Answer:
Second choice:


Fifth choice:


Step-by-step explanation:
Let's look at choice 1.


I'm going to subtract 1 on both sides for the first equation giving me
. I will replace the
in the second equation with this substitution from equation 1.

Expand using the distributive property and the identity
:




So this not the desired result.
Let's look at choice 2.


Solve the first equation for
by dividing both sides by 2:
.
Let's plug this into equation 2:



This is the desired result.
Choice 3:


Solve the first equation for
by adding 3 on both sides:
.
Plug into second equation:

Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:



Not the desired result.
Choice 4:


I'm going to solve the bottom equation for
since I don't want to deal with square roots.
Add 3 on both sides:

Divide both sides by 2:

Plug into equation 1:

This is not the desired result because the
variable will be squared now instead of the
variable.
Choice 5:


Solve the first equation for
by subtracting 1 on both sides:
.
Plug into equation 2:

Distribute and use the binomial square identity used earlier:



.
This is the desired result.
Step-by-step explanation:
steps are in picture above.
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If we were to foil
after experieence
we know
ax²+bx+c=0
and
in form
(ax+b)(cx+d)=0
if we expand it, we get
acx²+bcx+adx+bd=0
or
(ac)x²+(bc+ad)x+(bd)=0
compare to
ax²+bx+c=0
we notice that the middle terms (x terms) are
b=(bc+ad)
so
in form
(2x-1)(1x+5)
b=bc+ad=(-1*1+2*5)=-1+10=9
b=9
or you could just expand it