Solve the following system in three variables using linear combinations. 3x−2y+z=0 4x+y−3z=−9 9x−2y+2z=20

The solution of the system of equations is

x=2,y=7,z=8

Step-by-step explanation:

we have

3x-2y+z=0

isolate the variable z

z=2y-3x > equation A

4x+y-3z=-9 > equation B

9x-2y+2z=20 > equation C                  

substitute equation A in equation B and equation C

4x+y-3(2y-3x)=-9

4x+y-6y+9x=-9

13x-5y=-9 > equation D

9x-2y+2(2y-3x)=20

9x-2y+4y-6x=20

3x+2y=20 > equation E

Solve the system

13x-5y=-9 > equation D

3x+2y=20 > equation E

Multiply equation E by 2.5 both sides

2.5*(3x+2y)=20*2.5

7.5x+5y=50 > equation F

Adds equation D and equation F

13x-5y=-9

7.5x+5y=50

13x+7.5x=-9+50

20.5x=41

x=41/20.5

x=2

Find the value of y

substitute the value of x in the equation E

3(2)+2y=20

6+2y=20

2y=20-6

2y=14

y=7

Find the value of z

Substitute the value of x and the value of y in equation A

z=2y-3x

z=2(7)-3(2)

z=14-6

z=8

therefore

The solution of the system of equations is

x=2,y=7,z=8

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

~

isaiah walks 3 miles due north, turns and then 5 miles due east. how far is he from his starting point?

a^2 + b^2 = c^2

let c = distance from starting point.

(3)^2 + (5)^2 = c^2

9 + 25 = c^2

34 = c^2

sqrt{34} = sqrt{c^2}

sqrt{34} = c