To solve a system of equations, we need to somehow eliminate one of the variables to get a single equation in one variable. With linear systems of equations (no quadratic terms), there are several standard methods for this: substitution or Gaussian elimination are the most popular. In this case though,...

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To solve a system of equations, we need to somehow eliminate one of the variables to get a single equation in one variable. With linear systems of equations (no quadratic terms), there are several standard methods for this: substitution or Gaussian elimination are the most popular. In this case though, there is a quadratic term (the x^2 term in the second equation) that causes problems. This forces us to use substitution.

Let the first equation equal the second equation.

`x-1=x^2-6x+9` now move the left side to the right

`0=x^2-7x+10` factor the right side (and put the zero on the far right)

`(x-2)(x-5)=0` now each factor can equal zero

`x=2` or `x=5`

These values can be substituted back into the first equation to find y

`y=2-1=1` or `y=5-1=4`

**There are two solutions for the system, (x,y)=(2,1) and (x,y)=(5,4).**