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2 years ago

Find the solution of y=x+3 and y=2x+6

Mathematics

2 answers:

lisabon 2012 [21]2 years ago

3 0

Answer:

x = 3 and y = 0

Step-by-step explanation:

To find the solution where you have two equations is called a System of Equations.

What you do is you notice there are two equations and the let both equations equal each other. We can do this because both equations start with a y. These "y" work like puzzles and you can connect the "x+3" and "2x + 6" together. Then solve for it. Here's what i mean,

y = x+3

y = 2x+6

Now connecting them,

x+3 = 2x+6

Subtract x from both sides

3 = x + 6

Subtract 6 from both sides,

x = -3

Now solve for y,

y = -3 + 3 = 0

So our solution is x = 3 and y = 0.

olasank [31]2 years ago

3 0

Answer:

(x, y) = (-3, 0)

Step-by-step explanation:

Substitution or elimination work well for solving this system of equations.

<h3>Elimination</h3>

Subtract the first equation from the second.

(y) -(y) = (2x +6) -(x +3)

0 = x +3

-3 = x . . . . . . . subtract 3

y = (-3) +3 = 0 . . . . use the first equation to find y

(x, y) = (-3, 0)

<h3>Substitution</h3>

Use the second equation to substitute for y in the first.

2x +6 = x +3

x = -3 . . . . . . . . . subtract x+6 from both sides

y = (-3) +3 = 0 . . . . find y using the first equation

(x, y) = (-3, 0)

A graphing calculator answers these quickly.

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3 years ago

Read 2 more answers

Is x+y+1=0 a tangent of both y^2=4x and x^2=4y parabolas?

Lubov Fominskaja [6]

Answer:

yes

Step-by-step explanation:

The line intersects each parabola in one point, so is tangent to both.

For the first parabola, the point of intersection is ...

y^2 = 4(-y-1)

y^2 +4y +4 = 0

(y+2)^2 = 0

y = -2 . . . . . . . . one solution only

x = -(-2)-1 = 1

The point of intersection is (1, -2).

For the second parabola, the equation is the same, but with x and y interchanged:

x^2 = 4(-x-1)

(x +2)^2 = 0

x = -2, y = 1 . . . . . one point of intersection only

___

If the line is not parallel to the axis of symmetry, it is tangent if there is only one point of intersection. Here the line x+y+1=0 is tangent to both y^2=4x and x^2=4y.

_____

Another way to consider this is to look at the two parabolas as mirror images of each other across the line y=x. The given line is perpendicular to that line of reflection, so if it is tangent to one parabola, it is tangent to both.