Solving Linear Equations, with 3 unknowns
$begingroup$
$$6x-8y=24$$
$$frac{-2}{3}x+ frac{8}{9}y = m$$
I want to solve for all three of the variables. I did plot this into my calculator and the answer is $$frac{-8}{3}$$ I want to know the process in solving it.
linear-algebra fractions
asked Dec 5 '18 at 12:03
Mattking32Mattking32
11
$endgroup$
add a comment |
$begingroup$
$$6x-8y=24$$
$$frac{-2}{3}x+ frac{8}{9}y = m$$
I want to solve for all three of the variables. I did plot this into my calculator and the answer is $$frac{-8}{3}$$ I want to know the process in solving it.
linear-algebra fractions
asked Dec 5 '18 at 12:03
Mattking32Mattking32
11
$endgroup$
4
$begingroup$
Two linear equations will not determine $3$ unknowns uniquely. Furthermore, how is "$-frac{8}{3}$" an answer when asking for the values of $3$ unknowns?
$endgroup$
– Christoph
Dec 5 '18 at 12:07
$begingroup$
Do u know Gaussian elimination?
$endgroup$
– Cloud JR
Dec 5 '18 at 12:07
$begingroup$
@christoph well asked lol
$endgroup$
– Cloud JR
Dec 5 '18 at 12:08
add a comment |
$begingroup$
$$6x-8y=24$$
$$frac{-2}{3}x+ frac{8}{9}y = m$$
I want to solve for all three of the variables. I did plot this into my calculator and the answer is $$frac{-8}{3}$$ I want to know the process in solving it.
linear-algebra fractions
asked Dec 5 '18 at 12:03
Mattking32Mattking32
11
$endgroup$
$$6x-8y=24$$
$$frac{-2}{3}x+ frac{8}{9}y = m$$
I want to solve for all three of the variables. I did plot this into my calculator and the answer is $$frac{-8}{3}$$ I want to know the process in solving it.
linear-algebra fractions
linear-algebra fractions
asked Dec 5 '18 at 12:03
Mattking32Mattking32
11
asked Dec 5 '18 at 12:03
Mattking32Mattking32
11
asked Dec 5 '18 at 12:03
Mattking32Mattking32
11
asked Dec 5 '18 at 12:03
Mattking32Mattking32
11
asked Dec 5 '18 at 12:03
Mattking32Mattking32
11
11
4
$begingroup$
Two linear equations will not determine $3$ unknowns uniquely. Furthermore, how is "$-frac{8}{3}$" an answer when asking for the values of $3$ unknowns?
$endgroup$
– Christoph
Dec 5 '18 at 12:07
$begingroup$
Do u know Gaussian elimination?
$endgroup$
– Cloud JR
Dec 5 '18 at 12:07
$begingroup$
@christoph well asked lol
$endgroup$
– Cloud JR
Dec 5 '18 at 12:08
add a comment |
4
$begingroup$
Two linear equations will not determine $3$ unknowns uniquely. Furthermore, how is "$-frac{8}{3}$" an answer when asking for the values of $3$ unknowns?
$endgroup$
– Christoph
Dec 5 '18 at 12:07
$begingroup$
Do u know Gaussian elimination?
$endgroup$
– Cloud JR
Dec 5 '18 at 12:07
$begingroup$
@christoph well asked lol
$endgroup$
– Cloud JR
Dec 5 '18 at 12:08
4
4
$begingroup$
Two linear equations will not determine $3$ unknowns uniquely. Furthermore, how is "$-frac{8}{3}$" an answer when asking for the values of $3$ unknowns?
$endgroup$
– Christoph
Dec 5 '18 at 12:07
$begingroup$
Two linear equations will not determine $3$ unknowns uniquely. Furthermore, how is "$-frac{8}{3}$" an answer when asking for the values of $3$ unknowns?
$endgroup$
– Christoph
Dec 5 '18 at 12:07
$begingroup$
Do u know Gaussian elimination?
$endgroup$
– Cloud JR
Dec 5 '18 at 12:07
$begingroup$
Do u know Gaussian elimination?
$endgroup$
– Cloud JR
Dec 5 '18 at 12:07
$begingroup$
@christoph well asked lol
$endgroup$
– Cloud JR
Dec 5 '18 at 12:08
$begingroup$
@christoph well asked lol
$endgroup$
– Cloud JR
Dec 5 '18 at 12:08
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Since you has three unknowns, you need three equations to be able to solve for all three variables (your system of linear equations is underdetermined).
However, what you can do is solve for $m$.
Multiplying the second equation through by $-9$ we get:
$$ 6x - 8y = -9m$$
Aha! This looks a lot like our first equation: we already know $6x - 8y = 24$. Putting these together gives $-9m = 24$ and hence $$m = -frac{8}{3}$$
However, without another equation we cannot determine $x$ and $y$ (for example, both $x = 0, y=-3$ and $x = 4, y=0$ satisfy your system of equations. In fact, there will be infinitely many values of $x$ and $y$ satisfying the two equations).
answered Dec 5 '18 at 12:09
ODFODF
1,486510
$endgroup$
add a comment |
$begingroup$
From $frac{-2}{3}x+ frac{8}{9}y = m$ we get $-6x+8y=9m$. If we add this to the first equation we get $24+9m=0$, hence $m=frac{-8}{3}.$
It remains only one equation: $ 6x-8y=24$. There are infinitely many pairs $(x,y)$ which are solutions of this equation.
answered Dec 5 '18 at 12:09
FredFred
47.2k1849
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since you has three unknowns, you need three equations to be able to solve for all three variables (your system of linear equations is underdetermined).
