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3 years ago

Solve the system by substitution

Mathematics

2 answers:

IceJOKER [234]3 years ago

6 0

Answer:

In Point Form: (x,y,z)= (3,6,-1)

In Equation Form: x=3, y=6, z=-1

Step-by-step explanation:

 1) Solve for x 

2x+2z=4 (Subtract 2z from both sides)

-2z -2z

2x= 4-2z (Divide 2 from both sides)

/2 /2

x=2-z

2) Substitute 2-z for x in

-x-y-z=-8

-(2-z)-y-z=-8

-2+z-y-z=-8 (Combine Like Terms)

-2-y=-8 (Add 2 to both sides)

+2 +2

-y=-6 (Divide -1 from both sides)

/-1 /-1

y=6

3) Substitute 2-z for x & 6 for y in

-4x+4y+5z+=7

-4(2-z)+4(6)+5z=7 (Get rid of parentheses)

-8+4z+24+5z=7 (Combine Like Terms)

9z+16=7 (Subtract 16 from both sides)

-16 -16

9z+=-9 (Divide 9 from both sides)

/9 /9

z=-1

4) Now find x by substituting -1 for z in

x=2-z

x=2-(-1)

x=2+1

x=3

Therefore the answer is:

In Point Form: (x,y,z)= (3,6,-1)

In Equation Form: x=3, y=6, z=-1

2. Answers:

In Point Form: (x,y,z)= (1,1,0)

In Equation Form: x=1, y=1, z=0

Step-by-step explanation:

-2x + 2y + 3z = 0 (1)

-2x - y + z = -3 (2)

2x + 3y + 3z = 5 (3)

Solve (1) and (2)

Multiply 2 by 2

-2x + 2y + 3z = 0

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

-6x + 5 z = -6 (4)

Solve (2) and (3)

Multiply 2 by 3

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

2x + 3y + 3z = 5

-4x + 6z = -4 (5)

Solve (4) and (5)

Multiply (4) by 2 and (5) by -3

-12x + 10 z = -12

12x - 18z = 12

-6z = 0

z = 0

Then

-4x + 6(0) = -4

-4x = -4

x = -4/-4

x = 1

Finally

-2(1) - y + (0) = -3

-2 - y = -3

-y = -3 + 2

y = 1

Therefore the answer is:

In Point Form: (x,y,z)= (1,1,0)

In Equation Form: x=1, y=1, z=0

Andru [333]3 years ago

4 0

A solution on the photos

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