How do you solve y+2x=5y+2x=5 and 2x^2 - 3x - y =162x23xy=16?

2 Answers
Mar 10, 2016

A(-3.11)A(3.11)

B(7/2,-2)B(72,2)

Explanation:

In this way:

y+2x=5y+2x=5
2x^2 - 3x - y =162x23xy=16,

so:

y=5-2xy=52x

2x^2-3x-(5-2x)=16rArr2x^2-x-21=02x23x(52x)=162x2x21=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

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