# Please can some one help I had 10 problems to start I was able to get six of them but was stumped on the first four problems. I also added more cash for the help, I figured 5 bucks for four...

Please can some one help I had 10 problems to start I was able to get six of them but was stumped on the first four problems.

I also added more cash for the help, I figured 5 bucks for four problems.

Much Thanks!

Solve system of three equations by elimination, please show work.

1)

−x − 5y − 5z = 2

4x − 5y + 4z = 19

x + 5y − z = −20

2)

−4x − 5y − z = 18

−2x − 5y − 2z = 12

−2x + 5y + 2z = 4

3)

−x − 5y + z = 17

−5x − 5y + 5z = 5

2x + 5y − 3z = −10

4)

4x + 4y + z = 24

2x − 4y + z = 0

5x − 4y − 5z = 12

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1)

-x - 5y - 5z = 2 (Eq. 1)

4x - 5y +4z = 19 (Eq.2)

x+ 5y - z = -20 (Eq.3)

Comparing the equations, one can notice that the equations (1) and (3) contain the variables x and y with the opposite coefficients. Thus, adding these equations together will eliminate both x and y in one step and enable to find z:

Eq. 1: -x -5y - 5z = 2

Eq.3 x + 5y - z = -20

-------------------------- (Add)

-6z = -18; from here z = 3

Plugging z = 3 into Eq.1 results in

-x -5y - 5*3 = 2

-x - 5y = 17 (Eq.4)

Plugging z = 3 into Eq.2 results in

4x - 5y + 4*3 = 19

4x - 5y = 7 (Eq.5)

Equations 4 and 5 make up the system of equations in two variables, x and y. Notice that subtracting one from another will result in eliminating y:

Eq.4: -x - 5y = 17

Eq.5: 4x - 5y = 7

-------------------(Subtract)

-5x = 10

x = -2

Now y can be found by plugging x = -2 into either equation 4 or 5:

Equation 4: -(-2) - 5y = 17

-5y = 15

y = -3

**The answer is x = -2, y = -3, z = 3**

2)

-4x - 5y - z = 18 (Eq. 1)

-2x - 5y - 2z = 12 (Eq.2)

-2x+ 5y + 2z = 4 (Eq.3)

Comparing the equations, one can notice that the Eq. 2 and Eq.3 contain the variables y and z with the opposite coefficients. Thus, adding these equations together will eliminate both y and z in one step, which will enable solving for x:

Eq. 2: -2x -5y - 2z = 12

Eq.3 -2x + 5y +2z = 4

-------------------------- (Add)

-4x = 16; from here x = -4

Plugging x = -4 into Eq.1 results in

-4(-4) -5y - z = 18

- 5y - z = 2 (Eq.4)

Plugging x = -4 into Eq.2 results in

-2(-4) - 5y -2z = 12

- 5y - 2z = 4 (Eq.5)

Equations 4 and 5 make up the system of equations in two variables, y and z. Notice that subtracting one from another will result in eliminating y:

Eq.4: - 5y - z = 2

Eq.5: - 5y - 2z = 4

-------------------(Subtract)

z = -2

Now y can be found by plugging z = -2 into either equation 4 or 5:

Equation 4: - 5y - (-2) = 2

-5y = 0

y = 0

**The answer is x = -4, y = 0, z = -2**

3)

-x - 5y + z = 17 (Eq.1)

-5x - 5y + 5z = 5 (Eq.2)

2x + 5y -3z = -10 (Eq.3)

Notice that Eq.2 has a common factor of 5 in all its terms, so it can be simplified by dividing both sides by 5:

Eq. 2: -5x -5y + 5z = 5

-x - y + z = 1 (new Eq.2)

Eq.1 -x - 5y + z = 17

Subtracting Eq. 1 from new Eq. 2 yields

4y = -16; from here y = -4

Now plug in y = -4 new Eq.2 and Eq.3:

new Eq.2: -x - y + z = 1

-x - (-4) + z = 1

-x + z = -3 (Eq. 4)

Eq.3: 2x + 5y - 3z = -10

2x + 5(-4) - 3z = -10

2x - 3z = 10 (Eq. 5)

To solve the system of equations 4 and 5 by elimination, Eq.4 needs to be multiplied by 2 first, in order to eliminate x:

Eq.4: -x + z = -3

-2x + 2z = -6

Eq.5: 2x - 3z = 10

Adding these equations together yields -z = 4, or z = -4.

Now x can be found from Eq.4:

-x + (-4) = -3

-x = 1, or x = -1.

**Answer: x = -1, y = -4, z = -4.**

4)

4x + 4y + z = 24 (Eq.1)

2x - 4y + z = 0 (Eq.2)

5x - 4y - 5z = 12 (Eq.3)

There is no combination of any two of these equations that will eliminate two variables at once, but we can eliminate y first from equations 1 and 2 and then from equations 1 and 3:

Eq.1: 4x + 4y + z = 24

Eq.2: 2x -4y + z = 0

-------------------------(Add)

6x + 2z = 24

Divide both sides by 2:

3x + z = 12 (Eq.4)

Eq.1: 4x + 4y + z = 24

Eq.3: 5x - 4y -5z = 12

------------------------(Add)

9x -4z = 36 (Eq.5)

Multiply Eq.4 by 4 to get 12x + 4z = 48

Eq. 5: 9x - 4z = 36

Adding these together yields 21x = 84

x = 4

From Eq. 4, 3x + z = 12

Plugging in x = 4 results in

3(4) + z = 12

z = 0

If x = 4 and z = 0, y can be found from any of the original 3 equations. Consider Eq.2:

2x - 4y + z = 0

Plug in z = 0 and x = 4:

2*4 - 4y + 0 = 0

8 - 4y = 0

8 = 4y

y = 2

**Answer: x = 4, y = 2, z = 0**

1)

(a) −x − 5y − 5z = 2

(b) 4x − 5y + 4z = 19

(c) x + 5y − z = −20

1. Add equations a and c: -6z = -18 -> **z = 3**

2. Add equations b and c: 5x + 3z = -1

3. Substitute in the known z value: 5x + 9 = -1 -> **x = -2**4. Plug known values into c: -2 + 5y - 3 = -20 ->

**y = -3**

**----------------------------------------------------------------**

2)

(a) −4x − 5y − z = 18

(b) −2x − 5y − 2z = 12

(c) −2x + 5y + 2z = 4

1. Add b and c: -4x = 16 -> **x = -4**2.

**Subtract b from a: -2x + z = 6**

3. Plug in known x value: 8 + z = 6 ->

**z = -2**

4. Plug in known x and z into c: 8 + 5y + -4 = 4 ->

**y = 0**

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3)

(a) −x − 5y + z = 17

(b) −5x − 5y + 5z = 5

(c) 2x + 5y − 3z = −10

1. Simplify b by factoring out 5: -x - y + z = 1

2. subtract this simplified b from a: -4y = 16 -> **y = -4**

3. Add a and c: x - 2z = 7

4. Add b and c: -3x + 2z = -5

5. Add these two new equations: -2x = 2 -> **x = -1**

6. Plug known values into the equation from step 1: 1 + 4 + z = 1 -> **z = -4**

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