1.Diketahui ac+ad+bc+bd sm dg 68 dan c+d sm dg 4 .Berapa nilai dari a+b+c+d.?

ac+ad+bc+bd = 68
ac+bc+ad+bd = 68
c(a+b)+d(a+b) = 68
(c+d)(a+b) = 68
4(a+b) = 68
a+b = 17
Maka,
a+b+c+d = (a+b)+(c+d) = 17+4 = 21

ac + ad + bc + bd = 68
(a+b)c + (a+b)d = 68
(a+b)(c+d) = 68
karena nilai c+d = 4, maka
(a+b)(4) = 68
a+b = 68/4
a+b = 17
jadi nilai a+b+c+d = 17 + 4 = 21