SOLUTION: 3x+4y=5, 2x+y=1...If a system has an infinite number of solutions, use set-builder notation to write the solution set. If a system has no solution, state this.
This is what I ha
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Question 592587: 3x+4y=5, 2x+y=1...If a system has an infinite number of solutions, use set-builder notation to write the solution set. If a system has no solution, state this.
This is what I have......
3x+4y=5, 2x+y=1
3x+4y=5
3x-3x+4y=5-3x
4y=5-3x
4y=3x+5
y=3x+5/4
2x+y=1
2x-2x+y=1-2x
y=1-2x
1y=-2x+1/1
y=-2x+1 no solution
What am I doing wrong here?
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Your first error is at the step where you went from:
Correct
to
Incorrect (Sign error)
Should be:
Then you made another error when you divided by 4. You only divided one of the RHS terms by 4, so instead of:
It should have been (counting correcting the sign error from before)
I'll let you take it from there. Write back if you are still having trouble.
John
My calculator said it, I believe it, that settles it
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