How do you solve #y+2x=5# and #2x^2 - 3x - y =16#?

2 Answers
Mar 10, 2016

Answer:

#A(-3.11)#

#B(7/2,-2)#

Explanation:

In this way:

#y+2x=5#
#2x^2 - 3x - y =16#,

so:

#y=5-2x#

#2x^2-3x-(5-2x)=16rArr2x^2-x-21=0#,

#Delta=b^2-4ac=1-4(2)(-21)=1+168=169=13^2rArr#

#x_(1,2)=(-b+-sqrtDelta)/(2a)=(-(-1)+-13)/(2*2)=(1+-13)/4rArr#

#x_1=(1-13)/4=-12/4=-3# and #y_1=5-2*(-3)=11#
#x_2=(1+13)/4=14/4=7/2# and #y_2=5-2*7/2=-2#.

So:

#A(-3.11)#

#B(7/2,-2)#

Mar 10, 2016

Answer:

#x=7/2" or "x=-3#

Explanation:

Given:
#y+2x=5#........................(1)
#2x^2-3x-y=16#...........(2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Write equation (1) as:#" "y=-2x+5" "..............(1_a)#
Write equation (2) as:#" "y=2x^2-3x-16" ".......(2_a)#

Equate equation #(1_a)" to "(2_a)# through y

#=>" "-2x+5" "=" "y" "=" "2x^2-3x-16#

Collecting like terms and equation to 0

#" "2x^2-x-21=0#

Solving by quadratic formula method

Standard form#" "-> ax^2+bx+c=0#

Where:#" "x= (-b+-sqrt(b^2-4ac))/(2a)#

#a=2#
#b=-1#
#c=-21#

Thus we have:

#" "x=(+1+-sqrt((-1)^2-4(2)(-21)))/(2(2))#

#=>x=(1+-sqrt(169))/4#

#=>x=(1+-13)/4#

#color(blue)(=>x= 14/4 =7/2 " or "x=-12/4 = -3)#

Tony B