How do you simplify the expression #7x^2-4y+3x^2+5y+2# and evaluate it for x+3 and y=9?

1 Answer
Apr 27, 2017

Answer:

#y=-10x^2-2#
? #x=3# - #y=-92#
? #y=x+3# - No real solutions
#y=9# - No real solutions

Explanation:

Assuming this is all equal to 0, we add like terms
#10x^2+y+2=0#

We can then isolate one of the variables, usually #y#.
#y=-10x^2-2#

To evaluate it for #x=3#, just plug it into the function to get
#y(3)=-10(3)^2-2=-10(9)-2=-90-2=-92#

To solve for #y=x+3# we get
#x+3=-10x^2-2#
#10x^2+x+5=0#

We can use the quadratic formula to get the solutions
#\frac{-(1)^2\pm\sqrt{(1)^2-4(10)(5)}}{2(10)}#
#\frac{-1\pm\sqrt{1-200}}{20}#
#\frac{-1\pm\sqrt{-199}}{20}#
This results in undefined so there are no real solutions.

For #y=9# we do the following
#9=-10x^2-2#
#-10x^2=11#
#x^2=11/-10#
#x=\sqrt{-11/10}#
This is also undefined, so there are no real solutions.