SOLUTION: 2x+3y=8
3x+2y=7
I have to determind how many solutions exit.
Use either elimination or substitution to find solutions.
I have tried and my teacher told me I had every questio
Algebra.Com
Question 476503: 2x+3y=8
3x+2y=7
I have to determind how many solutions exit.
Use either elimination or substitution to find solutions.
I have tried and my teacher told me I had every question wrong. I am lost please help.
Found 2 solutions by mananth, ccs2011:
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
2 x + 3 y = 8 .............1
3 x + 2 y = 7 .............2
Eliminate y
multiply (1)by -2
Multiply (2) by 3
-4 x -6 y = -16
9 x + 6 y = 21
Add the two equations
5 x = 5
/ 5
x = 1
plug value of x in (1)
2 x + 3 y = 8
2 + 3 y = 8
3 y = 8 -2
3 y = 6
y = 2
x=1 ,y=2
m.ananth@hotmail.ca
Answer by ccs2011(207) (Show Source): You can put this solution on YOUR website!
For this problem I think elimination is the best way to find solution.
Step 1: Determine which variable you want to eliminate
Step 2: Find the least common multiple of the 2 coefficients.
Step 3: Multiply each equation by appropriate scale factor such that the coefficients will be equal but have opposite signs.
Step 4: Add the equations
Step 5: Solve the resulting single-variable equation.
Step 6: Substitute value back into one of the original equations to solve for 2nd variable.
----------------------------------------
Step 1: Eliminate x
Step 2: Least common multiple of 2 and 3 is 6.
Best way to do this is multiply them together...2*3=6
Step 3: Goal is to have 6x on top and -6x on bottom
Multiply top equation by 3... 2x*3 = 6x
Multiply bottom equation by -2... 3x*-2 = -6x
Equations change to:
Step 4: Add equations, just add the coefficients in each column
Result:
Step 5: solve for y
Step 6: Pick 1 of the original equations and replace y with 2
Then there is one unique solution of x =1 , y = 2.
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