Solve the system of linear equations
10x – 3y + 2z = 0
19x - 5y – z = 0

answer:

for 10x-3y+2z=0 is; x=3/10y-1/5z

                              y=10/3x+2/3z

                              z=-5x+3/2y

now for 19x-5y-z=0; x=5/19y+1/19z

                                  y=19/5x-1/5z

                                    z=19x-5y

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susie has 7 apples

< 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)