There are 3 ways using matrices and linear algebra :
Method number 1 - Gauss-Jordan Elimination
Write the linear system in the form #Ax=b# where
#A=[(4,-1),(5,-3)]# is the coefficient matrix of the system,
#x=[(x),(y)]# is the column vector of variables, and
#b=[(-3),(-23)]# is the column vector of solutions.
Now augment the coefficient matrix with the column vector of solutions and perform elementary row operations until you obtain the #2xx2# identity matrix #I_2# on the left hand side and then what remains on the right hand side will be the solution set for #x#. I will attach a sketch at the end to show these steps.
Eventually we get :
#[(4,-1),(5,-3)]|[(-3),(-23)] -> [(1,0),(0,1)]|[(2),(11)]#
and hence #x=2 and y=11#.
Method number 2 - Using the inverse matrix method
Since #Ax=b# it implies that #x=A^(-1)b#.
We may find #A^(-1)# by using elementary row operations in augmenting the coefficient matrix A with the identity matrix #I_2#, alternatively, we may use the formula :
#A^(-1)=1/(det(A))*adj(A)#, where #det(A)=Delta#.
#=1/(-12-(-5))*[(-3,1),(-5,4)]#
#=[(3/7,-1/7),(5/7,-4/7)]#
Hence #x=A^(-1)b=[(3/7,-1/7),(5/7,-4/7)][(-3),(-23)]=[(2),(11)]#
and so #x=2 and y=11#.
Method number 3 - Kramer's Rule
Find the determinants of the matrices formed by replacing each column vector in the coefficient matrix with the column vector of solutions :
#Delta_x=|(-3,-1),(-23,-3)|=(-3)(-3)-(-1)(-23)=-14#.
#Delta_y=|(4,-3),(5,-23)|=-77#.
Then : #x=Delta_x/Delta=-14/-7=2#
and #y=Delta_y/Delta=-77/-7=11#.
