## Precalculus (6th Edition) Blitzer

The solution of the given system can be obtained by using the Cramer’s rule, $\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} \\ {{a}_{2}} & {{b}_{2}} \\ \end{matrix} \right|=\left| \begin{matrix} {{c}_{1}} \\ {{c}_{2}} \\ \end{matrix} \right|$ \begin{align} & x=\frac{{{D}_{x}}}{D}\text{ and }y=\frac{{{D}_{y}}}{D}\text{ } \\ & D=\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} \\ {{a}_{2}} & {{b}_{2}} \\ \end{matrix} \right|={{a}_{1}}\cdot {{b}_{2}}-{{b}_{1}}\cdot {{a}_{2}} \\ & {{D}_{x}}=\left| \begin{matrix} {{c}_{1}} & {{b}_{1}} \\ {{c}_{2}} & {{b}_{2}} \\ \end{matrix} \right|={{c}_{1}}\cdot {{b}_{2}}-{{b}_{1}}\cdot {{c}_{2}} \\ & {{D}_{y}}=\left| \begin{matrix} {{a}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{c}_{2}} \\ \end{matrix} \right|={{a}_{1}}\cdot {{c}_{2}}-{{c}_{1}}\cdot {{a}_{2}} \end{align} So, \begin{align} & x=\frac{{{c}_{1}}\cdot {{b}_{2}}-{{b}_{1}}\cdot {{c}_{2}}}{{{a}_{1}}\cdot {{b}_{2}}-{{b}_{1}}\cdot {{a}_{2}}}\text{ (i)} \\ & y=\frac{{{a}_{1}}\cdot {{c}_{2}}-{{c}_{1}}\cdot {{a}_{2}}}{{{a}_{1}}\cdot {{b}_{2}}-{{b}_{1}}\cdot {{a}_{2}}}\text{ (ii)} \\ \end{align} Now consider the first equation as a sum of two equation, i.e. ${{a}_{1}}x+{{b}_{1}}y={{c}_{1}}$ As a sum of two equations, \begin{align} & {{a}_{3}}x+{{b}_{3}}y={{c}_{3}} \\ & {{a}_{4}}x+{{b}_{4}}y={{c}_{4}} \\ \end{align} Where, \begin{align} & {{a}_{1}}={{a}_{3}}+{{a}_{4}}\text{ }(iii) \\ & {{b}_{1}}={{b}_{3}}+{{b}_{4}}\text{ }(iv) \\ & {{c}_{1}}={{c}_{3}}+{{c}_{4}}\text{ }(v) \\ \end{align} So, equation 1 can be written as, $({{a}_{3}}+{{a}_{4}})x+({{b}_{3}}+{{b}_{4}})y={{c}_{3}}+{{c}_{4}}$ Now the solution of two resulting equations is, $\left| \begin{matrix} {{a}_{3}}+{{a}_{4}} & {{b}_{3}}+{{b}_{4}} \\ {{a}_{2}} & {{b}_{2}} \\ \end{matrix} \right|=\left| \begin{matrix} {{c}_{3}}+{{c}_{4}} \\ {{c}_{2}} \\ \end{matrix} \right|$ \begin{align} & x=\frac{{{D}_{x}}}{D}\text{ and }y=\frac{{{D}_{y}}}{D}\text{ } \\ & D=\left| \begin{matrix} {{a}_{3}}+{{a}_{4}} & {{b}_{3}}+{{b}_{4}} \\ {{a}_{2}} & {{b}_{2}} \\ \end{matrix} \right|=({{a}_{3}}+{{a}_{4}})\cdot {{b}_{2}}-({{b}_{3}}+{{b}_{4}})\cdot {{a}_{2}} \\ & {{D}_{x}}=\left| \begin{matrix} {{c}_{3}}\text{+}{{\text{c}}_{4}} & {{b}_{3}}\text{+}{{\text{b}}_{4}} \\ {{c}_{2}} & {{b}_{2}} \\ \end{matrix} \right|=({{c}_{3}}\text{+}{{\text{c}}_{4}})\cdot {{b}_{2}}-({{b}_{3}}\text{+}{{\text{b}}_{4}})\cdot {{c}_{2}} \\ & {{D}_{y}}=\left| \begin{matrix} {{a}_{3}}+{{a}_{4}} & {{c}_{3}}+{{c}_{4}} \\ {{a}_{2}} & {{c}_{2}} \\ \end{matrix} \right|=({{a}_{3}}+{{a}_{4}})\cdot {{c}_{2}}-({{c}_{3}}+{{c}_{4}})\cdot {{a}_{2}} \end{align} So, \begin{align} & x=\frac{({{c}_{3}}\text{+}{{\text{c}}_{4}})\cdot {{b}_{2}}-({{b}_{3}}\text{+}{{\text{b}}_{4}})\cdot {{c}_{2}}}{({{\text{a}}_{3}}\text{+}{{\text{a}}_{4}})\cdot {{b}_{2}}-({{b}_{3}}\text{+}{{\text{b}}_{4}})\cdot {{a}_{2}}}\text{ (vi)} \\ & y=\frac{({{\text{a}}_{3}}\text{+}{{\text{a}}_{4}})\cdot {{c}_{2}}-({{c}_{3}}\text{+}{{\text{c}}_{4}})\cdot {{a}_{2}}}{({{\text{a}}_{3}}\text{+}{{\text{a}}_{4}})\cdot {{b}_{2}}-({{b}_{3}}\text{+}{{\text{b}}_{4}})\cdot {{a}_{2}}}\text{ (vii)} \\ \end{align} Clearly, from equations $(i),(ii),(iii),(iv),(v),(vi) and (vii)$, the first equation of the system is replaced by the sum of the two equations; the resulting system has the same solution as the original system. So, if the first equation of the system is replaced by the sum of the two equations, the resulting system has the same solution as the original system.