Clifland

2021-09-01

The system below has one solution: $x=1,y=-1,$ and $z=2$.
Solve the systems provided by
(a) Equations 1 and 2,
(b) Equations 1 and 3, and
(c) Equations 2 and 3.
(d) How many solutions does each of these systems have?
$4x-2y+5z=16$ — Equation 1
$x+y=0$ Equation2
$-x-3y+2z=6$ Equation3

lamanocornudaW

Consider a system containing equation 1 and 2.

$4x-2y+5z=16$ . . . (1)

$x+y=0$ . . . (2)

From equation (2)

$y=-x$

Plug this in equation (1)

$4x-2\left(-x\right)+5z=16$

$4x+2x+5z=16$

$6x+5z=16$

$x=\frac{16}{6}-\frac{5x}{6}$

$x=\frac{8}{3}-\frac{5x}{6}$

The solution of the system is $\left(x,y,z\right)=\left(x,-x,\frac{8}{3}-\frac{5x}{6}\right)$

This system has infinitely many solutions.

Step 2: (b)

Consider a system containing equation 1 and 3.

$4x-2y+5z=16$ . . . (1)

$-x-3y+2z=6$ . . . (3)

From equation (3)

$x=6+3y-2z$

Plug the value of x in equation (1)

$4\left(6+3y-2z\right)-2y+5z=16$

$24+12y-8z-2y+5z=16$

$10y-3z=-8$

$10y+8=3z$

$z=\frac{10y}{3}+\frac{8}{3}$

The solution of the given system is $\left(6-3y+2z,y,10\frac{y}{3}+\frac{8}{3}\right)$, y can be any real number.

So this system have infinitely many solutions.

Step 3

Consider a system containing equation 2 and 3.

$x+y=0$ . . . (1)

$-x-3y+2z=6$ . . . (3)

From (1)

$y=-x$

$-x-3\left(-x\right)+2z=6$

$-x+3x+2z=6$

$2x+2z=6$

$x+z=3$

$z=3-x$

The given system of equations have solutions $\left(x,y,z\right)=\left(x.-x,3-x\right)$ as x can be any real number.

Therefore, this system have infinitely many solutions.

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