2/y+3/x=7/xy,1/y+9/x=11/xy by substitution method
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Answer:
How can we solve 2/y+3/x=7/xy, 6/y+9/x=11/xy?
2/y + 3/x = 7/xy
6/y + 9/x = 11/xy
Solve for x in the first equation:
2/y + 3/x = 7/xy
Multiply both sides by xy:
2x + 3y = 7
Subtract 3y from both sides, and then divide by 2:
2x = -3y + 7
x = (-3y+7)/2
Now solve for x in the second equation:
6/y + 9/x = 11/xy
Multiply both sides by xy:
6x + 9y = 11
subtract 9y from both sides, and then divide by 6:
6x = -9y+11
x = (-9y+11)/6
This means:
(-3y+7)/2 = (-9y+11)/6
Multiply both sides by 6 to eliminate the denominators:
3(-3y+7) = (-9y+11)
-9y+7 = -9y+11
Ordinarly from here you would add 9y to both sides, but in this case it actually eliminates y from both sides of the equation. You end up with:
7 NOT = 11
This tells us there are no solutions to this system of equations. This is further evidenced by the original equations themselves:
2/y + 3/x = 7/xy
6/y + 9/x = 11/xy
There’s a pattern between the two equations: the numerators in the first two terms are products of the numerators in the first equation and the number 3. The problem is that the numerator of the term on the opposite side of the equation is not multiplied by 3. If it were, it would look like this:
6/y + 9/x = 21/xy
Because all three terms are not of the same proportion there are no solutions to this system of equations. They each describe their terms differently so they cannot share any terms (which is what happens in a system of equations). Thus, these equations are not of the same system!
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