Find the solution to the following system of equations using the addition method. 3x + 2y = 7 -3x + 3y = 8

x = 1/3 and y = 3

Step-by-step explanation:

It is given the system of equations

3x + 2y = 7   (1)

-3x + 3y = 8    (2)

To find the solution

Add eq(1) and eq (2) we get,

3x + 2y = 7   (1)

-3x + 3y = 8    (2)

 0 + 5y = 15

y = 15/5 = 3

Substitute the value of y in eq(1)

3x + 2y = 7   (1)

3x + 2*3 = 7

3x + 6 = 7

3x = 7 - 6 = 1

x = 1/3

Therefore

x = 1/3 and y = 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)