How can one halfx − 5 = one thirdx + 6 be set up as a system of equations? 2y + x = −10 3y + x = 18 2y + 2x = −10 3y + 3x = 18 2y − x = −10 3y − x = 18 2y − 2x = −10 3y − 3x = 18

2y-x= -10

3y-x = 18

Step-by-step explanation:

The correct option is . 2y-x = -10 ,  3y-x = 18

one half x − 5 = 1/2 (x-5)

one third x + 6 = 1/3 *(x+6)

1/2 x-5 = 1/3x+6 =y

y= 1/2x-5

y = 1/3x+6

Now,

y=x/2 -5     equation 1

y = x/3 +6    equation 2

By taking L.C.M of the first equation we get:

y=x/2 -5

y= x-10/2

Now multiply both terms by 2.

2y=x-10

2y-x= -10

Now lets solve second equation:

Take L.C.M of the second equation:

y = x/3 +6

y=x+18/3

Multiply both sides by 3

3y= x+18

3y-x = 18

Therefore the system of equations we get is:

2y-x= -10

3y-x = 18

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3