Solve the system of 3 equations by elimination. State as an ordered triple

Question

Solve the system by elimination

x+y+2z=2

3x+4y-z=25

4x-y+3z=-7

Answer 1

x + y + 2z = 2

3x + 4y - z = 25

4x - y + 3z = -7

Multiply the first equation by -3 and add the result to the second equation.

Multiply the first equation by -4 and add the result to the third equation.

x + y + 2z = 2

y - 7z = 19

-5y - 5z = -15

Multiply the second equation by 5 and add the result to the third equation.

x + y + 2z = 2

y - 7z = 19

-40z = 80

So, from the third equation, we see that z = -2.

Back substitute into the second equation: y - 7(-2) = 19. So, y = 5.

Back substitute into the first equation: x + 5 + 2(-2) = 2. So, x = 1.

Solution: (1, 5, -2)

Answer 2

Answer 3

x + y + 2z = 2

3x +4y-z = 25

4x -y +3z = -7'

(x,y,z) = (1, 5, -2) is the solution.

if you could graph in 3 dimensions the intersection point would then be = (1,5,-2). although it's very difficult to sketch 3D graphs

3 equations 3 unknowns. solve by elimination

add the 1st & 3rd to eliminate y

5x +5z =-5, divide by 5

x +z = -1

multiply the 1st by -4 and add the 2nd to also eliminate y

-4x -4y -8z = -8

3x +4y-z = 25

-x -9z = 17 add x+z=-1 to eliminate x and solve for z

-8z = 16, divide by -8

z = 16/-8 = -2

x = -1 -z =-1 +2 = 1

x =1, z = -2

y = 2-2z-x = 2+4 -1 = 5

(x,y,z) = (1,5,-2) check the answer, plug it into each of the 3 equations

x +y +2z = 2

3x +4y-z = 25

3 +20 +2 which does = 25,

other posted answers all have (1,5, -2), the same solution

check that answer, plug it into the 3 equations

x+y +2z =2

1 +5+2(-2) = 2, it does satisfy the 1st equation

3x +4y-z =25

3 +20 +2 = 25. it does satisfy the 2nd equation

4x -y +3z = -7

4 -5 -6 does =-7

it satisfies all 3 equations

Answer 4

x + y + 2z = 2 (1^st equation)

3x + 4y - z = 25 (2^nd equation)

4x - y + 3z = -7 (3^rd equation)

The idea behind the elimination method is to cancel variables to reduce the number of variables in the system which makes it easier to solve for each variable. Firstly, we can take any 2 equations first and try to cancel one of the variable. Lets take the first and second equation:

x + y + 2z = 2

2[3x + 4y - z = 25] Lets multiply 2^nd equation by 2 to get -2z so we can cancel out the z

_______________

x + y + 2z = 2

6x + 8y - 2z =50

___________ Now combine them

7x + 9y = 52 (Lets call this 4^th Equation)

Now, we can take any other combination of equations except for equation 1 and 2 since we already chose those at first. Though we do know that we have to make them cancel out z again so it lines up with equation 4 and we are able to eliminate another variable.

So, lets take equations 1 and 3:

3[x + y + 2z = 2] Multiply 1^st equation by 3

-2[4x - y + 3z = -7] Multiply 2^nd equation by -2

_______________ It cancels out the z's

3x + 3y + 6z = 6

-8x + 2y - 6z = 14

________________ Combine them

-5x + 5y = 20 (Lets call this 5^th equation)

We can now take the 4^th and 5^th equations and try to cancel out one of the variables so that we are able to solve for a variable:

5[7x + 9y = 52] Multiply it by 5 to cancel out x

7[-5x + 5y = 20] Multiply it by 7 to cancel out x

_____________

35x + 45y = 260

-35x + 35y = 140

_____________ Combine them

80y = 400 Divide by 80 on both sides to solve for y

y = 5

Now that we got the value of y, we can substitute it into the 4^th equation or the 5^th equation to solve for x. Lets choose the 4^th equation:

7x + 9y = 52 Plug in the value of y=5

7x + 9(5) = 52 Simplify

7x + 45 = 52 Subtract by 45 on both sides

7x = 7 Divide by 7 on both sides to isolate x

x = 1

We now have the value of x and y. We can take those values and plug them into either the 1^st, 2^nd or the 3^rd equation to get the value of z. Lets take the 1^st equation since it seems to be the most simple one to solve:

x + y + 2z = 2 Plug in x = 1 and y = 5

1 + 5 + 2z = 2 Add like terms

6 + 2z = 2 Subtract 6 on both sides

2z = -4 Divide by 2 on both sides to isolate z

z = -2

We have finally solved for all variables.

x = 1; y = 5; z = -2

Answer 5

We are given:

x + y + 2z = 2

3x + 4y - z = 25

4x - y + 3z = -7

First, I’ll eliminate z.

From the first equation:

x + y + 2z = 2

→ 2z = 2 - x - y

→ z = (2 - x - y)/2

Now substitute z into the other two equations.

Substitute into the 2nd equation:

3x + 4y - ((2 - x - y)/2) = 25

Multiply both sides by 2:

6x + 8y - (2 - x - y) = 50

Simplify:

6x + 8y - 2 + x + y = 50

→ 7x + 9y = 52 …(1)

Substitute into the 3rd equation:

4x - y + 3((2 - x - y)/2) = -7

Multiply both sides by 2:

8x - 2y + 3(2 - x - y) = -14

Simplify:

8x - 2y + 6 - 3x - 3y = -14

→ 5x - 5y + 6 = -14

→ 5x - 5y = -20

→ x - y = -4 …(2)

Now solve the two equations (1) and (2):

From (2): x = y - 4

Substitute into (1):

7(y - 4) + 9y = 52

7y - 28 + 9y = 52

→ 16y = 80

→ y = 5

Now plug y = 5 into x = y - 4:

x = 1

Find z using the first equation:

x + y + 2z = 2

1 + 5 + 2z = 2

→ 2z = -4

→ z = -2

Final Answer:

(x, y, z) = (1, 5, -2)

Answer 6

x+y+2z=2

3x+4y-z=25

4x-y+3z=-7

Pick any two equations to start with. You'll elimination one variable with them. I will start with the first and last equation. Add then together to eliminate the y.

x+y+2z=2

4x-y+3z=-7

5x +5z=-5

Pick any other combination of equations, but know that you have to eliminate the SAME variable again. So I will multiply the bottom equation by 4 and add it to the 2nd equation, to again eliminate the y.

3x+4y-z=25

16x-4y+12z=-28

19x + 11z = -3

Take the two new equations and work those as a system. You can do either elimination or substitution, but I would go with substitution this time, because it's easy to take the 5x +5z=-5 and solve for x (or do the same with y).

5x +5z=-5

5x = -5z - 5

x =- z -1

Substitute the x = -z-1 into the other equation.

19(-z-1)+11z=-3

-19z-19+11z=-3

-8z = 16

z = -2.

Substitute z = -2 into x =- z -1

x =-(-2)-1 = 1

Substitute both of those into x+y+2z=2 to get

1+y+2(-2)=2

-3+y=2

y=5

Solution: (1, 5, -2)