SOLUTION: How do you solve {{{y=3x + 1}}} {{{ x=3y + 1}}}. Using substitution. Our teacher has given us this practice with no review from her. I don't know where to start using substitutio
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-> SOLUTION: How do you solve {{{y=3x + 1}}} {{{ x=3y + 1}}}. Using substitution. Our teacher has given us this practice with no review from her. I don't know where to start using substitutio
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Question 75916This question is from textbook Prentice Hall Algebra
: How do you solve . Using substitution. Our teacher has given us this practice with no review from her. I don't know where to start using substitution. This question is from textbook Prentice Hall Algebra
You can put this solution on YOUR website! 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
:
The 2nd equation:
x = 3y + 1
:
Now substitute (3x+1) for y in the above equation:
x = 3(3x+1) + 1
:
Multiply what's inside the brackets and you have:
x = 9x + 3 + 1
x = 9x + 4
:
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?
You can put this solution on YOUR website! substitution means taking a quantity from one equation and substituting its equivalent into another equation
In your case, taking the y from the first equation and substituting into the second equation gives x=3(3x+1)+1...every "y" in the second equation is replaced by its equivalent from the first equation "3x+1"... so x=9x+3+1...or x=-1/2
noting the symmetry of the two equations (they are identical, with the variables switched) means that y will also be -1/2. You could confirm this by practicing a substitution from the second equation to the first
