Question 75916
How do you solve y = 3x + 1: x = 3y + 1. Using substitution. 
:
y = (3x+1); than means that where you see "y" in the 2nd equation you can
substitute (3x+1) for y
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The 2nd equation:
x = 3y + 1
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Now substitute (3x+1) for y in the above equation:
x = 3(3x+1) + 1
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Multiply what's inside the brackets and you have:
x = 9x + 3 + 1
x = 9x + 4
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Subtract x from both sides, subtract 4 from both sides and you have:
-4 = 9x - x
-4 = 8x
x = -4/8
x = -1/2
:
Remember that y = (3x+1)
x = -1/2 so substitute -1/2 for x:
y = 3(-1/2) + 1
y = -3/2 + 1
y = -3/2 + 2/2
y = -1/2 also
:
Check our solutions by substituting for both x and y in the 2nd equation:
x = 3y + 1. 
-1/2 = 3(-1/2) + 1
-1/2 = -3/2 + 2/2
-1/2 = -1/2; equality reigns
:
How about this? Make sense to you?