Question

1)A System of equations is shown below . What is the solution to the system of equations? 5x+2y=-15 2x-2y=-6

Answer 1

x= -3 and y= 0

Explanation:

5x+2y=-15

2x-2y=-6

7x =-21

x= -3

Putting value of x in equation 1

5(-3) +2y=-15

-15+2y= -15

2y= 0

y= 0

This can be solved with the help of matrices

In matrix form the above equations can be written in the form

`\left[\begin{array}{ccc}5&2\\2&-2\/\end{array}\right]`
`\left[\begin{array}{ccc}x\\y\\\end{array}\right]` =
`\left[\begin{array}{ccc}-15\\-6\\\end{array}\right]`

Let

`\left[\begin{array}{ccc}5&2\\2&-2\/\end{array}\right]` = A
`\left[\begin{array}{ccc}x\\y\\\end{array}\right]` = X and
`\left[\begin{array}{ccc}-15\\-6\\\end{array}\right]`= B

Then AX= B

or X= A⁻¹ B

where A⁻¹= adj A/ ║A║ where mod A≠ 0

adj A=
`\left[\begin{array}{ccc}-2&-2\\-2&5\/\end{array}\right]`

║A║= ( 5*-2- 2*2)= -10-4= -14≠0

X= A⁻¹ B

`\left[\begin{array}{ccc}x\\y\\\end{array}\right]` =- 1/14
`\left[\begin{array}{ccc}-2&-2\\-2&5\/\end{array}\right]`
`\left[\begin{array}{ccc}-15\\-6\\\end{array}\right]`

`\left[\begin{array}{ccc}x\\y\\\end{array}\right]` =- 1/14
`\left[\begin{array}{ccc}-2*-15&+ -2*-6\\-2*-15&+ 5*-6\\\end{array}\right]`

`\left[\begin{array}{ccc}x\\y\\\end{array}\right]` =- 1/14
`\left[\begin{array}{ccc} 30&+12\\30&+-30\\\end{array}\right]`

`\left[\begin{array}{ccc}x\\y\\\end{array}\right]` =- 1/14
`\left[\begin{array}{ccc}42\\0\\\end{array}\right]`

`\left[\begin{array}{ccc}x\\y\\\end{array}\right]` =
`\left[\begin{array}{ccc}-42/14\\0/-14\\\end{array}\right]`

`\left[\begin{array}{ccc}x\\y\\\end{array}\right]` =
`\left[\begin{array}{ccc}-3\\0\\\end{array}\right]`

From here x= -3 and y= 0

Solution Set = [(-3,0)]