Well if you're wanting to use substitution, you first have to end up with one term on either side of the equation. Use the second one as it's easiest.
So: x-2y=11, so find -2y as there is a -2y in the first equation.
then that becomes -2y=11-x. then sub that into equation 1, and you get:
-x+11-x=-13, which equals to -2x+11=-13, which is -2x=-24, so therefore
x=12. then chuck the x into any of the equations to find what y equals.
hope this helps!
Answer:
Sally is not right
Step-by-step explanation:
Given the two sequences which have their respective 
terms as following:
Sequence A. 
Sequence B. 
As per Sally, there exists only one number which is in both the sequences.
To find:
Whether Sally is correct or not.
Solution:
For Sally to be correct, we need to put the terms of the respective sequences as equal and let us verify that.
When we talk about terms, 
here is a whole number not a fractional number.
But as per the statement as stated by Sally is a fractional number, only then the two sequences can have a number which is in the both sequences.
Therefore, no number can be in both the sequences A and B.
Hence, Sally is not right.
(0,30), (1,20)
(20-30)/(1-0) = -10/1 = -10 slope
We know y intercept is 30
Equation: y = -10x + 30
Answer:
Eq: (x+a/2)²+(y+1)²=(a²-8)/4
Center: O(-a/2, -1)
Radius: r=0.5×sqrt(a²-8)
Mandatory: a>2×sqrt(2)
Step-by-step explanation:
The circle with center in O(xo,yo) and radius r has the equation:
(x-xo)²+(y-yo)²=r²
We have:
x²+y²+ax+2y+3=0
But: x²+ax=x²+2(a/2)x+a²/4-a²/4= (x+a/2)²-a²/4
And
y²+2y+3=y²+2y+1+2=(y+1)²+2
Replacing, we get:
(x+a/2)²-a²/4+(y+1)²+2=0
(x+a/2)²+(y+1)²=a²/4-2=(a²-8)/4
By visual inspection we note that:
- center of circle: O(-a/2, -1)
- radius: r=sqrt((a²-8)/4)=0.5×sqrt(a²-8). This means a²>8 or a>2×sqrt(2)
