Question

How do you solve the system of equations `3x-2y=-1` and `6y + 3 = 7x`?

Answer 1

`x=-3`

`y=-4`

Explanation:

`3x-2y=-1`

Let this be equation number `1`

`3x-2y=-1->1`

`6y+3=7x`

`3=7x-6y`

`7x-6y=3`

Let this be equation number `2`

`7x-6y=3->2`

Multiply Equation number `1` with `3`
That is, `=>(3x-2y)*3=-1*3`

`=>3x*3-2y*3=-3`

`=>9x-6y=-3`

Let this be equation number `3`

#9x-6y=-3->3

Now subtract equation number `2` with equation number `3`
That is,

`(7x-6y=3)-(9x-6y=-3),`we have to subtract the right-hand side and left-hand side separately
So it becomes,

`7x-6y-(9x-6y)=3-(-3)`

`7x-6y-9x+6y=3+3`

`7x-9x-6y+6y=6`

`7x-9x=6`

`-2x=6`

`x=6/-2`

`x=-6/2`

`x=-3`

Put the value of `x` in equation number `1` `(`you can take any of the given equation`)`
Then equation number `1` becomes

`3*(-3)-2y=-1`

`=>-9-2y=-1`

`=>-2y=-1+9`

`=>-2y=8`

`y=8/-2`

`y=-8/2`

`y=-4`

`x=-3` `&` `y=-4`

Answer 2

The solution is `((x),(y))=((-3),(-4))`

Explanation:

We perform the Gauss Jordan elimination with the augmented matrix

`M=((3,-2,:,-1),(7,-6,:,3))`

We perform the row operations

`R2larrR2-7/3R1`, `=>`, `((3,-2,:,-1),(0,-4/3,:,16/3))`

`R2larr(R2)/(-3/4)`, `=>`, `((3,-2,:,-1),(0,1,:,-4))`

`R1larrR1+2R2`, `=>`, `((3,0,:,-9),(0,1,:,-4))`

`R1larr(R1)/3`, `=>`, `((1,0,:,-3),(0,1,:,-4))`