Try this option:
if x+y=-9 is I (one)
and x+2y=-25 is II (two), then
if II-I, then (x+2y)-(x+y)=-25+9 ⇔ y=-16.
Answer:
When we have a quadratic equation:
a*x^2 + b*x + c = 0
There is something called the determinant, and this is:
D = b^2 - 4*a*c
If D < 0, then the we will have complex solutions.
In our case, we have
5*x^2 - 10*x + c = 0
Then the determinant is:
D = (-10)^2 - 4*5*c = 100 - 4*5*c
And we want this to be smaller than zero, then let's find the value of c such that the determinant is exactly zero:
D = 0 = 100 - 4*5*c
4*5*c = 100
20*c = 100
c = 100/20 = 5
As c is multiplicating the negative term in the equation, if c increases, then we will have that D < 0.
This means that c must be larger than 5 if we want to have complex solutions,
c > 5.
I can not represent this in your number line, but this would be represented with a white dot in the five, that extends infinitely to the right, something like the image below:
34.I
for this problem you just multiply the 20 by 2=40
and then you add 40+3=43
which will equal

35.G
if you divide

so G

36. C

so if you subtract

will equal

The correct answer is Sometimes

To solve x:
1. Substract 19 in both sides of the equation:

2. Multiply both sides of the ewuation by 10/9:

<h2>Then, the value of x is 60 (x=60)</h2>