# What is the x-coordinate of the solution to the system of equations? 3x – 2y = –30 6x + 3y = 3

Answer: I think it is x= -4 because if you use elimination you would multiply all the values in the first equation so 3x-2y=-30 then multiply each value by two so it would be 6x-4y=-60. Then you would subtract the equations on top of each other to find the y value. you would do 6x-6x=0, -4y-3y=-7y, and -60-3=-63. So the remaining numbers after subtraction would be =7y=-63. Then divide each side by negative 7 and you would get y=9. Next take that number and substitute it into the first equation so it would be 3x-2(9)=-30. Then solve for x. 3x-18=-30, add 18 to both sides. 3x=-12, divide each side by 3. Then you would get x=-4. To check if this works plug in both the y and x value into the other equation (6x+3y=3) and solve to see if they are equal. 6(-4)+3(9)=3, -24=27=3, 3=3. that is how you check the answers are correct.
Answer:  I believe the right answer is 4y=6x-6 Have a nice day

## Related Questions

Describe the locus in space. points 4 mm from

Question 34 options:

a sphere of radius 4 cm

two planes parallel to , each 4 mm from

an endless cylinder with radius 4 mm and centerline

two lines parallel to , each 4 mm from

Parameterize the plane that contains the three points (−3,1,−2), (−6,−10,−4) and (15,5,20). (Use s and t for the parameters in your parameterization)

The plane that contains the points (-3, 1, -2), (-6, -10, -4) and (15, 5, 20) is described by .

Let be , and , the equation of the plane in parametric form is:

(1)

Where and are parameters and:

(2)

(3)

If we know that , and , then the parametric form is:

The plane that contains the points (-3, 1, -2), (-6, -10, -4) and (15, 5, 20) is described by .

We kindly invite to check this question on parametric formulas: brainly.com/question/23070611

First find any two vectors in the plane, and take their cross product. The cross product will be perpendicular to both of them.

To find any vectors in the plane, take any two of the given points at a time, and treating them as vectors, subtract them. These vector differences are parallel to the plane we want to find.

(-3, 1, -2) - (-6, -10, -4) = (3, 11, 2)

(-3, 1, -2) - (15, 5, 20) = (-18, -4, -22)

Take the cross product to get the normal vector to the plane:

(3, 11, 2) x (-18, -4, -22) = (-234, 30, 186)

Let (x, y, z) be any point on the plane. The vector

(x, y, z) - (-3, 1, -2) = (x + 3, y - 1, z + 2)

runs parallel to the plane, so it's perpendicular to the plane's normal, which means their dot product is 0. This gives us the Cartesian equation for the plane,

(-234, 30, 186) • (x + 3, y - 1, z + 2) = 0

-234(x + 3) + 30(y - 1) + 186(z + 2) = 0

-234x + 30y + 186 = 360

Every coefficient has a GCD of 6, so the equation is equivalent to

-39x + 5y + 31z = 60

Finally - and this is the easiest step - write this in parametric form. Let x = s and y = t, then

z = (60 + 39s - 5t)/31

So the plane can be parameterized by

P(s, t) = (s, t, (60 + 39s - 5t)/31)

where s and t are any real numbers.

12 ounces is roughly the same

Roughly the same as what?

Step-by-step explanation:

A. 400 GRAMS

B. 120 GRAMS

C. 356 GRAMS

D. 340 GRAMS

PENN FOSTER