Let x,y be two different numbers
suppose x^2=y^2
then x^2-y^2=0
which yields (x+y)(x-y)=0
so either x=y or x=-y
In any case, x and y must be the same value
also when a vairable is squared like y=x^2
we must note that there are 2 possible solutions
x=(+/-)sqrt(y)
Answer:
roam
Step-by-step explanation:
free points
We have that
<span>[6x+5]=1+2*(3x+2)
[6x+5]=1+2*3x+2 --------> is not correct ------->1+ 2*(3x+2)=1+2*3x+2*2
then
</span>[6x+5]=1+6x+4---------------> [6x+5]=[6x+5]
<span>this equation is an identity, all real numbers are solutions.</span>
Answer:
y = 1/2x - 6
Step-by-step explanation:
y2 - y1 / x2 - x1
-2 - (-5) / 8 - 2
3 / 6
= 1/2
y = 1/2x + b
-5 = 1/2(2) + b
-5 = 1 + b
-6 = b
4th answer is the answer for theis question because they are equivalent
