# `2x + y + 3z = 1, 2x + 6y + 8z = 3, 6x + 8y + 18z = 5` Solve the system of linear equations and check any solutions algebraically. (1)   `2x+y+3z=1`

(2)    `2x+6y+8z=3`

(3)    `6x+8y+18z=5`

` `

Using Equations (1) and (2) eliminate the x. Multiply equation (2) by -1.

` 2x+1y+3z=1`

`-2x-6y-8z=-3`

----------------------------

` -5y-5z=-2`

(4) `-5y-5z=-2`

Using Equations (2) and (3) eliminate the x. Multiply equation (2) by -3.

`-6x-18y-24z=-9`

` 6x+8y+18z=5`

--------------------------------

` -10y-6z=-4`

(5) `-10y-6z=-4`

Using equations (4) and (5) eliminate the y. Multiply equation 4 by -2.

` 10y+10z=4`

`-10y-6z=-4`

-----------------------

`4z=0`

`z=0`

` `

Use equation (4) to solve for y.

`-5y-5z=-2`

`-5y-5(0)=-2`

`y=2/5`

Use equation (1) to solve for x.

`2x+y+3z=1`

`2x+2/5+3(0)=1`

`x=3/10`

The solution set is (3/10, 2/5, 0).

You may check your answers by plugging in the x, y, z values into the original equations.

(1)  `2(x)+y+3z=1`

`2(3/10)+(2/5)+3(0)=1`

`6/10+2/5=1`

`6/10+4/10=1`

`1=1`

(2) `2x+6y+8z=3`

`2(3/10)+6(2/5)+8(0)=3`

`6/10+12/5=3`

`6/10+24/10=3`

`30/10=3`

`3=3`

(3)  `6x+8y+18z=5`

`6(3/10)+8(2/5)+18(0)=5`

`18/10+16/5=5`

`18/10+32/10=5`

`50/10=5`

`5=5`

Approved by eNotes Editorial Team EQ1:  `2x+y+3z=1`

EQ2:  `2x+6y+8z=3`

EQ3: `6x+8y+18z=5`

To solve this system of equation, let's apply elimination method. In elimination method, a variable/variables should be removed in order to come up with another equation.

In this system of equation, let's eliminate variable x using EQ1 and EQ2.

`2x + y + 3z=1`

`2x +6y + 8z=3`

To do so, multiply EQ2 by -1.

`-1 * (2x + 6y+8z)=3*(-1)`

`-2x - 6y-8z=-3`

`2x + y + 3z = 1`

`+`     `-2x - 6y -8z=-3`

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

`-5y - 5z=-2 `        Let this be EQ4.

Eliminate x variable again to come up with another equation that has y and z only. Use EQ1 and EQ3.

`2x+y+3z=1`

`6x+8y+18z=5`

So multiply EQ1 by -3.

`-3(2x+y+3z)=1*(-3)`

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

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

`+`       `6x + 8y+18z=5`

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

`5y + 9z = 2 `          Let this be EQ5.

So the two new equations are:

EQ4:  `-5y-5z = -2`

EQ5:   `5y+ 9z = 2`

Since there two equations still contain more than two variables, let's eliminate another variable. Let's remove the y. To do so, add these two equations.

`-5y - 5z =-2`

`+`       `5y +9z=2`

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

`4z=0`

Then, solve for z.

`4z=0`

`z=0/4`

`z=0`

Then, plug-in this to either EQ4 or EQ5 to get the value of y.

`5y+9z=2`

`5y+9(0)=2`

`5y=2`

`y=2/5`

Now that the values of y and z are known, solve for x. To do so, plug-in these two values to either EQ1, EQ2 or EQ3.

`2x+y+3z=1`

`2x+2/5+3*0=1`

`2x+2/5=1`

`2x=1-2/5`

`2x=3/5`

`x=(3/5)/2`

`x=3/10`

Therefore, the solution is `(3/10, 2/5,0)` .

-6x-3y-9z=-3

Approved by eNotes Editorial Team You may use reduction method, hence, you may subtract the first equation from the second, such that:

`5y + 5z = 2`

You may multiply the first equation by -3 and then you may add it to the third:

`-6x - 3y - 9z + 6x + 8y + 18z = -3+5`

`5y + 9z = 2 => 5y = 2 - 9z`

You may replace` 2 - 9z` ` for 5y` in the equation `5y + 5z = 2` , such that:

`2 - 9z + 5z = 2 => -4z = 0 => z = 0 => 5y = 2 => y = 2/5`

Replace` 2/5` for y and 0 for z in equation `2x + y + 3z = 1` , such that:

`2x + 2/5 + 0 = 1 => 2x = -2/5 => x = -1/5`

Hence, evaluating the solutions to the system yields `x = -1/5, y = 2/5, z = 0.`

Posted on