Question

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

Answer 1

`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`

Answer 2

`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`

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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`.