solve the system of equations below by graphing both equations with a pencil and paper what is the solution y=2x-4 y=-x+5
2 answers:
Answer:
The coordinate of the points on the graph as shown , where ( x ,y ) = ( 3 , 2 )
Step-by-step explanation:
Given as :
The two algebraic equation is
y = 2 x - 4 .......A and y = - x + 5 ........B
The given two algebraic equation can be solved by putting the one variable to the other equation
Now , put the value of y from Eq A into the Eq B
I.e 2 x - 4 = - x + 5
or, 2 x + x = 5 + 4
or, 3 x = 9
∴ x =
I.e x = 3
So The value of x = 3
Again, put the value of x in the equation A
I.e y = 2 x - 4
or, y = 2 × 3 - 4
∴ y = 6 - 4
I.e y = 2
So The value of y = 2
So , The coordinate of points on the graph as ( x , y ) = ( 3 , 2 )
Hence The coordinate of the points on the graph as shown , where ( x ,y ) = ( 3 , 2 ) Answer
You might be interested in
Answer:
The statement is now presented as:
In other words, this mathematical statement can be translated as:
<em>There is a point (h, k) in the set of real ordered pairs so that a circumference centered at (h,k) and with a radius r implies a equivalent circumference centered at (h+4,-k) and with a radius r. </em>
Step-by-step explanation:
Let the coordinates of the center of the circle, which must be transformed into
by operations of translation and reflection. From Analytical Geometry we understand that circles are represented by the following equation:
Where is the radius of the circle, which remains unchanged in every operation.
Now we proceed to describe the series of operations:
1) <em>Center of the circle is translated 4 units to the right</em> (+x direction):
(Eq. 1)
Where is the translation vector, dimensionless.
If we know that and
, then:
2) <em>Reflection over the x-axis</em>:
(Eq. 2)
Where is the reflection point, dimensionless.
If we know that and
, the new point is:
And thus, and
. The statement is now presented as:
In other words, this mathematical statement can be translated as:
<em>There is a point (h, k) in the set of real ordered pairs so that a circumference centered at (h,k) and with a radius r implies a equivalent circumference centered at (h+4,-k) and with a radius r. </em>
<em />