SOLUTION: i need to solve these two systems of equations: 3x+4y=12 and 2x+y=13 we have to solve these equations using: graphing, substitution, elimination, matrix and the cramers rule met

Algebra ->  Matrices-and-determiminant -> SOLUTION: i need to solve these two systems of equations: 3x+4y=12 and 2x+y=13 we have to solve these equations using: graphing, substitution, elimination, matrix and the cramers rule met      Log On


   

Question 395961: i need to solve these two systems of equations: 3x+4y=12 and 2x+y=13
we have to solve these equations using: graphing, substitution, elimination, matrix and the cramers rule methods.
i know that all the answers will be the same, and when i worked out this problem according to the elimination method i got the point of intersection as (-8,9) but when i was trying to graph it i was getting weird numbers and couldn't set up the graph.
-thanks so much if you could help me out :) kristianne
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
3x+4y=12
2x+y=13
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y = (-3/4)x + 3
y = -2x+13
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Graph:
graph%28400%2C300%2C-10%2C10%2C-10%2C10%2C%28-3%2F4%29x%2B3%2C-2x%2B13%29
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It doesn't look like (-8,9) is the solution.
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Let's look at elimination:
3x+4y=12
2x+y=13
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Multiply thru the bottom equation by 4 to get:
3x+4y = 12
8x+4y = 52
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Subtract the top from the bottom and solve for "x":
5x = 40
x = 8
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Now, solve for "y":
3x+4y = 12
3*8 + 4y = 12
24 + 4y = 12
4y = -12
y = -3
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Looks like the solution is (8,-3)
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Recommendation: Check your answers.
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Cheers,
Stan H.