However, what you can do is solve for $m$.
Multiplying the second equation through by $-9$ we get:
$$ 6x - 8y = -9m$$
Aha! This looks a lot like our first equation: we already know $6x - 8y = 24$. Putting these together gives $-9m = 24$ and hence $$m = -frac{8}{3}$$
However, without another equation we cannot determine $x$ and $y$ (for example, both $x = 0, y=-3$ and $x = 4, y=0$ satisfy your system of equations. In fact, there will be infinitely many values of $x$ and $y$ satisfying the two equations).
answered Dec 5 '18 at 12:09
ODFODF
1,486510
$endgroup$
add a comment |
$begingroup$
Since you has three unknowns, you need three equations to be able to solve for all three variables (your system of linear equations is underdetermined).
However, what you can do is solve for $m$.
Multiplying the second equation through by $-9$ we get:
$$ 6x - 8y = -9m$$
Aha! This looks a lot like our first equation: we already know $6x - 8y = 24$. Putting these together gives $-9m = 24$ and hence $$m = -frac{8}{3}$$
However, without another equation we cannot determine $x$ and $y$ (for example, both $x = 0, y=-3$ and $x = 4, y=0$ satisfy your system of equations. In fact, there will be infinitely many values of $x$ and $y$ satisfying the two equations).
answered Dec 5 '18 at 12:09
ODFODF
1,486510
$endgroup$
add a comment |
$begingroup$
Since you has three unknowns, you need three equations to be able to solve for all three variables (your system of linear equations is underdetermined).
However, what you can do is solve for $m$.
Multiplying the second equation through by $-9$ we get:
$$ 6x - 8y = -9m$$
Aha! This looks a lot like our first equation: we already know $6x - 8y = 24$. Putting these together gives $-9m = 24$ and hence $$m = -frac{8}{3}$$
However, without another equation we cannot determine $x$ and $y$ (for example, both $x = 0, y=-3$ and $x = 4, y=0$ satisfy your system of equations. In fact, there will be infinitely many values of $x$ and $y$ satisfying the two equations).
answered Dec 5 '18 at 12:09
ODFODF
1,486510
$endgroup$
Since you has three unknowns, you need three equations to be able to solve for all three variables (your system of linear equations is underdetermined).
However, what you can do is solve for $m$.
Multiplying the second equation through by $-9$ we get:
$$ 6x - 8y = -9m$$
Aha! This looks a lot like our first equation: we already know $6x - 8y = 24$. Putting these together gives $-9m = 24$ and hence $$m = -frac{8}{3}$$
However, without another equation we cannot determine $x$ and $y$ (for example, both $x = 0, y=-3$ and $x = 4, y=0$ satisfy your system of equations. In fact, there will be infinitely many values of $x$ and $y$ satisfying the two equations).
answered Dec 5 '18 at 12:09
ODFODF
1,486510
answered Dec 5 '18 at 12:09
ODFODF
1,486510
answered Dec 5 '18 at 12:09
ODFODF
1,486510
answered Dec 5 '18 at 12:09
ODFODF
1,486510
1,486510
add a comment |
add a comment |
$begingroup$
From $frac{-2}{3}x+ frac{8}{9}y = m$ we get $-6x+8y=9m$. If we add this to the first equation we get $24+9m=0$, hence $m=frac{-8}{3}.$
It remains only one equation: $ 6x-8y=24$. There are infinitely many pairs $(x,y)$ which are solutions of this equation.
answered Dec 5 '18 at 12:09
FredFred
47.2k1849
$endgroup$
add a comment |
$begingroup$
From $frac{-2}{3}x+ frac{8}{9}y = m$ we get $-6x+8y=9m$. If we add this to the first equation we get $24+9m=0$, hence $m=frac{-8}{3}.$
It remains only one equation: $ 6x-8y=24$. There are infinitely many pairs $(x,y)$ which are solutions of this equation.
answered Dec 5 '18 at 12:09
FredFred
47.2k1849
$endgroup$
add a comment |
$begingroup$
From $frac{-2}{3}x+ frac{8}{9}y = m$ we get $-6x+8y=9m$. If we add this to the first equation we get $24+9m=0$, hence $m=frac{-8}{3}.$
It remains only one equation: $ 6x-8y=24$. There are infinitely many pairs $(x,y)$ which are solutions of this equation.
answered Dec 5 '18 at 12:09
FredFred
47.2k1849
$endgroup$
From $frac{-2}{3}x+ frac{8}{9}y = m$ we get $-6x+8y=9m$. If we add this to the first equation we get $24+9m=0$, hence $m=frac{-8}{3}.$
It remains only one equation: $ 6x-8y=24$. There are infinitely many pairs $(x,y)$ which are solutions of this equation.
answered Dec 5 '18 at 12:09
FredFred
47.2k1849
answered Dec 5 '18 at 12:09
FredFred
47.2k1849
answered Dec 5 '18 at 12:09
FredFred
47.2k1849
answered Dec 5 '18 at 12:09
FredFred
47.2k1849
47.2k1849
add a comment |
add a comment |
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4
$begingroup$
Two linear equations will not determine $3$ unknowns uniquely. Furthermore, how is "$-frac{8}{3}$" an answer when asking for the values of $3$ unknowns?
$endgroup$
– Christoph
Dec 5 '18 at 12:07
$begingroup$
Do u know Gaussian elimination?
$endgroup$
– Cloud JR
Dec 5 '18 at 12:07
$begingroup$
@christoph well asked lol
$endgroup$
– Cloud JR
Dec 5 '18 at 12:08