Question

y = x2 − 6x + 8
2y + x = 4
The pair of points representing the solution set of this system of equations is

Answer 1

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.

Step-by-step 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:

$\frac{-3}{-2}=x$

$\frac{3}{2}=x$

$x=\frac{3}{2}$

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

2y=-x+4

Divide both sides by 2:

$y=\frac{-x+4}{2}$

Plug in 3/2 for x:

$y=\frac{\frac{-3}{2}+4}{2}$

Multiply top and bottom by 2:

$y=\frac{-3+8}{4}$

$y=\frac{5}{4}$

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).

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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

$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

{Plug in our numbers:

$x=\frac{11 \pm \sqrt{(-11)^2-4(2)(12)}}{2(2)}$

$x=\frac{11 \pm \sqrt{121-96}}{4}$

$x=\frac{11 \pm \sqrt{25}}{4}$

$x=\frac{11 \pm 5}{4}$

$x=\frac{11+5}{4} \text{ or } \frac{11-5}{4}$

$x=\frac{16}{4} \text{ or } \frac{6}{4}$

$x=4 \text{ or } \frac{3}{2}$

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).