## Precalculus (6th Edition) Blitzer

The system of equations are $3x-y=1$ and $x+2y=12$. Solution of the equations is $\left( x,y \right)=\left( 2,5 \right)$.
Let us assume the first number is x and the second number is y. Subtract the second number from three times the first number to get 1. This gives the equation as: $3x-y=1$ And addition of the first number and twice the second number gives 12. This gives the equation as: $x+2y=12$ Multiply the equation $3x-y=1$ by 2 $6x-2y=2$ And add equations $x+2y=12$ and $6x-2y=2$ to obtain: \begin{align} & x+2y+6x-2y=12+2 \\ & 7x=14 \\ & x=2 \end{align} Put the value of x in the equation $3x-y=1$ to obtain: \begin{align} & 6-y=1 \\ & y=5 \end{align} Hence, the systems of equation are $3x-y=1$ and $x+2y=12$. Solution of the equations is $\left( x,y \right)=\left( 2,5 \right)$.