# Algebra 1

bobpursey, In this problem you helped me with earlier:

Solve using any appropriate method.

y = 2× - 5,

y = ½× + 1

Substitution, substitute for y.

2×-5=½ × + 1

3/2 x=6

x=4

y= 2*4-5=3

How did you choose which method you were going to use between substitution and elimination? Could this have done with the elimination method? Thanks

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1. Since both equations were in the form y = ...
he simply replaced the y value of the first equation with the y value of the second.
Some texts call this the "comparison" method
e.g.
if a = b
and c = b , then a = c

yes, you could use elimination after some doctoring of the equations
y = 2x - 5 ---> 2x - y = 5
y = (1/2)x + 1 ---> 2y = x + 2 --> x - 2y = -2

new first by 2: --> 4x - 2y = 10
new 2nd : ------> x - 2y = -2
subtract them:
3x = 12
x = 4
back in the original first:
y = 2(4) - 5 = 3

x=4 , y = 3

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