Help me the question about algebra

➡️do not inconsequential and must with and must be accompanied by steps for the process

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9514 1404 393

  (x, y, z) = (8√7/105, 2√7/21, 4√7/35)

Step-by-step explanation:

The algebra is straightforward, if tedious.

We'll define X = x^2, Y = y^2, Z = z^2, -1/16 = a, -1/25 = b, -1/36 = c. Then the equations become ...

  √X = √(Y +a) +√(Z +a) . . . . [eq1]

  √Y = √(Z +b) +√(X +b) . . . . [eq2]

  √Z = √(X +c) +√(Y +c) . . . . [eq3]

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As with any radical equation, you square both sides, isolate the remaining radical, then square both sides again. Each squaring introduces extraneous solutions.

The equations are symmetrical, so we can do the tedious stuff on the first one, then make the appropriate substitutions to see the other two equations.

  X = Y +a + 2√((Y+a)(Z+a)) +Z +a

  X - Y - Z -2a = 2√((Y+a)(Z+a)) . . . . isolate the remaining radical

  (X -Y -Z -2a)^2 = 4(Y+a)(Z+a) . . . . square both sides again

  X^2 +Y^2 +Z^2 +4a^2 -2XY -2XZ +2YZ -4aX +4aY +4aZ = 4YZ +4aY +4aZ +4a^2

Subtracting the right side gives ...

  X^2 +Y^2 +Z^2 -2XY -2XZ -2YZ -4aX = 0 . . . . [eq4]

Then the remaining equations are ...

  X^2 +Y^2 +Z^2 -2XY -2XZ -2YZ -4bY = 0 . . . . [eq5]

  X^2 +Y^2 +Z^2 -2XY -2XZ -2YZ -4cZ = 0 . . . . [eq6]

Subtracting [eq6] from [eq5] gives a relation between Y and Z:

  -4bY +4cZ = 0

  Z = bY/c . . . . [eq7]

Substituting this into [eq4], we have ...

  X^2 +Y^2 +(bY/c)^2 -2XY -2X(bY/c) -2Y(bY/c) -4aX = 0

  X^2 +Y^2(1 -2b/c +(b/c)^2) -2XY(1 +b/c) -4aX = 0 . . . . [eq8]

Subtracting [eq5] from [eq4] gives a relation between X and Y:

  -4aX +4bY = 0

  Y = aX/b . . . . [eq9]

Substituting this into [eq8], we have ...

  X^2 +(aX/b(1 -b/c))^2 -2X(aX/b)(1 +b/c) -4aX = 0

Dividing by X* gives a linear equation in X:

  X(1 +(a/b(1 -b/c))^2 -2a/b(1+b/c)) -4a = 0

  X = 4a/(1 +(a/b(1 -b/c))^2 -2a/b(1+b/c))

Substituting the above values for a, b, c, we have ...

  X = 64/1575

  x = √X = √(64/1575) = 8/(15√7)

  Y = X(25/16)

  y = x(5/4) = 2/(3√7)

  Z = Y(36/25)

  z = y(6/5) = 4/(5√7)

The solution is ...

* At this point the equation is of the form pX^2 +qX = 0 or X(pX +q) = 0. One solution is X=0, which is extraneous. The solution of interest is X = -q/p.

ruth is 6 years old

step-by-step explanation:

6 + 24 = 30,   so 6/30 = .2  

if you do 1/5 it also equals .2, so that is the answer.

answer: the answer is ann