How do you determine whether a linear system has one solution, many solutions, or no solution when given 5x+ 4y= -18 and 2x+3y=-24?

Question

Eva Abramson · Accepted Answer

These have one solution

One can answer this in several ways! Method 1 We can try directly solving the system of equations #5x+ 4y= -18# #2x+3y=-24# by multiplying the second one by #4/3# and subtracting from the first one : #(5-2 times 4/3)x=-18+24times 4/3 qquad implies# #7/3 x = 14 qquad implies qquad x = 6# which in turn implies # 5times 6+4y=-18 implies 4y = -48 implies y = -12# Thus, there is only one solution #x=6,y = -12# Method 2 A bit more sophisticated method involves the coefficient matrix #A = ((5,4),(2,3))# The determinant of this matrix is #det A = |(5,4),(2,3)| = 5times 3-2 times 4 = 7 ne 0# Hence this matrix is non-singular and so #A^-1# exits. Thus the system has a unique solution # ((x).(y)) = ((5,4),(2,3))^-1 ((-18),(-24))# Carrying out the explicit calculation will lead to the same answer as the first method (however, since all we need here is whether the solution exists or is unique, this last step is unnecessary - the existence of #A^-1# is enough to say that a unique answer exists) Method 3 The two equations can be represented graphically by two straight lines, and the question of whether solutions exist then boils down to whether the lines intersect. Since the slope of the line representing the first equation is #-5/4# , and that of the second is #-2/3#, the two lines are not parallel. So, they will intersect at exactly one point, and hence the solution is unique. Note

Kennedy Adams · Answer

undefined You can determine the number of solutions of a linear system by analyzing the coefficients of the variables and the constants in the equations. If the system has the same number of equations as variables and the determinant of the coefficient matrix is non-zero, it has a unique solution. If the determinant is zero and the system is consistent (all equations are not contradictory), it has infinitely many solutions. If the determinant is zero and the system is inconsistent (at least one equation is contradictory), it has no solution. For the given system, the determinant is non-zero, so it has one unique solution.