1) and 2

subtracting equation 2 from equation 1
3a+b = 2
-2a+b = 13
__-_______________
we have 5a = -11
a=
plugging a=
in equation (1),we get
3(
) +b =2
b = 2 -3(
)
b = 2+
b=
therefore equation is obtained by plugging
a =
and b = 
we get equation as
y =
x + 