How do you solve the system of equations 3x-2y=-13x−2y=−1 and 6y + 3 = 7x6y+3=7x?

2 Answers
Aswin K.
Aug 10, 2017

x=-3x=−3

y=-4y=−4

Explanation:

3x-2y=-13x−2y=−1

Let this be equation number 11

3x-2y=-1->13x−2y=−1→1

6y+3=7x6y+3=7x

3=7x-6y3=7x−6y

7x-6y=37x−6y=3

Let this be equation number 22

7x-6y=3->27x−6y=3→2

Multiply Equation number 11 with 33
That is, =>(3x-2y)*3=-1*3⇒(3x−2y)⋅3=−1⋅3

=>3x*3-2y*3=-3⇒3x⋅3−2y⋅3=−3

=>9x-6y=-3⇒9x−6y=−3

Let this be equation number 33

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

Now subtract equation number 22 with equation number 33
That is,

(7x-6y=3)-(9x-6y=-3),(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-6y-9x+6y=3+37x−6y−9x+6y=3+3

7x-9x-6y+6y=67x−9x−6y+6y=6

7x-9x=67x−9x=6

-2x=6−2x=6

x=6/-2x=6−2

x=-6/2x=−62

x=-3x=−3

Put the value of xx in equation number 11 ((you can take any of the given equation))
Then equation number 11 becomes

3*(-3)-2y=-13⋅(−3)−2y=−1

=>-9-2y=-1⇒−9−2y=−1

=>-2y=-1+9⇒−2y=−1+9

=>-2y=8⇒−2y=8

y=8/-2y=8−2

y=-8/2y=−82

y=-4y=−4

x=-3x=−3 && y=-4y=−4

Narad T.
Aug 10, 2017

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))

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