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.