What is x-2y=0 and then also 2x-5y=-4

2 answers:

4 0

$x-2y=0\:/(-2) \\ 2x-5y=-4 \\ \\ =>-2x+4y=0 \\ 2x+-5y=-4 \\ \\ =>-y=-4 \\ y=4 \\ \\ x-2.4=0 \\ =>x-8=0 \\ x=8 \\ \\ \framebox[1.1\width]{Good Luck!} \par$

seraphim [82]3 years ago

3 0

x + 2y = 0 ⇒ 2x + 4y = 0
2x - 5y = -4 ⇒ 2x - 5y = -4
 9y = 4
 9 9
 y = 4/9
 x + 2(4/9) = 0
 x + 8/9 = 0
 -8/9 -8/9
 x = -8/9
 (x, y) = (-8/9, 4/9)

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Find an equation of the plane. The plane that passes through the line of intersection of the planes x − z = 3 and y + 4z = 1 and

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Answer:

• 3x +y +z = 10

Step-by-step explanation:

The cross product of the normals of the intersecting planes will give a vector that is parallel to the line of intersection.

(1, 0 -1) × (0, 1, 4) = (1, -4, 1) . . . . using an appropriate calculator

Since this line is in the desired plane, it will be perpendicular to the desired plane's normal vector. The reference perpendicular plane likewise has its normal vector perpendicular to that of the desired plane. Hence the normal vector of the desired plane is perpendicular to both (1, -4, 1) and (1, 1, -4). It can be found from the cross product of these:

(1, -4, 1) × (1, 1, -4) = (15, 5, 5)

We can divide this by 5 to find our desired plane's normal vector: (3, 1, 1). To finish the equation of our plane, we need to know a point on the line of intersection. If we suppose z=0 at that point, then the remaining coordinates are ...

x -0 = 3 ⇒ x=3

y+4·0 = 1 ⇒ y=1

Using point (3, 1, 0) to find the constant in our plane's formula

3x +y +z = constant

we see it is ...

3·3 +1·1 +1·0 = constant = 10

The equation of the desired plane is ...

3x +y +z = 10

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