[tex] the \ point \ of \ intersection \ of \ a \ two lines :\\\begin{cases} 2x-3y=12 \\ 2y-x=-7 \ \ \cdot 2\end{cases}\\ \\\begin{cases} 2x-3y=12 \\ -2x+4y=-14 \end{cases}\\+-------- \\y=-2\\ \\2x-3\cdot (-2)=12\\2x+6=12\\2x=12-6\\2x=6\ \ /:2\\x=3 \\\\\begin{cases} x=3 \\y=-2 \end{cases}\\ \\y=5x-7 \\ \\slope: \ m=5 \\ \\ The \ slope \ of \ lines \ parallel \ is \ the \ same : \\ m_{1}=5\\ \\ use \ y = m_{1}x + b \ and \ the \ point \ (3,-2) \ to \ find \ b \\ \\ -2=5\cdot 3+b \\-2=15+b\\ b=-2-15\\b=-[/tex]
