Answer:
3x+y=8 and 3x+y=9
Step-by-step explanation:
Assume these two equations:
a1x+b1y+c1=0
a2x+b2y+c2=0
If (a1/a2) = (b1/b2) ≠ (c1/c2) than those two would have no solution.
In these case, take a look at the last option:

Re-write 'em as a standard form:

So (a1/a2) = (b1/b2) ≠ (c1/c2) is true here:

And they do not have any collision point
Answer:
or 
Step-by-step explanation:
You need to complete the square before you can take the square root of both sides.

Subtract 10 from both sides.

To complete the square, you need to add the square of half of the x-term coefficient to both sides.
The x-term coefficient is 7. Half of that is 7/2. Square it to get 49/4. Now we add 49/4 to both sides of the equation.



Now we use the square root property, if
, then



or
or 
or 
Answer:
(1, -1)
Step-by-step explanation:
we isolate y in the first equation
2x+y=1
-2x. -2x
y=1-2x
and now we fill y into second equation
4x-2(1-2x)=6
4x-2+4x=6
8x-2=6
+2 +2
8x=8
/8 /8
x=1
and now to find y we insert x
2(1)+y=1
2+y=1
-2. -2
y= -1
hopes this helps
Answer:
24.84
Step-by-step explanation:
use Pythagoras theorem
h=√p²+b² = √(19)²+(16)² = √617 =24.84