Question

solve each system
x+2/6 - y+4/3+z/2=0
x+1/2+y-1/2-z/4=9/2
x-5/4+y+1/3+z-2/2=19/4

Answer 1

I was taught to use the elimination method with these ones.
\[\frac{ x+2 }{ 6 }-\frac{ y+4 }{ 3 }+\frac{ z }{ 2 }=0\]
Get rid of the fractions by multipling the denominator by it's HCF (Highest Common Factor).

Answer 2

Do that for all of the three equations. Tell me what you get when you did that.

Answer 3

So for the first one 6 is highest so I would get
X+2+2(y+4)+3z=0 ?

Answer 4

-2(y+4)*

Answer 5

remember that minus sign. You don't wave a magic wand and make it disappear.

Answer 6

Right!!

Answer 7

But that's good.

Answer 8

So what would you get for the second one.

Answer 9

?*

Answer 10

4 would be highest so
2(x+1)+2(y-1)-z=18 ?

Answer 11

How come you got a minus sign for z this time?

Answer 12

Wrote second one wrong should be
x+1/2+y-1/2-z/4=9/2

Answer 13

Ah okay. Then well done correct.

Answer 14

Do the same thing for the third equation.

Answer 15

After that just expand all three equations and collect the like terms. Tell me what you get for each equation.

Answer 16

? Lol kinda lost now

Answer 17

Show me the equatons that had no fractions.

Answer 18

equations*

Answer 19

\[x+2-2(y+4)+3z=0\]
\[x+2-2y-8+3z=0\]
\[x-2y+3z=-2+8\]
\[x-2y+3z=6[1]\]
There's your first equation.
I placed a number inside some brackets to indicate that this is your first equation. (ie. [1])

Answer 20

Are you able to put the other two equations in that format please? Show me what you get after doing that.

Answer 21

@ninab731

Answer 22

2x+2y-z=17 ?

Answer 23

Close but not quite.

Answer 24

Expand all the brackets
\[2x+2+2y-2-z=18\]
\[2x+2y-z=18-2+2\]
\[2x+2y-z=18[2]\]
Second equation done.

Answer 25

Try and do the third equation.

Answer 26

3x+4y+6z=47

Answer 27

Oops I goofed

Answer 28

Not quite.

Answer 29

Multiply by 12 and expand.
\[3x-15+4y+4+6z-12=57\]
\[3x+4y+6z=57+15-4+12\]
\[3x+4y+6z=34[3]\]

Answer 30

Third equation done.

Answer 31

Now Let me get all three equations listed here.
\[x-2y+3z=6[1]\]
\[2x+2y-z=18[2]\]
\[3x+4y+6z=32[3]\]

Answer 32

What you want is to get rid of one variable and then after that, you try and get rid of the last variable.

Answer 33

The easiest variable to get rid of is the y. Because all you hav e to do is multipl by 2 in the first 2 equations and then once you've done that, you can subtract each equation by another equation in order to get rid of the y.

Answer 34

So my first step would be multiplying the first equation by 2.
It's the same equation but the numbers are just twice as big.
\[[1]\times 2\]
\[2(x-2y+3z)=2(6)\]

Answer 35

Do you get me so far?

Answer 36

Got it

Answer 37

Okay now we expand. Even though this equation is practically the same as the first, we will rename it and say that this is the fourth equation [4].
\[2x-4y+3z=12[4]\]

Answer 38

Okay, I multiplied by 2 so then I turn the -2y into -4y. The 4 in the -4y is what we want because there's a 4y in equation [3].

Answer 39

SO if we add equation [4] with equation [3] we can ELIMINATE the 4y and -4y so then there will only be two variables. You get me so far?

Answer 40

Let me show you.

Answer 41

\[[4]+[3]\]
\[(2x−4y+3z)+(3x+4y+6z)=(12)+(32)\]
We collect like terms and then I will call this equation [5].
\[5x+9z=44[5]\]

Answer 42

Do you see what I did there?

Answer 43

@ninab731

Answer 44

Ya

Answer 45

Okay. Try and do the same for equation 2.

Answer 46

Multiply both sides by 2 to get a 4y. and then subtract the equation with equation 3.

Answer 47

@ninab731

Answer 48

If then I multiply by -2 I get -4x-4y+2z=-36

Answer 49

Don't multiply by a negative. I told you multiply by 2.

Answer 50

4x+4y-2z=36

Answer 51

Okay. Equation three is this right? And we want to get rid of the 4y. \[3x+4y+6z=32[3]\]
Would be subtract equation two with equation three or would be add them both together?

Answer 52

Would we*

Answer 53

Subtract

Answer 54

Good!! Excellent.

Answer 55

X-8z=2 ?

Answer 56

Well done.

Answer 57

Perfect!

Answer 58

Now we have two equations with two variables.
\[5x+9z=44[5]\]
I just made this equation [6].
\[x-8z=2[6]\]

Answer 59

Now we want to get rid of another variable.

Answer 60

We can use substitution because we can already make x the subject in equation 6 without dividing or anything like that.

Answer 61

So
\[x=8z+2\]
Right?

Answer 62

Ah ok

Answer 63

sub it into equation 5.

Answer 64

and you can find z

Answer 65

And then once you do that. sub that z-value into either equation 5 or 6.

Answer 66

40z+10+9z=44
49z+10=44

Answer 67

Yep good job.

Answer 68

You can find x by putting what you get for z into one of the two equations that have two variables (ie. [5] or [6])

Answer 69

But then 44-10=34 and when u divide 49 get decimal

Answer 70

Do you have the correct answers that say differently?

Answer 71

You sub in the fraction.

Answer 72

Don't round up as a decimal. use the fraction.

Answer 73

Suppose to be (4,8,6)

Answer 74

rewrite all three equations. because you said equation two had a minus. Can you rewrite the three equations you were given because you must of done something.

Answer 75

Equations are
x+2/6 - y+4/3 + z/2 =0
x+1/2 + y-1/2 - z/4 =9/2
x-5/4 + y+1/3 + z-2/2 =19/4

Answer 76

\[4x+4y-2z=36[1]\]
\[2x-4y+6z=12[2]\]
\[3x+4y+6z=80[3]\]
\[[1]-[3]\]
\[(4x+4y-2z)-(3x+4y+6z)=(36)-(80)\]
\[x-8z=-44[4]\]
\[[2]+[3]\]
\[(2x-4y+6z)+(3x+4y+6z)=(12)+(80)
\[5x+12z=92[5]\]
\[x=8z-44\]
\[5(8z-44)+12z=92\]
\[40z-220+12z=92\]
\[52z=312\]
\[z=6\]

Answer 77

if three are unknowns n two given syatem then syatem has no solution