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3 years ago

Solve the system. x – 4y – z = 21 6x – 3y – z = –4 –x +2y –5z = 19 Answer: x = , y = , z =

Mathematics

1 answer:

FrozenT [24]3 years ago

7 0

Consider the given equations:

$-x+2y-5z = 19$ (Equation 1)

$6x-3y-z= -4$ (Equation 2)

$x-4y-z=21$ (Equation 3)

Adding equations 1 and 3, we get

$-x+2y-5z+x-4y-z = 19+21$

$-2y-6z = 40$

So, we get $-y -3z = 20$ (Equation 4)

Multiplying equation 3 by '6', we get

$6x - 24y-6z = 126$ (Equation 5)

Subtracting equation 5 from equation 2, we get

$(6x-3y-z)-(6x - 24y-6z) = -4-126$

$6x-3y-z-6x + 24y+6z) = -4-126$

$21y+5z = -130$ (Equation 6)

Multiplying equation 4 by '21' and adding it to equation 6, we get

$-21y-63z+21y+5z = 420-130$

$-58z = 290$

So, z = -5

Since, $-y-3z = 20$ $-y+15=20$ $y=-5$

So, y=-5

Now, $x-4y-z=21$ $x-4(-5)-(-5)=21$ $x+20+5=21$ $x+25=21$ $x= -4$

So, x = -4

Therefore, x = -4, y= -5 and z= -5 are the solutions to the given equations.

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Add 3 and 7. then multiply by 3.

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Answer:

Step-by-step explanation:

First, add 3 and 7

3 + 7 =10

The multiply by 3

10*3= 30

Hope this helps!

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3 years ago

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1.Solve the equation. Check for extraneous solutions.

elena55 [62]

Okay so I did the math.

1. Two solutions were found :
z=-1
z=2/3
Step 1 :Rearrange this Absolute Value Equation

Absolute value equalitiy entered
|2z-3| = 4z-1

Step 2 :Clear the Absolute Value Bars

Clear the absolute-value bars by splitting the equation into its two cases, one for the Positive case and the other for the Negative case.

The Absolute Value term is |2z-3|

For the Negative case we'll use -(2z-3)

For the Positive case we'll use (2z-3)

Step 3 :Solve the Negative Case

      -(2z-3) = 4z-1

     Multiply
      -2z+3 = 4z-1

     Rearrange and Add up
      -6z = -4

     Divide both sides by 6
      -z = -(2/3)

     Multiply both sides by (-1)
      z = (2/3)
     Which is the solution for the Negative Case

Step 4 :Solve the Positive Case

      (2z-3) = 4z-1

     Rearrange and Add up
      -2z = 2

     Divide both sides by 2
      -z = 1

     Multiply both sides by (-1)
      z = -1
     Which is the solution for the Positive Case
Step  5  :Wrap up the solution

z=2/3
 z=-1

2. -6 < x < 26/3

Step 1 :Rearrange this Absolute Value Inequality

Absolute value inequalitiy entered
|3x-4|+5 < 27

Another term is moved / added to the right hand side.

|3x-4| < 22

Step 2 :Clear the Absolute Value Bars

Clear the absolute-value bars by splitting the equation into its two cases, one for the Positive case and the other for the Negative case.

The Absolute Value term is |3x-4|

For the Negative case we'll use -(3x-4)

For the Positive case we'll use (3x-4)

Step 3 :Solve the Negative Case

-(3x-4) < 22

 Multiply
 -3x+4 < 22

 Rearrange and Add up
 -3x < 18

 Divide both sides by 3
 -x < 6

 Multiply both sides by (-1)
 Remember to flip the inequality sign
 x > -6
 Which is the solution for the Negative Case

Step 4 :Solve the Positive Case

(3x-4) < 22

 Rearrange and Add up
 3x < 26

 Divide both sides by 3
 x < (26/3)

 Which is the solution for the Positive Case

Step 5 :Wrap up the solution

-6 < x < 26/3

Solution in Interval Notation

(-6,26/3)

HOPE THIS HELPS :D

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3 years ago

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3 years ago

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Use undetermined coefficient to determine the solution of:y"-3y'+2y=2x+ex+2xex+4e3x

Kitty [74]

First check the characteristic solution: the characteristic equation for this DE is

r ² - 3r + 2 = (r - 2) (r - 1) = 0

with roots r = 2 and r = 1, so the characteristic solution is

y (char.) = C₁ exp(2x) + C₂ exp(x)

For the ansatz particular solution, we might first try

y (part.) = (ax + b) + (cx + d) exp(x) + e exp(3x)

where ax + b corresponds to the 2x term on the right side, (cx + d) exp(x) corresponds to (1 + 2x) exp(x), and e exp(3x) corresponds to 4 exp(3x).

However, exp(x) is already accounted for in the characteristic solution, we multiply the second group by x :

y (part.) = (ax + b) + (cx ² + dx) exp(x) + e exp(3x)

Now take the derivatives of y (part.), substitute them into the DE, and solve for the coefficients.

y' (part.) = a + (2cx + d) exp(x) + (cx ² + dx) exp(x) + 3e exp(3x)

… = a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)

y'' (part.) = (2cx + 2c + d) exp(x) + (cx ² + (2c + d)x + d) exp(x) + 9e exp(3x)

… = (cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

Substituting every relevant expression and simplifying reduces the equation to

(cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

… - 3 [a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)]

… +2 [(ax + b) + (cx ² + dx) exp(x) + e exp(3x)]

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

… … …

2ax - 3a + 2b + (-2cx + 2c - d) exp(x) + 2e exp(3x)

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

Then, equating coefficients of corresponding terms on both sides, we have the system of equations,

x : 2a = 2

1 : -3a + 2b = 0

exp(x) : 2c - d = 1

x exp(x) : -2c = 2

exp(3x) : 2e = 4

Solving the system gives

a = 1, b = 3/2, c = -1, d = -3, e = 2

Then the general solution to the DE is

y(x) = C₁ exp(2x) + C₂ exp(x) + x + 3/2 - (x ² + 3x) exp(x) + 2 exp(3x)

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2 years ago

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olga55 [171]

Answer:

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2 years ago

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