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f(a+1) = 10
2(a+1)-3 = 10
2a+2-3 = 10
2a = 10-2+3
2a = 11
a = 11/2
f(x) = 2x-3 ⇒rumus fungsi
f(a+1) =2(a+1) - 3 = 10
Gunakan persamaan
2(a+1) - 3 = 10
2a+2 - 3 = 10
2a+2 = 13
2a = 11
a = 5,5
Pembuktian
f(5,5+1)= f(6,5)
f(6,5) = 2.(6,5) - 3
= 13 - 3
= 10
Jadi, nilai a yang memenuhi adalah 5,5
Maaf kalau salah