Solve
5/x+y - 2/x-y = -1 and 15/x+y + 7/x-y = 10
Using cross-multiplication method.

x = 3 and y = 2

Step-by-step explanation:

Start by multiplying the first equation by "-3" so we can cancel the term with denominator (x+y) when we combine:

which now added term by term to the second equation gives:

Now we solve for x-y by cross multiplying:

Now we use this substitution for x  back in the first equation to solve for the unknown y:

and now that we know that y = 2, we use the substitution equation to solve for x:

< 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)