What is the solution of the system of equations?
{3x+2y+z=7}
{5x+5y+4z=3}
{3x+2y+3z=1}
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OpenStudy (vishweshshrimali5):
Do you know the matrix method of solving such linear equations ?
OpenStudy (vishweshshrimali5):
@yelahn you there ?
OpenStudy (anonymous):
i didnt think i had to use a matrix method. it just says to solve the system...
OpenStudy (vishweshshrimali5):
Its okay.
OpenStudy (vishweshshrimali5):
Lets solve it in a very easy way
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OpenStudy (anonymous):
okay. where do i start?
OpenStudy (vishweshshrimali5):
First of all select any two equations.
OpenStudy (anonymous):
euation 1 &2
OpenStudy (vishweshshrimali5):
Ok
OpenStudy (anonymous):
do i add the two equations or subtract??
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OpenStudy (vishweshshrimali5):
That now depends.
From these 2 equations you have to remove at least one variable by solving them.
OpenStudy (anonymous):
soo remove x?
OpenStudy (vishweshshrimali5):
Okay
OpenStudy (vishweshshrimali5):
Now, you can see that the coefficient of x in eq. 1 and 2 are not the same.
So, we are going to multiply eq. 1 and 2 by numbers such that the coefficients of x will become same.
OpenStudy (vishweshshrimali5):
For example:
If I multiply eq. 1 by 5 on both sides and eq. 2 by 3 on both sides, both will have 15 x. Right ?
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OpenStudy (anonymous):
right
OpenStudy (anonymous):
what do you mean by both sides??
OpenStudy (vishweshshrimali5):
I meant multiply 5 in LHS and in RHS
OpenStudy (anonymous):
woahhh slow down. whats LHS and RHS
OpenStudy (vishweshshrimali5):
LHS = left hand side of = sign
RHS = right hand side of = sign
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OpenStudy (anonymous):
so the equations would be
15x+2y+z=35
15x+5y+4z=15
???
OpenStudy (vishweshshrimali5):
Did you multiply 5 by 2y and z in eq. 1?
OpenStudy (anonymous):
no
OpenStudy (vishweshshrimali5):
What I meant by multiplying 5 on both sides is this:
(3x+2y+z) = 7
=> 5*(3x+2y+z) = 5*7
OpenStudy (anonymous):
okay so
15x+10y+5z=35
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OpenStudy (vishweshshrimali5):
Yes
OpenStudy (anonymous):
and eq2 is
15x+15y+12z=9
OpenStudy (vishweshshrimali5):
Very nice
OpenStudy (anonymous):
and then 15x cancels out?
OpenStudy (vishweshshrimali5):
Yes subtract both equations and 15x would cancel out.
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OpenStudy (anonymous):
-5y+-7z=26
OpenStudy (vishweshshrimali5):
Yelahn, this : -+ or +- is not correct in maths, use instead -(+) or +(-). Ok?
-5y + (-7x) = 26 okay ?
OpenStudy (anonymous):
oh. my mistake. i meant-5y-7z=26
OpenStudy (vishweshshrimali5):
Its ok
OpenStudy (anonymous):
now what do i do
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OpenStudy (vishweshshrimali5):
Now, you are going to find out the value of y in terms of z from this equation.
OpenStudy (anonymous):
how/ thats seems impossible
OpenStudy (vishweshshrimali5):
See:
You had this equation:
-5y - 7z = 26
=> -5y = 26 + 7z
=> 5y = -26 - 7z
=> y = 1/5 * (-26-7z)
OpenStudy (vishweshshrimali5):
Did you get it ?
OpenStudy (anonymous):
no i cant figure it out
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OpenStudy (vishweshshrimali5):
Which step ?
OpenStudy (anonymous):
y=1/5*(-26-7z)
OpenStudy (vishweshshrimali5):
What I did was that I divided by 5 on both sides . Ok ?
OpenStudy (anonymous):
i got
y=-5.2-1.4z
OpenStudy (vishweshshrimali5):
Correct
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OpenStudy (vishweshshrimali5):
Now, similarly select any other pair of equations.
OpenStudy (anonymous):
should this one be simplified more? how can y=-5.2-1.4z
OpenStudy (anonymous):
y=-6.6
OpenStudy (vishweshshrimali5):
How ?
OpenStudy (anonymous):
no nvm. ignore that
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OpenStudy (vishweshshrimali5):
k
OpenStudy (vishweshshrimali5):
So, now select any other pair of equations.
OpenStudy (anonymous):
2 and 3
OpenStudy (vishweshshrimali5):
Ok
Now from these equation eliminate x.
OpenStudy (anonymous):
why are you eliminating x again
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OpenStudy (vishweshshrimali5):
I am doing this because after eliminating x, I am going to ask you to derive another relationship between y and z and then equate this relationship to the previous one.