## Precalculus (6th Edition) Blitzer

The system of equations has a unique solution and the solution is $\left\{ \left( \frac{41}{7},\frac{36}{7} \right) \right\}$.
Let us consider the system of equations: $4x-3y=8$ $2x-5y=-14$ And for equations ${{a}_{1}}x+{{b}_{1}}y={{c}_{1}}$ and ${{a}_{2}}x+{{b}_{2}}y={{c}_{2}}$. If, $\frac{{{a}_{1}}}{{{a}_{2}}}\ne \frac{{{b}_{1}}}{{{b}_{2}}}$ then the system of equations has a unique solution. Here, ${{a}_{1}}=4,\text{ }{{a}_{2}}=2,\text{ }{{b}_{1}}=-3\text{ and }{{b}_{2}}=-5$. Therefore, $\frac{4}{2}\ne \frac{-3}{-5}$ Or, $2\ne \frac{3}{5}$ Thus the system of equations has a unique solution. Multiply the first equation $2x-5y=-14$ by 2 to obtain the coefficient of x equal to 4, so that variable x is eliminated, $-4x+10y=28$ Add equation $4x=3y+8\text{ and }2x=-14+5y$: \begin{align} & 4x-3y-4x+10y=8+28 \\ & 7y=36 \\ & y=\frac{36}{7} \end{align} Put the value of $y=\frac{36}{7}$ in equation $2x-5y=-14$ as follows: \begin{align} & 2x-5y=-14 \\ & 2x-5\left( \frac{36}{7} \right)=-14 \\ & 2x-\frac{180}{7}=-14 \\ & 2x=-14+\frac{180}{7} \end{align} So, further simplify it to get: \begin{align} & 2x=-14+\frac{180}{7} \\ & =\frac{-98+180}{7} \\ & =\frac{82}{7} \end{align} It implies that: \begin{align} & x=\frac{82}{7\times 2} \\ & =\frac{82}{14} \\ & =\frac{41}{7} \end{align} It is, $x=\frac{41}{7}$ Now, verify the solution by substituting $x=\frac{41}{7}$ and $y=\frac{36}{7}$ into the equation $4x=3y+8$ as follows: $\begin{matrix} 4x=3y+8 \\ 4\left( \frac{41}{7} \right)\overset{?}{\mathop{=}}\,3\left( \frac{36}{7} \right)+8 \\ \frac{164}{7}\overset{?}{\mathop{=}}\,\frac{108}{7}+8 \\ \frac{164}{7}=\frac{164}{7} \\ \end{matrix}$ It is true. Now, put $x=\frac{41}{7}$ and $y=\frac{36}{7}$ in the equation $2x-5y=-14$, to verify it as follows: \begin{align} & 2x-5y=-14 \\ & 2\left( \frac{41}{7} \right)-5\left( \frac{36}{7} \right)=-14 \\ & \frac{82}{7}-\frac{180}{7}=-14 \\ & \frac{-98}{7}=-14 \end{align} The result comes out to be: $-14=-14$ So, both equation (I) and (II) are satisfied by the solution, $x=\frac{41}{7}$ and $y=\frac{36}{7}$. Hence, the system of equations has a unique solution and the solution is $\left\{ \left( \frac{41}{7},\frac{36}{7} \right) \right\}$.