The system below has one solution: x = 1, y = −1, and z = 2.Solve the systems provided by

Clifland

Clifland

Answered question

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?
4x2y+5z=16 — Equation 1
x+y=0 Equation2
x3y+2z=6 Equation3

Answer & Explanation

lamanocornudaW

lamanocornudaW

Skilled2021-09-02Added 85 answers

Consider a system containing equation 1 and 2.

4x2y+5z=16 . . . (1)

x+y=0 . . . (2)

From equation (2)

y=x

Plug this in equation (1)

4x2(x)+5z=16

4x+2x+5z=16

6x+5z=16

x=1665x6

x=835x6

The solution of the system is (x,y,z)=(x,x,835x6)

This system has infinitely many solutions.

Step 2: (b)

Consider a system containing equation 1 and 3.

4x2y+5z=16 . . . (1)

x3y+2z=6 . . . (3)

From equation (3)

x=6+3y2z

Plug the value of x in equation (1)

4(6+3y2z)2y+5z=16

24+12y8z2y+5z=16

10y3z=8

10y+8=3z

z=10y3+83

The solution of the given system is (63y+2z,y,10y3+83), 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)

x3y+2z=6 . . . (3)

From (1)

y=x

x3(x)+2z=6

x+3x+2z=6

2x+2z=6

x+z=3

z=3x

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

Therefore, this system have infinitely many solutions.

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