(0,a) , (b,0) , (2,d) and (e,7) lie on y=2x+1, find a, b, d and e

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The discussion focuses on finding the coordinates of points (0,a), (b,0), (2,d), and (e,7) that lie on the line defined by the equation y=2x+1. The value of a is determined to be 1, as calculated from the point (0,a). The value of b is found by solving the equation 0=2b+1, resulting in b being -0.5. The value of d is calculated as d=2(2)+1, yielding d as 5. Lastly, the value of e is derived from the equation 7=2e+1, resulting in e being 3.

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A STRAIGHT LINE has equation y=2x+1. The point coordinates (0,a) , (b,0) , (2,d) and (e,7) lie on this line. Find the values of a,b,d, and e.
 
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Parthyy said:
A STRAIGHT LINE has equation y=2x+1. The point coordinates (0,a) , (b,0) , (2,d) and (e,7) lie on this line. Find the values of a,b,d, and e.
You have an equation y = 2x + 1 and several ordered pairs on that line.

So for the first, (0, a) we have
[math]a = 2(0) + 1 = 1[/math]

Thus a =1. Can you finish?

-Dan
 
Parthyy said:
A STRAIGHT LINE has equation y=2x+1. The point coordinates (0,a) , (b,0) , (2,d) and (e,7) lie on this line. Find the values of a,b,d, and e.
Each pair represents (x, y). The line is given as y= 2x+ 1 so (0, a) must satisy a= 2(0)+ 1. similarly, (b, 0) must satisfy 0= 2b+ 1. Solve that equation for b. (2, d) must satisfy d= 2(2)+ 1. (e, 7) must satisfy 7= 2e+ 1. Solve that equation for e.
 

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