In either case we will need to rewrite the given equations into matrix form:

#((3,3,color(blue)(15)),(2,3,color(blue)(-5)))#

**Gauss-Jordan**

Divide row 1 (by #3# so that the leading term is 1

#color(white)("XXX")((1,1,5),(2,3,-5))#

Subtract #2xx# this new row 1 from row 2 to reduce its leading term to 0

#color(white)("XXX")((1,1,5),( 0,1,-15 ))#

Subtract this new row 2 from row 1:

#color(white)("XXX")((1,0,20),(0,1,-15))#

The first column is #x#, the second is #y# and the third is the value they are equal to.

**Cramer's Rule**

#D=((3,3),(2,3))color(white)("XX")D_x=((color(blue)(15),3),(color(blue)(-5),3))color(white)("XX")D_y=((3,color(blue)(15)),(2,color(blue)(-5)))#

The determinants:

#color(white)("XXX")abs(D) = 3xx3-2xx3=3#

#color(white)("XXX")abs(D_x)= 15xx3-(-5)xx3=60#

#color(white)("XXX")abs(D_y)=3xx(-5)-2xx15=-45#

#x=abs(D_x)/abs(D) = 60/3=20#

#y=abs(Dy)/abs(D) = (-45)/3=-15#