Chapter 11.1, Problem 44AYU

To determine

To find: The solution of the given system of equations.

Answer to Problem 44AYU

Solution:

  $x=\frac{56}{13},y=-\frac{7}{13},z=\frac{35}{13}$

Explanation of Solution

Given:

  $2x+y-3z=0$

  $-2x+2y+z=-7$

  $3x-4y-3z=7$

Calculation:

The system of equations is:

  $2x+y-3z=0$

  $-2x+2y+z=-7$

  $3x-4y-3z=7$

Multiplying the second equation by 3 and then adding to the third equation we get,

  $3(-2x+2y+z)+(3x-4y-3z)=-21+7$

  $-6x+6y+3z+3x-4y-3z=-14$

  $-3x+2y=-14$

  $3x-2y=14$$3x=2y+14$

  $x=\frac{1}{3}(2y+14)$

Substituting this in the first equation we get,

  $\frac{2}{3}(2y+14)+y-3z=0$

  $\frac{4}{3}y+\frac{28}{3}+y-3z=0$

  $3z=\frac{7}{3}y+\frac{28}{3}$

  $z=\frac{1}{9}(7y+28)$

Substituting these expressions for $x$ and $z$ in the second equation we get,

  $-\frac{2}{3}(2y+14)+2y+\frac{1}{9}(7y+28)=-7$

  $-\frac{4}{3}y-\frac{28}{3}+2y+\frac{7}{9}y+\frac{28}{9}=-7$

  $\frac{13}{9}y=-7+\frac{28}{2}-\frac{28}{9}$

  $\frac{13}{9}y=-\frac{7}{9}$

  $y=-\frac{7}{13}$

Therefore we get,

  $x=\frac{1}{2}(2y+14)$$x=\frac{1}{3}\left[2\left(-\frac{7}{12}\right)+14\right]$

  $x=\frac{1}{2}\left(-\frac{14}{12}+14\right)$

  $x=\frac{1}{2}\times \frac{169}{13}$

  $x=\frac{56}{13}$

And

  $z=\frac{1}{9}\left[7\left(-\frac{7}{13}\right)+28\right]$

  $z=\frac{1}{9}\left(-\frac{49}{12}+28\right)$

  $z=\frac{1}{9}\times \frac{315}{13}$

  $z=\frac{35}{12}$

Thus $x=\frac{56}{13},y=-\frac{7}{13},z=\frac{35}{13}$ is the solution of the given system of equations.