Option D:
y = 2x – 1; y is twice x minus 1.
Solution:
Option A: y = 2x; y is double x
substitute x values in y = 2x, we get
For x = 1, y = 2(1) = 2
so, y = 2x is false.
Option B: y = 3x; y is triple x
substitute x values in y = 3x, we get
For x = 1, y = 3(1) = 3
so, y = 3x is false.
Option C: y = 2x + 1; y is twice x plus 1.
substitute x values in y = 2x + 1, we get
For x = 1, y = 2(1) + 1 = 3
so, y = 2x + 1 is false.
Option D: y = 2x – 1; y is twice x minus 1.
substitute x values in y = 3x, we get
For x = 1, y = 2(1) – 1 = 1
For x = 2, y = 2(2) – 1 = 3
For x = 3, y = 2(3) – 1 = 5
For x = 4, y = 2(4) – 1 = 7
For x = 5, y = 2(5) – 1 = 9
so, y = 2x – 1 is true.
Hence the expression for the table is y = 2x – 1; y is twice x minus 1.
Each value of y is double that of corresponding value of x, minus 1.
