# `2x - 2y - 6z = -4, -3x + 2y + 6z = 1, x - y - 5z = -3` Solve the system of linear equations and check any solutions algebraically.

EQ1:   `2x-2y-6z=-4`

EQ2:   `-3x + 2y + 6z=1`

EQ3:    `x-y-5z=-3`

To solve this system of equation, let's apply elimination method. In this method, a variable or variables should be removed in order to get the value of the other variable.

Let's eliminate y. To do so, add EQ1 and EQ2.

`2x-2y-6z=-4`

`+`      `-3x+2y+6z=1`

`----------------`

` `

`-x = -3`

Then, solve for x.

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

`x=3`

Isolate y again.Consider EQ1 and EQ3.

`2x-2y-6z=-4`

`x-y-5z=-3`

To be able to eliminate y, multiply EQ3 by -2. Then, add the two equations.

`2x-2y-6z=-4`

`+` `-2x+2y+10z=6`

`---------------`

`4z=2`

And, solve for z.

`(4z)/4=2/4`

`z=1/2`

Now that the values of the two variables are known, let's solve for the remaining variable. Let's plug-in x=3 and z=1/2 to EQ1.

`2x - 2y -6z = -4`

`2(3)-2y-6(1/2)=-4`

`6-2y-3=-4`

`3-2y=-4`

`3-3-2y=-4-3`

`-2y=-7`

(-2y)/(-2)=(-7)/(-2)

`y=7/2`

Therefore, the solution is  `(3, 7/2, 1/2)` .

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(1) `2x-2y-6z=-4`

(2) `-3x+2y+6z=1`

(3) `x-y-5z=-3`

``

Use equations (1) and (2) to eliminate the y and z variables.

(1) `2x-2y-6z=-4`

(2) `-3x+2y+6z=1`

___________________

`-x=-3`

`x=3`

``

Divide equation (1) by a -2. Use equations (1) and (3) to eliminate the x and y variables.

(1) `-x+y+3z=2`

(3) `x-y-5z=-3`

________________

`-2z=-1`

`z=1/2`

Solve for y by substituting the x and z variables in to equation (3).

(3) `x-y-5z=-3`

`(3)-y-5(1/2)=-3`

`3-y-(5/2)=-3`

`y=3-(5/2)+3`

`y=7/2`

The solution set for the given system of equations is `(3, 7/2, 1/2).`

` `

Check your answer by substituting the values for x, y, and z in the given equations.

(1) `2x-2y-6z=-4`

`2(3)-2(7/2)-6(1/2)=-4`

`6-7-3=-4`

`-4=-4`

(2) `-3x+2y+6z=1`

`-3(3)+2(7/2)+6(1/2)=1`

`-9+7+3=1`

`1=1`

(3) `x-y-5z=-3`

`3-(7/2)-5(1/2)=-3`

`3-(7/2)-(5/2)=-3`

`-3=-3`

Approved by eNotes Editorial Team