Question

How do you solve the following system?: `5x +5y =27, 9x -5y = 0`

Answer 1

By substitution.

Explanation:

We have two equations, `5x+5y=27` and `9x-5y=0`. In order to use substitution, we need to solve for one variable. The easiest one to solve for would be `9x-5y=0`

`9x-5y=0`
`9x=5y`

Now looking at the equation we need to substitute in for, it would be easier for us to plug something in with a denominator of 5 so they would cancel out and we can do a little less math. So we shall solve for y in this equation.

`y = 9/5x`

Now we plug this y into the other equation.

`5x+5(9/5x)=27` and simplifying this would give us
`5x+9x=27` now we need to combine like terms
`14x=27` and finally
`x=27/14`

Now we plug this into the other equation.

`9(27/14) - 5y = 0`

`243/14 - 5y = 0`

`243/14 = 5y`

`243/70 = y`

My calculator isn't working right now so I can't quite give you decimal values but those fractions should be correct.