2. solve the following system. show all your work. -3x + 2y + z = -6 x + 3y + 2z = 5 4x + 4y + 3z = 13 is there a step i'm skipping because every time i add the first 2 equations, i keep getting -2x + 5y + 3z with no variables eliminated.

-3x + 2y + z = -6

(3)x + 3y + 2z = 5(3) We want to get rid of any variable within 2 equations. The easiest one would be by multiplying both sides of eq#2 by 3 and add it to eq#1.

-3x+2y+z=-6

3x+9y+6z=15

Add them together to get 11y+7z=9

(-4)x+3y+2z=5(-4) Now,multiply the original eq#2 by -4 to cancel out x in the last equation

-4x-12y-8z=-20

4x + 4y + 3z = 13

add them together to get -8y-5z=-7

Now multiply the eq 11y+7z=9 by 5 and the equation -8y-5z=-7 by 7 to cancel out the z variable when we add them together.

Once we multiply them, we get

-56y-35z=-49

55y+35z=45

add them together to get -y=-4 and we know y=4

substitute y into any equation here that has y and any other 1 variable like -8y-5z=-7 and get z=-5

Finally, we know y=4 and z=-5, we can substitue that into any of the original eqs like  -3x + 2y + z = -6 and we get x=-3

To check, just substitute these values into any of the equations above

location: point 0

step-by-step explanation:

well, -8 to 10 on a number line is 18 spaces.

1/9 of 18 is 2.

so, 4/9 of 18 must be 8.

8 spaces from -8 (a) is 0.

so, 4/9 of the way from -8 to 10 is 0!

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !