Asked by slomo
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
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
Answers
Answered by
Reiny
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
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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