Answer:
(arranged from top to bottom)
System #3, where x=6
System #1, where x=4
System #7, where x=3
System #5, where x=2
System #2, where x=1
Step-by-step explanation:
System #1: x=4

To solve, start by isolating your first equation for y.

Now, plug this value of y into your second equation.

System #2: x=1

Isolate your second equation for y.

Plug this value of y into your first equation.

System #3: x=6

Isolate your first equation for y.

Plug this value of y into your second equation.

System #4: all real numbers (not included in your diagram)

Plug your value of y into your second equation.

<em>all real numbers are solutions</em>
System #5: x=2

Isolate your second equation for y.

Plug in your value of y to your first equation.

System #6: no solution (not included in your diagram)

Isolate your first equation for y.

Plug your value of y into your second equation.

<em>no solution</em>
System #7: x=3

Plug your value of y into your second equation.

One expression could be three x to the power of two. Let me know if that's right :)
Answer:
As the lines are not intersecting nor parallel, they must be skew.
Step-by-step explanation:
Question is incomplete, we consider the nearest match available online.
Parametric equations of two lines are:
L₁ : x=4t+2 , y = 3 , z =-t+1
L₂: x=2s+2 , y= 2s+5 , z = s+1
If lines are parallel then parametric coordinates must be equal scalar multiple of each other which s not true here.

If lines are intersecting then parametric coordinates must be equal for some value of t and s.

Hence the lines are not intersecting nor parallel, they must be skew.
4/26 which simplified to 2/13
so 2/13 is the answer