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OpenStudy (anonymous):

What is the solution of the system of equations? {3x+2y+z=7} {5x+5y+4z=3} {3x+2y+3z=1}

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

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??

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 ?

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

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

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.

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

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

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

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

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

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.

OpenStudy (vishweshshrimali5):

So first eliminate x from this pair of equations.

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