Question

What is the solution to the system of equations?

-3x-3y+2z=-7

z=1

-2x-3y+z=-6


A.(2, 1, –1)


B.(2, 1, 1)


C.(2, –1, 1)


D.(–2, 1, 1)

Answer 1

Hello!

The answer is:

The correct option is B.(2, 1, 1)

Why?

We can solve the system of equations by using the reduction method. The reduction method consists of reducing the variables in order to be able to calculate the other variables to finally calculate all the variables.

We are given the equations:

I.

`-3x-3y+2z=-7`

II.

`z=1`

II.

`-2x-3y+z=-6`

Since the second equation is already solved, let's work with the first and third one, so, calculating we have:

`\left \{ {{-3x-3y+2z=-7} \atop {-2x-3y+z=-6}} \right.`

Now, multiplying the first equation by -1 in order to reduce the variable "y", we have:

`\left \{ {{3x+3y-2z=7} \atop {-2x-3y+z=-6}} \right\\\\x-z=1`

Then, substituting "z" into the obtained equation:

`x-1=1\\x=1+1=2`

Now, substituting "x" and "z" into the first equation, we have:

`-3x-3y+2z=-7`

`-3*(2)-3y+2*(1)=-7`

`-6-3y+2=-7`

`-3y-4=-7`

`-3y=-7+4`

`-3y=-3`

`y=(-3)/(-3)=1`

Hence, we have that the solutions are:

`x=2\\y=1\\z=1`

So, the correct option is B.(2, 1, 1)

Have a nice day!

Answer 2

B(2,1,1)

Explanation:

Given:

-3x-3y+2z=-7

z=1

-2x-3y+z=-6

Let -3x-3y+2z=-7 be equation i, z=1 be equation ii and -2x-3y+z=-6 be equation iii

Solving the system of simultaneous equation by substituting value of z from equation ii to i , we get:

-3x-3y+2=-7

-3x-3y=-7-2

-3x-3y=-9 -------iv

Solving the system of simultaneous equation by substituting value of z from equation ii to iii, we get:

-2x-3y+1=-6

-2x-3y=-6-1

-2x-3y=-7

re-arranging the above equation, we get

3y=-2x+7

substituting value of 3y from above in equation iv, we get

-3x-(-2x+7)=-9

-3x+2x-7=-9

-x=-9+7

-x=-2

x=2

Now putting x=1 from above in equation v, we get

3y=-2(2) +7

3y=-4+7

3y=3

y=3/3

y=1

Hence the solution of system of given equations is (2,1,1) !