What is the solution set of the system #4x+y=12# and #15x+y=45#?

2 Answers
Apr 15, 2018

#x=3, y=0#

Explanation:

Equation (1): #4x+y=12#
Equation (2): #15x+y=45#

#(2)-(1): 11x=33 =>x=3#

Substitute #x=3# into (1):
#4(3)+y=12#
#12+y=12#
#therefore y=0#

Apr 15, 2018

# x = 3 #
# y = 0 #

Explanation:

There are different ways to solve this. The easiest way (probably) is to add or subtract each equation with each other and get rid of either the #x# or the #y# terms.

In this case it is easiest to take the second equation and subtract it by the first:

#4 x + y = 12 # Equation (1)
#15 x + y = 45 # Equation (2)

Equation (2) - Equation (1)

#15 x - 4 x + y - y = 45 - 12#

Simplify

# 11 x = 33 #
# x = 3 #

Then substitute this answer into Equation (1) to get #y#:

#4 * 3 + y = 12 #
# y = 0 #


Another (more complex) way to answer is to use matrix multiplication .

The system of euquations can be rewritten in matrix form:

#4 x + y = 12 # Equation (1)
#15 x + y = 45 # Equation (2)

Become

# ((4,1),(15,1)) ((x),(y)) = ((12),(45)) #

Now we need to inverse the left hand matrix to isolate the #x# and #y# (the #X# matrix in the notation below) i.e.

# A X = B => A^(-1) A X = A^(-1) B => X = A^(-1) B#

where #A = ((4,1),(15,1)) #, #X = ((x),(y)) # and # B = ((12),(45))#

To solve this system we need to get the inverse of #A#. For a 2x2 matrix the rule is:

If # M = ((a,b),(c,d))# then
#M^(-1) = (1 / ( a d - b c) ) ((d,-b),(-c,a)) #,

For #A^(-1)# we get:

#A^(-1) = (1 / ( 4 * 1 - 15 * 1 ) ) ((1,-1),(-15,4)) #,
#A^(-1) = (- 1 / 11) ((1,-1),(-15,4)) #
#A^(-1)= ((-1/11,1/11),(15/11,-4/11)) #

Now we can solve the problem as we can substitute this into the equation

#X = A^(-1) B#

#((x),(y)) = ((-1/11,1/11),(15/11,-4/11)) ((12),(45))#

We multiply out the matrices and get the following result:

#((x),(y)) = ((-1/11 * 12 + 1/11 * 45),(15/11 * 12 + -4/11 * 45 )) = ((3),(0))#

This gives us the answer for #x# and #y#.