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Y = x2 − 6x + 8

2y + x = 4
The pair of points representing the solution set of this system of equations is

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Answer:

The two intersections are (3/2 , 5/4) and (4,0).

I put two ways to do it. You can pick your favorite of these are try another route if you like.

Explanation:

The system is:


y=x^2-6x+8


2y+x=4.

I don't know how good at factoring you are but the top equation consists of polynomial expression that has a factor of (x-4). I see that if I solve 2y+x=4 for 2y I get 2y=-x+4 which is the opposite of (x-4) so -2y=x-4.

So anyways, factoring x^2-6x+8=(x-4)(x-2) because -4+(-2)=-6 while -4(-2)=8.

This is the system I'm looking at right now:


y=(x-4)(x-2)


-2y=x-4

I'm going to put -2y in for (x-4) in the first equation:


y=-2y(x-2)

So one solution will occur when y is 0.

Now assume y is not 0 and divide both sides by y:


1=-2(x-2)

Distribute:


1=-2x+4

Subtract 4 on both sides:


-3=-2x

Divide both sides by -2:


(-3)/(-2)=x


(3)/(2)=x


x=(3)/(2)

Now let's go back to one of the original equations:

2y=-x+4

Divide both sides by 2:


y=(-x+4)/(2)

Plug in 3/2 for x:


y=((-3)/(2)+4)/(2)

Multiply top and bottom by 2:


y=(-3+8)/(4)


y=(5)/(4)

So one solution is at (3/2 , 5/4).

The other solution happened at y=0:

2y=-x+4

Plug in 0 for y:

2(0)=-x+4

0=-x+4

Add x on both sides:

x=4

So the other point of intersection is (4,0).

-------------------------------------------------------

The two intersections are (3/2 , 5/4) and (4,0).

Now if you don't like that way:


y=x^2-6x+8


2y+x=4

Replace y in bottom equation with (x^2-6x+8):


2(x^2-6x+8)+x=4

Distribute:


2x^2-12x+16+x=4

Subtract 4 on both sides:


2x^2-12x+16+x-4=0

Combine like terms:


2x^2-11x+12=0

Compare this to
ax^2+bx+c=0


a=2


b=-11


c=12

The quadratic formula is


x=(-b \pm √(b^2-4ac))/(2a)

{Plug in our numbers:


x=(11 \pm √((-11)^2-4(2)(12)))/(2(2))


x=(11 \pm √(121-96))/(4)


x=(11 \pm √(25))/(4)


x=(11 \pm 5)/(4)


x=(11+5)/(4) \text{ or } (11-5)/(4)


x=(16)/(4) \text{ or } (6)/(4)


x=4 \text{ or } (3)/(2)

Using 2y+x=4 let's find the correspond y-coordinates.

If x=4:

2y+4=4

Subtract 4 on both sides:

2y=0

Divide both sides by 2:

y=0

So we have (4,0) is a point of intersection.

If x=3/2

2y+(3/2)=4

Subtract (3/2) on both sides:

2y=4-(3/2)

2y=5/2

Divide 2 on both sides:

y=5/4

The other intersection is (3/2 , 5/4).

User Koustav Chanda
by
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