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a.
f(3)=3a+b
6=3a+b -->persamaan 1
f(0)=0(a)+b
4=b
substitusi
6=3a+4
2=3a
a=2/3
f(x)=2/3 x + 4
b.
f(1)=a+b
15=a+b -->persamaan 1
f(0)=0(a)+b
10=b
substitusi
15=a+10
15-10=a
a=5
f(x)=5x+10
c.
f(1)=a+b
2=a+b
b=2-a -->persamaan 1
f(2)=2a+b
3=2a+b
3=2a+(2-a)
3-2=2a-a
a=1
substitusi
b=2-1
b=1
f(x)=x+1