Mathematics, 18.07.2019 10:10, jaz5380

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

Answers: 1

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answered: gransazer863

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).

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

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).

answered: Guest

A= r2 × 3.14 11×11 ↓ ↓ 121 × 3.14 = 379.94 → 379.9

answered: Guest

proofs involving isosceles triangles often require special consideration because an isosceles triangle has several distinct properties that do not apply to normal triangles.(more about triangle types) therefore, when you are trying to prove that two triangles are congruent, and one or both triangles, are isosceles you have a few theorems that you can use to make your life easier.

step-by-step explanation:

an isosceles triangle has two congruent sides and two congruent angles. the congruent angles are called the base angles and the other angle is known as the vertex angle. ∠bac and ∠bca are the base angles of the triangle picture on the left. the vertex angle is ∠abc

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Y= x2 − 6x + 8
2y + x = 4
the pair of points representing the solution set of this syst...