Tolong yah.... yang dlingkari soalnya bserta jalannya

Fungsi linier
(dapat diselesaikan dengan eliminasi/substitusi)

2)
g(x) = px + q
g(1) = 6 --> p(1) + q = 6 --> p + q = 6 ....(1)
g(2)= 10--> p(2) + q =10--> 2p + q = 10..(2)
.
persamaan (1) dan (2) dieliminasikan
p+ q = 6
2p + q = 10
..................(kurangkan)
-p = -4
p= 4
p + q = 6 --> 4 + q = 6
q = 6-4
q = 2
a) bentuk fungsi  g(x)= px + q --> g(x) = 3x + 2
b) g(6) = 3(6) + 2 = 18+ 2  = 20

3) f(x)= ax+ b
f(1) = 4
f(2)= 2
kurangkan
(1- 2) a = 4-2
-a = 2
a = -3
.
b = 4 - a
b = 4 -(-3)|B = 7
f(x)= -3x + 7
f(7) = -3(7) + 7 = - 14
f(3) = -3(3) +  7 = - 2

6.f(x)= ax + b

a) nilai a dan b
f(3) = 15
f(5)= 20
...kurangkan.,..
(3-5)a = 15 -20
-2a= - 5
a = 5/2

b = 15 - 3a
b = 15- 3(5/2)
b = 15 - 15/2
b = 15/2

b) bentuk fungsi --> f(x)= 5/2 x + 15/2  atau f(x)= 1/2 ( 5x + 15)

c) nilai f(-11)
f(-11) = 1/2 ( 5(11) + 15)
f(-11) = 1/2 (55+15) = 1/2 (70)
f(-11) = 35

7. f(x) = px + q
f(-2) = 7
f(3) = - 3
...kurangkan
(-2-3) p = 7-(-3)
-5p = 10
p = -2

q = -3 - 3p
q = -3- 3(-2)
q= - 3 + 6
q= 3
a) bentuk fungsi f(x)= -2 x + 3
b) f(4n) = -2(4n) + 3
f(4n) = -8n + 3