Solving Systems of Equations in Three Variables?

Question

Yeah Well i dont get how to do this one particular one
x - 4y + 3z = -27
2x + 2y - 3z = 22
4z = -16

Answer 1

x-4y+3z= -27.....(1)
2x+2y-3z=22......(2)
4z= -16........(3)
From eqn (3) we get
z= -16/4= -4
Plugging the value of z in eqn1 and 2 ,we get
x-4y-12= -27
or,x-4y= -15.........(4),and
2x+2y+12=22
or2x+2y=22-12=10.....(5)
Multiplying eqn 5 by2 and adding it to eqn 4
4x+4y=20
x-4y=-15
(adding) 5x=5 or,x=1
Plugging the value of x in eqn 5,we get
2+2y=10
or2y=10-2=8
or y=8/2=4
therefore,x=1,y=4 and z= -4 ans

Answer 2

x = 1
y = 4
z = -4

First solve the third one: 4z = -16 so z = -4.

Then substitute for "z" in each of the other two and combine like terms to get:

x - 4y + 3*(-4) = -27 and 2x + 2y - 3*(-4) = 22 and so:
x - 4y = -15 and 2x + 2y = 10 or x + y = 5.

Then solve the second one for x and substitute into the first one:

x + y = 5 gives: y = 5 - x and so the first equation becomes:

x - 4*(5 - x) = -15 and so: x - 20 + 4x = -15 which gives: 5x = 5 and so: x = 1.

Then substitute back into the interim second equation: x + y = 5 so: 1 + y = 5 and so: y = 4.

Answer 3

Since the last equation is almost already solved for z, then just divide this one by 4 to get z = -4 Now substitute this value into the other two equations, and you will have an easily solvable system in two variables.

x - 4y +3(-4) = -27 which becomes x - 4y = -15
2x + 2y -3(-4) = 22 which becomes 2x + 2y = 10

Now this system is easily solved for y by multiplying the first equation by -2 to get -2x + 8y = 30
Add to the second equation to get 10y = 40, and divide by 10 to get y = 4
Find x by substituting this value of y into the 2nd equation to get 2x + 8 = 10, so 2x = 2 and x = 1

So x = 1, y = 4 and z = -4

Answer 4

get z alone by dividing 4z and -16 so z=-4
plug -4 in for z in one of the other equations and solve for either x or y.
ex: 2x+2y+12=22
2x+2y= 10
2x= 10-2y
x=5-y
then u plug 5-y in for x and -4 for z in the last equation that u haven't worked on yet.
ex: 5-y-4y-12= -27
-5y-7= -27
-5y= -20
y=4
put 4 in for y and -4 in for z in the equation u solved second.
ex: 2x+8+12=22
2x+20= 22
2x= 2
x=1

Answer 5

since 4z = -16, z = -4. that makes the other equations

x - 4y + 3(-4) = -27
2x + 2y - 3(-4) = 22, which become

x - 4y = -15
2x + 2y = 10

x - 4y = -15
4x + 4y = 20

5x = 5
x = 1

1 - 4y = -15
-4y = -16
y = 4

Answer 6

x - 4y + 3z = -27 -------- (a)
2x + 2y - 3z = 22 ------- (b)
4z = -16 ------------------- (c)

first of all,dun panic =)

from (c) z =-16/4 => -4 -sub into (a) & (b)
(a)
=>x - 4y + 3(-4) = -27
=> x - 4y -12 = -27
=> x - 4y = -15
=>x =4y-15 -------- (d)

(b)
=>2x + 2y - 3(-4) = 22
=>2x + 2y + 12 = 22
=>2x + 2y = 10 ------- (e)

sub (d) into (e)

2(4y-15) + 2y = 10
8y-30+2y=10
10y=10+30
y=40/10 => 4 ---(f)

sub (f) into (d) or (e)
x =4(4)-15 or 2x + 2(4) = 10
x=16-15 or 2x=10-8
x = 1 or x= (10-8)/2 =>x=1

x=1, y=4 & z=(-4)

Answer 7

you're able to desire to isolate between the variables and plug it into considered one of the different equations. Then repeat to get an equation with only one variable. case in point, isolate 'b' interior the 1st equation: b = (-a million/7)*(9a+30) then you definately can plug this expression into the 2d equation so: (-8/7)*(9a+30) +5c = eleven. This equation is now in terms of 'a' and 'c' only like the 0.33. Isolate between the variables and replace to resolve. case in point, the 0.33 equation might properly be arranged to examine: c = (a million/10)*(seventy 4+3a). in case you plug that 'c' into the recent 2d equation, you get: (-8/7)*(9a+30) +(a million/2)*(seventy 4+3a) = eleven. resolve for 'a' and then back replace on your different variables.

Answer 8

z = -4
;;
x + y =5
x - 4y = -15
::

x = 1
y = 4
z = -4

Answer 9

i like the first answer