Solve for x, y, and z 2x - y + 3z= 10 -3x + y - 2z= -11 -4x + y - 4z = -15 a) (-2, -23, -3) b) (2, -9, -1) c) (2, -3, 1) d) (2, -6, 0)

2x - y + 3z= 10  > equation 1

-3x + y - 2z= -11   > equation 2

-4x + y - 4z = -15  > equation 3

Add first and second equation

2x - y + 3z= 10  

-3x + y - 2z= -11    

-x + z = -1 > equation 4

Add first and third equation

2x - y + 3z= 10  

-4x + y - 4z = -15

-2x - z = -5 > equation 5

Now add equation 4  and 5

-x + z = -1

-2x - z = -5

-3x = -6

Divide both sides by -3

so x= 2

We use equation 4  and solve for z

-x + z = -1

-2 + z = -1

Add 2 on both sides

z = 1

Now plug in x=2  and z=1 in any one of the original equation

2x - y + 3z= 10

2(2) - y + 3(1)= 10

4 - y +3 = 10

7 - y = 10

subtract 7 on both sides

-y = 3

y = -3

x=2, y=-3, z= 1

Solution is (2,-3, 1)

solve by linear combination

use 1st equation to cancel y's in 2nd and 3rd equations

add 1st equation to 2nd equation to get

-x+z=-1

add 1st equation to 3rd equation to get

-2x-z=-5

now add the 2 new equations -x+z=-1 and -2x-z=-5 to get

-3x=-6

x=2

sub back

-x+z=-1

-2+z=-1

z=1

sub back

-3x+y-2z=-11

-3(2)+y-2(1)=-11

-6+y-2=-11

y=-3

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

C is answer

[2, 2.5)

(2.5, 3]

step-by-step explanation:

i think its these two not sure [2, 2.5)

(2.5, 3]

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)