## Precalculus (6th Edition) Blitzer

The systems of equations are $x+y=2$ and $x-y=8$ and the solution to the equations is $\left( x,y \right)=\left( 5,-3 \right)$.
Let us assume the two numbers are x and y. And the sum of the numbers is 2. This gives the equation as: $x+y=2$ And the difference between the numbers is −1. This gives the equation as: $x-y=8$ Add equations $x+y=2$ and $x-y=8$ to obtain: \begin{align} & x+y+x-y=2+8 \\ & 2x=10 \\ & x=5 \end{align} Substitute the value of x in equation (I) \begin{align} & 5+y=2 \\ & y=-3 \end{align} Hence, the system of equations are $x+y=2$ and $x-y=8$ and the solution to the system of equations is $\left( x,y \right)=\left( 5,-3 \right)$.
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