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3 years ago

Solve the system 2x+2y+z=-2 -x-2y+2z=-5 2x+4y+z=0

1 answer:

8 0

The values of (x,y,z) are (3,-1,-2) , if the given are equations $2x+2y+z=- 2$ , $-x-2y+2z=-5$ and $2x+4y+z=0$ .

Step-by-step explanation:

The given is,

$2x+2y+z=- 2$ .......................................(1)

$-x-2y+2z=-5$ ......................................(2)

$2x+4y+z=0$ .........................................(3)

Step:1

Equation (2) is multiplied by -1 ( Eqn(2) × -1 )

$x+2y-2z=5$ .............................(4)

Subtract the equation (1) and (4)

$2x+2y+z=- 2$

$x+2y-2z=5$

( - )

$(2x-x)+(2y-2y)+(z+2z)=(-2-5)$

$x+3z=-7$ ......................(5)

Step:2

Equation (2) is multiplied by -2 ( Eqn(2) × -2)

$2x+4y-4z=10$ ........................(6)

Subtract equation (6) and (3),

$2x+4y-4z=10$

$2x+4y+z=0$

( - )

$(2x-2x)+(4y-4y)+(-4z-z)= (10-0)$

$-5z=10$

$z = - \frac{10}{5}$

z = -2

From the equation (5),

$x+3z=-7$

$x+(3)(-2)=-7$

$x = -7+6$

x = -1

From equation (1),

$2x+2y+z=- 2$

Substitute x and z values,

$(2)(-1)+2y+(-2)=-2$

$2y - 4=2$

$2y=4-2$

$2y=2$

$y=\frac{2}{2}$

y = 1

Step:3

Check for solution,

$-x-2y+2z=-5$

Substitute x,y and z values,

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

$1-2-4=-5$

-5 = -5

Result:

The values of (x,y,z) are (3,-1,-2) , if the given are equations $2x+2y+z=- 2$ , $-x-2y+2z=-5$ and $2x+4y+z=0$ .

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PLEASE ANSWER

Anni [7]

x is the independent variable.

y is the dependent variable.

-3 is the rate of change (slope).

-7 is the initial value.

Step-by-step explanation:

The form of the linear relation is y = m x + b, where

m is the rate of change (slope)
b is the initial value(value of y at x = 0)
x is the independent variable
y is the dependent variable

∵ The equation of the line is y = -3 x - 7

- Compare it with the form above

∴ m = -3

∵ m is the rate of change

∴ The rate of change is -3

∴ b = -7

∵ b is the initial value

∴ The initial value is -7

∵ y depends on x

∴ x is the independent variable

∴ y is the dependent variable

x is the independent variable.

y is the dependent variable.

-3 is the rate of change (slope)

-7 is the initial value.

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What is the initial velocity of the boiling water? H=-16t^2+170t

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3 years ago

Kwans The Kenningy account balance was $20 his ending balance is zero dollars what integer represents the changing in his accoun

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The integer number that represents the changing in his account balance from beginning to end is of -20.

<h3>Which numbers belong to the set of integer numbers?</h3>

Non-decimal numbers, either positive or negative numbers, also zero, belong to the set, which can be represented by:

Z = {..., -3, -2, -1, 0, 1, 2, 3, ...}

A change can be represented by the <u>final value subtracted by the initial value</u>. For this problem, we have that:

The initial balance is of $20.

The final balance is of $0.

Hence the change in the balance is:

0 - 20 = -20.

And the integer is -20.

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1 year ago

mrs lim had some money she spent 4/7 of it on necklace and 1/6 of her remainig money on a bracelet she then had $990 left how mu

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Answer:

she spent $1782

Step-by-step explanation:

first we have to find out her total money then minus 990

to get that 990 /5 then multipliedby 6 is 1188

1188 over 3 then times by 7 is 2772

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Help!! 50 points and brainliest!

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Answer:

Second choice:

$x=2t$

$y=4t^2+4t-3$

Fifth choice:

$x=t+1$

$y=t^2+4t$

Step-by-step explanation:

Let's look at choice 1.

$x=t+1$

$y=t^2+2t$

I'm going to subtract 1 on both sides for the first equation giving me $x-1=t$ . I will replace the $t$ in the second equation with this substitution from equation 1.

$y=(x-1)^2+2(x-1)$

Expand using the distributive property and the identity $(u+v)^2=u^2+2uv+v^2$ :

$y=(x^2-2x+1)+(2x-2)$

$y=x^2+(-2x+2x)+(1-2)$

$y=x^2+0+-1$

$y=x^2$

So this not the desired result.

Let's look at choice 2.

$x=2t$

$y=4t^2+4t-3$

Solve the first equation for $t$ by dividing both sides by 2:

$t=\frac{x}{2}$ .

Let's plug this into equation 2:

$y=4(\frac{x}{2})^2+4(\frac{x}{2})-3$

$y=4(\frac{x^2}{4})+2x-3$

$y=x^2+2x-3$

This is the desired result.

Choice 3:

$x=t-3$

$y=t^2+2t$

Solve the first equation for $t$ by adding 3 on both sides:

$x+3=t$ .

Plug into second equation:

$y=(x+3)^2+2(x+3)$

Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:

$y=(x^2+6x+9)+(2x+6)$

$y=(x^2)+(6x+2x)+(9+6)$

$y=x^2+8x+15$

Not the desired result.

Choice 4:

$x=t^2$

$y=2t-3$

I'm going to solve the bottom equation for $t$ since I don't want to deal with square roots.

Add 3 on both sides:

$y+3=2t$

Divide both sides by 2:

$\frac{y+3}{2}=t$

Plug into equation 1:

$x=(\frac{y+3}{2})^2$

This is not the desired result because the $y$ variable will be squared now instead of the $x$ variable.

Choice 5:

$x=t+1$

$y=t^2+4t$

Solve the first equation for $t$ by subtracting 1 on both sides:

$x-1=t$ .

Plug into equation 2:

$y=(x-1)^2+4(x-1)$

Distribute and use the binomial square identity used earlier:

$y=(x^2-2x+1)+(4x-4)$

$y=(x^2)+(-2x+4x)+(1-4)$

$y=x^2+2x+-3$

$y=x^2+2x-3$ .

This is the desired result.

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3 years ago

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