Diketahui rumus fungsi f(x) =ax + b, f(2) = 1 dan f(–3) =–9. Nilai f(5) adalah ... A.-6 B.7 C.9 D.11

f(x) = ax + b
F(2) = a(2) + b = 1
F(-3) = a(-3) + b = -9

lalu eliminasikan

 2a + b = 1
-3a + b = -9
------------------- -
5a = 10
a = 10/5 = 2

lalu ambil salah satu persamaan

2a + b = 1
2.2 + b = 1
4 + b = 1
b = 1-4 = -3

f(5) = 2.5 + -3
= 10 - 3
= 7 (b)

2*f(x)=ax+b
f(2)=a(2)+b=1
f(2)=2a+b=1
*f(x)=ax+b
f(-3)=a(-3)+b=-9
f(-3)=-3a+b=-9

eliminasi

2a+b=1
-3a+b=-9
————— -
5a=10
a=10/5
a=2

ambill salah satu dr persamaan yg kita gabungkan td
*2a+b=1
2(2)+b=1
4+b=1
b=1-4
b=-3

*f(5)=2.5+(-3)
= 10-3
=7