Which equation exemplifies the commutative property of addition???

14, 16, and 20 using elimination method showing work. Thanks so much

Nady [450]

14) x=0, y=3, z=-2

Solution Set (0,3,-2)

16) x=1, y=1 and z=1

Solution set = (1,1,1)

20) x = -263/31, y=164/31 ,z=122/31

Solution set (-263/31, 164/31 ,122/31)

Step-by-step explanation:

14)

$x-y+2z=-7\\y+z=1\\x=2y+3z$

Rearranging and solving:

$x-y+2z=-7\,\,\,eq(1)\\y+z=1\,\,\,eq(2)\\x-2y-3z=0\,\,\,eq(3)$

Eliminate y:

Adding eq(1) and eq(2)

$x-y+2z=-7\,\,\,eq(1)\\ 0x+y+z=1\,\,\,eq(2)\\-------\\x+3z=-6\,\,\,eq(4)$

Multiply eq(2) with 2 and add with eq(3)

$0x+2y+2z=2\,\,\,eq(2)\\\\x-2y-3z=0\,\,\,eq(3)\\--------\\x-z=2\,\,\,eq(5)$

Eliminate x:

Subtract eq(4) and eq(5)

$x+3z=-6\,\,\,eq(4)\\x-z=2\,\,\,eq(5)\\-\,\,\,+\,\,\,\,\,\,-\\---------\\4z=-8\\z= -2$

So, value of z = -2

Now putting value of z in eq(2)

$y+z=1\\y+(-2)=1\\y-2=1\\y=1+2\\y=3$

So, value of y = 3

Now, putting value of z and y in eq(1)

$x-y+2z=-7\\x-(3)+2(-2)=-7\\x-3-4=-7\\x-7=-7\\x=-7+7\\x=0$

So, value of x = 0

So, x=0, y=3, z=-2

S.S(0,3,-2)

16)

$3x-y+z=3\\\x+y+2z=4\\x+2y+z=4$

Let:

$3x-y+z=3\,\,\,eq(1)\\x+y+2z=4\,\,\,eq(2)\\x+2y+z=4\,\,\,eq(3)$

Eliminating y:

Adding eq(1) and (2)

$3x-y+z=3\,\,\,eq(1)\\x+y+2z=4\,\,\,eq(2)\\---------\\4x+3z=7\,\,\,eq(4)$

Multiply eq(1) by 2 and add with eq(3)

$6x-2y+2z=6\,\,\,eq(1)\\x+2y+z=4\,\,\,eq(3)\\---------\\7x+3z=10\,\,\,eq(5)$

Now eliminating z in eq(4) and eq(5) to find value of x

Subtracting eq(4) and eq(5)

$4x+3z=7\,\,\,eq(4)\\7x+3z=10\,\,\,eq(5)\\-\,\,\,-\,\,\,\,\,\,\,\,\,\,-\\-----------\\-3x=-3\\x=-3/-3\\x=1$

So, value of x = 1

Putting value of x in eq(4) to find value of x:

$4x+3z=7\\4(1)+3z=7\\4+3z=7\\3z=7-4\\z=3/3\\z=1$

So, value of z = 1

Putting value of x and z in eq(2) to find value of y:

$x+y+2z=4\\1+y+2(1)=4\\1+y+2=4\\y+3=4\\y=4-3\\y=1$

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

Solution set = (1,1,1)

20)

$x+4y-5z=-7\\3x+2y+2z=-7\\2x+y+5z=8$

Let:

$x+4y-5z=-7\,\,\,eq(1)\\3x+2y+2z=-7\,\,\,eq(2)\\2x+y+5z=8\,\,\,eq(3)$

Solving:

Eliminating z :

Adding eq(1) and eq(3)

$x+4y-5z=-7\,\,\,eq(1)\\2x+y+5z=8\,\,\,eq(3)\\---------\\3x+5y=1\,\,\,eq(4)$

Multiply eq(1) with 2 and eq(2) with 5 and add:

$2x+8y-10z=-14\,\,\,eq(1)\\15x+10y+10z=-35\,\,\,eq(2)\\----------\\17x+18y=-49\,\,\,eq(5)$

Eliminate y:

Multiply eq(4) with 18 and eq(5) with 5 and subtract:

$54x+90y=18\\85x+90y=-245\\-\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\\-------\\-31x=158\\x=-\frac{263}{31}$

So, value of x = -263/31

Putting value of x in eq(4)

$3x+5y=1\\3(-\frac{263}{31})+5y=1\\-\frac{789}{31}+5y=1 \\5y=1+\frac{789}{31}\\5y=\frac{820}{31}\\y=\frac{820}{31*5}\\y=\frac{164}{31}$

Now putting x = -263/31 and y=164/31 in eq(1) and finding z:

We get z=122/31

So, x = -263/31, y=164/31 ,z=122/31

Solution set (-263/31, 164/31 ,122/31)

Keywords: Solving system of Equations

Learn more about Solving system of Equations at:

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Step-by-step explanation:

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the sum of three numbers is twenty. the second number is four times the first and the sum of the first and third is eight. what

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Let's call the three numbers x, y, and z.

The numbers are 3, 12, and 5.
x + y + z = 20
y = 4x
x + z = 8
We can rewrite the first equation because y is 4x, so
x + 4x + z = 20
5x + z = 20
And now we can use simultaneous equation rules to solve

5x + z = 20
x + z = 8

5x + z = 20
5x + 5z = 40
(subtract)
-4z + -20
÷ -4
z = 5

x + z = 8
x + 5 = 8
x = 3

y = 4x
y = 4 × 3
y = 12

I hope this helps!

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A room has an area of 60 square meters and a perimeter of 32 meters. What are the length and width of the room?

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