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System of Equations...3
- Thread starter nycmathguy
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where did you find that one?
solutions are a bit complicated, you have two equations and four unknown variables
it takes a lot to get solutions of two variables expressed in terms of two other variables
solution looks like this:
or
solutions are a bit complicated, you have two equations and four unknown variables
it takes a lot to get solutions of two variables expressed in terms of two other variables
solution looks like this:
or
Last edited:
- Joined
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where did you find that one?
solutions are a bit complicated, you have two equations and four unknown variables
it takes a lot to get solutions of two variables expressed in terms of two other variables
solution looks like this:
or
All the system of equations threads come from David Cohen's Precalculus With Unit Circle Trigonometry Edition 4 textbook.
Instructions:
Solve each system of equations. If
there are no solutions in a particular case, say so. In cases in which there are literal (rather than numerical) coefficients, specify any restrictions that your solutions impose on those coefficients.
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(2b)/x-3/y=7ab..........eq.1
(4a)/x-(5a)/(by)=3a^2......eq.2
---------------------------------------
(2b)/x-3/y=7ab..........eq.1, solve for x
(2b)/x=7ab+3/y
x=(2b)/(7ab+3/y)............eq.1a
substitute x in eq.2
(4a)/((2b)/(7ab+3/y))-(5a)/(by)=3a^2......eq.2, solve for y
14a^2 + (6a)/(by)-(5a)/(by)=3a^2
(6a)/(by)-(5a)/(by)=3a^2-14a^2
(6a-5a)/(by)=-11a^2
a/(by)=-11a^2
a/(-11a^2)= by
-1/(11a)= by
y=-1/(11ab)
go to
x=(2b)/(7ab+3/y)............eq.1a, substitute y
x=(2b)/(7ab+3/(-1/(11ab)))
x=(2b)/(7ab-3(11ab))
x=(2b)/(7ab-33ab)
x=(2b)/(-26ab)
x=-1/(13a)
restrictions:
(4a)/x-(5a)/(by)=3a^2......eq.2
---------------------------------------
(2b)/x-3/y=7ab..........eq.1, solve for x
(2b)/x=7ab+3/y
x=(2b)/(7ab+3/y)............eq.1a
substitute x in eq.2
(4a)/((2b)/(7ab+3/y))-(5a)/(by)=3a^2......eq.2, solve for y
14a^2 + (6a)/(by)-(5a)/(by)=3a^2
(6a)/(by)-(5a)/(by)=3a^2-14a^2
(6a-5a)/(by)=-11a^2
a/(by)=-11a^2
a/(-11a^2)= by
-1/(11a)= by
y=-1/(11ab)
go to
x=(2b)/(7ab+3/y)............eq.1a, substitute y
x=(2b)/(7ab+3/(-1/(11ab)))
x=(2b)/(7ab-3(11ab))
x=(2b)/(7ab-33ab)
x=(2b)/(-26ab)
x=-1/(13a)
restrictions:
- Joined
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- Messages
- 5,386
- Reaction score
- 422
(2b)/x-3/y=7ab..........eq.1
(4a)/x-(5a)/(by)=3a^2......eq.2
---------------------------------------
(2b)/x-3/y=7ab..........eq.1, solve for x
(2b)/x=7ab+3/y
x=(2b)/(7ab+3/y)............eq.1a
substitute x in eq.2
(4a)/((2b)/(7ab+3/y))-(5a)/(by)=3a^2......eq.2, solve for y
14a^2 + (6a)/(by)-(5a)/(by)=3a^2
(6a)/(by)-(5a)/(by)=3a^2-14a^2
(6a-5a)/(by)=-11a^2
a/(by)=-11a^2
a/(-11a^2)= by
-1/(11a)= by
y=-1/(11ab)
go to
x=(2b)/(7ab+3/y)............eq.1a, substitute y
x=(2b)/(7ab+3/(-1/(11ab)))
x=(2b)/(7ab-3(11ab))
x=(2b)/(7ab-33ab)
x=(2b)/(-26ab)
x=-1/(13a)
restrictions:
This is not your typical system of equations. This challenging problem is perfect to see if students understand the concept of solving system of equations. Well-done.
