Question

5 circle the table that shows an additive pattern (y = x + a).

x	y
2	18
4	36
6	54

x	y
6	12
8	16
10	20

x	y
10	17
15	22
20	27

x	y
20	60
30	90
40	120

Explanation:

Step1: Analyze the first table

For \(x = 0,y=0\); if \(y=x + a\), then \(0=0 + a\), so \(a = 0\). But for \(x = 2\), if \(a = 0\), \(y\) should be \(2\), but \(y = 18\). So it's not \(y=x + a\).

Step2: Analyze the second table

For \(x = 4,y = 8\); if \(y=x + a\), then \(8=4 + a\), \(a = 4\). For \(x = 6\), \(y=6 + 4=10
eq12\). So it's not \(y=x + a\).

Step3: Analyze the third table

For \(x = 5,y = 12\); if \(y=x + a\), then \(12=5 + a\), \(a = 7\). For \(x = 10\), \(y=10 + 7=17\); for \(x = 15\), \(y=15 + 7=22\); for \(x = 20\), \(y=20+7 = 27\). It satisfies \(y=x + 7\).

Step4: Analyze the fourth table

For \(x = 10,y = 30\); if \(y=x + a\), then \(30=10 + a\), \(a = 20\). For \(x = 20\), \(y=20 + 20=40
eq60\). So it's not \(y=x + a\).

Answer:

The third table (with \(x = 5,y = 12\); \(x = 10,y = 17\); \(x = 15,y = 22\); \(x = 20,y = 27\)) shows the additive pattern \(y=x + a\) (where \(a = 7\)).