Consider the given equations.
6x+7y−11=0
5x+2y−13=0
By using cross multiplication, we get
6 7 −11
5 2 −13
x7×(−13)−2×(−11)=y−11×5−(−13)×6=16×2−7×5⟹x−91+22=y−55+78=112−35
x−69=y23=1−23
x=3,y=−1
Hence, the value of x and y is 3,−1.
Solve each of the following systems of eqautions by the method of cross-multiplication:
3x+2y+25=0
2x+y+10=0