Simultaneous Equations?

Question

Can u explain how to solve the following three three simultaneous equations?

2x-y+z=2
x-y-z=1
-x+2y+3z=1

Answer 1

Multiply the second equation by -2:
2x-y+z=2
-2x+2y+2z=-2
==========
y+3z=0

x-y-z=1
-x+2y+3=1
========
y+2z=2

y+3z=0
y+2z=2
======
3y+9z=0
-3y-6z=-6
========
3z=-6
z=-2

3y+9(-2)=0
3y-18=0
3y=18
y=6

x-6-(-2)=1
x-6+2=1
x-4=1
x=5

The solution set is (5,6,-2)

Check:
2(5)-(6)-2=2
10-6-2=2
10-8=2
2=2

5-6-(-2)=1
-1+2=1
1=1

-5+2(6)+3(-2)=1
-5+12-6=1
-5+6=1
1=1

Answer 2

p = a + 5 164 = 8a + 9p replace p = a + 5 into the backside equation - - - - - - - - - - - - - - - - - - - 164 = 8a + 9p 164 = 8a + 9(a + 5) 164 = 8a + 9a + 40 5 The distributive belongings on the suitable part 164 = 17a + 40 5 amassing like words 164 - 40 5 = 17a + 40 5 - 40 5 Subtracting - 40 5 from the two factors of the equation 119 = 17a 119/17 = 17a/17 Dividing the two factors of the equation with the aid of 17 7 = a Insert the a cost into the equation - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - fixing for p P = a + 5 p = 7 + 5 p = 12 Insert the p fee into the equation 164 = 8a + 9p 164 = 8(7) + 9(12) 164 = fifty six +108 164 = 164 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - the respond is: a = 7 p = 12

Answer 3

i'll give each equation a letter to identify it.

2x-y+z=2 (A)
x-y-z=1 (B)
-x+2y+3x=1 (C)

2x-y+z=2 (A)
-x+y+z=-1 (-B)
x+2z=1 (A+(-B)=D)

2x-2y-2z=2 (2B)
-x+2y+3z=1 (C)
x+z=3 (2B+C=E)

-x-2z=-1 (-D)
x+z=3 (E)
z=-2

x-2=3 (E with z as -2)
x=5

5-y+2=1 (B with x as 5 and z as -2)
y=6

So x=5, y=6 and z=-2

Answer 4

This is a bit random but I think it will get there.
Add the first two:
3x-2y=3 ........ i
so 3x-3=2y
Use in the third one:
-x+(3x-3)+3z=1
so 2x+3z=4 ....... ii
so z=1/3(4-2x)
Use in first one:
2x-y+1/3(4-2x)=2
or 6x-3y+4-2x=6
or 4x-3y=2 ....... iii
Multiply i by 4 and iii by 3 and subtract them from each other:
12x-8y=12 (i.e. i x 4) and
12x-9y=6 (i.e. iii x 3) ... now subtract ...
y= 6 ..... now use iii
4x-18=2
x= 5 ... now use ii
10+3z=4
z= -2

Answer 5

x=5,y=6,z=(-2)

Answer 6

2x+y+z = 2 .Do the same thing to both sides.
i.e. -z
2x+y = 2-z
-y
2x = 2-z-y
/2
x = (2-z-y)/2


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x-y-z=1
-y-z
x=1-y-z

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-x+2y+3z=1
-1
-x+2y+3z-1=0
+x
2y+3z-1=x
or x=2y+3z-1

Answer 7

ok im not all that gr8 with math but this might help:
2x-y+z=2
2x-y-2=-z
z=-2x+y+2
y=-2x+2-z
x=y-z
x =2x+2-z +2x-y-2

ok thats all i could figure out
hope it helped

Answer 8

add all three

Answer 9

For me YES....

Answer 10

no