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Nana76 [90]
4 years ago
10

Which two points satisfy y = -x^2 + 2x + 4 and x + y = 4?

Mathematics
2 answers:
earnstyle [38]4 years ago
4 0
X + y = 4 that two points will be satisfy but the more easy top do is second one

ELEN [110]4 years ago
4 0

Answer:

Two points that satisfy the system propose are P1  = (3, 1) and P2 = (0, 4).

Step-by-step explanation:

Despite this equation system could be solve in different ways, as many things in Math do, we can take advantage of the negative symbol of equation   <em>y = -x^{2}+ 2x + 4</em>. Yes, instead of solving this using the quadratic formula, and doing so many other steps, and with the high probability of making a mistake observing that y = -x^{2} , we can solve this system using the well known logical formula that <em>if x = b and y = b, then x = y</em>.

<h3>First Step: the way of solving this system</h3><h3 />

This equation system could be solve in the following way:

First, we have these equations:

x+ y = 4  ( 1 )

y = -x^{2}+ 2x + 4   ( 2 )

Equation ( 2 ) can be rewritten as:

-y = x^{2} - 2x - 4;

And Equation ( 1 ) can be rewritten as:

x + y = 4; y = 4 - x, but also  -y = x - 4;

So,

-y = x^{2} - 2x - 4, and

-y = x - 4;  

That is, the <em>logical formula</em> previously described above.

<h3>Second Step: solving the system in the way proposed</h3><h3 />

If -y = x^{2} - 2x - 4, and

  -y = x - 4;  then,

x^{2} - 2x - 4 = x - 4 .

x^{2} - 2x - 4 -( x - 4 ) = 0 .

x^{2} - 2x - 4 -x + 4 ) = 0 .

x^{2} - 2x -x + 4 - 4  = 0, where ( 4 - 4 = 0 ), that is

x^{2} - 3x = 0;

x^{2} - 3x = 0, where x is a <em>common factor</em>, so

x * (x - 3)   = 0.

For this product to be true,

x  = 0  or

x - 3  = 0  ⇒    x  =  3 .

Having these values for x (two values, x = 0 and x = 3), we can find the other values for y, that is:

We know that  x + y = 4 (Equation 1), then

For <em>x = 3</em>;

3 + y = 4 , and

y = 4 - 3 = 1;  <em>y = 1</em> .

So,  P1 = ( 3, 1 );

For <em>x = 0</em>;

0 + y = 4 or <em>y = 4 </em>.

So, P2 = ( 0, 4 ).

<h3>Proving these results </h3><h3 />

Using equations ( 1 ) and ( 2 ):

For P1 = ( 3, 1 ),

3 + 1 = 4 ( 1 )

1 = -(3)^{2} + 2*(3) + 4  ( 2 ) or

1 = -(9) + 6 + 4 = -9 + 10 or

1 = 1 .

For P2 = ( 0, 4 ),

0 + 4 = 4 ( 1 )

4 = -(0)^{2} + 2*(0) + 4 ( 2 ) or

4 = 0 + 0 + 4 = 4 or

4 = 4 .

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