Answer to Question #311226 in Algebra for james

Question

A. Elimination Method; 1. 2x-5y = 9; 3x+4y = 6 2. 3x+4y= 3; 4x-5y=5

Answer 1

$1. \\ 2x-5y=9\\ 3x+4y=6\\ \left \{\begin{matrix} 4(2x-5y)=4\cdot9 \\ 5(3x+4y)=5\cdot6 \end{matrix}\right.\\ \left \{\begin{matrix} 8x-20y=36 \\ 15x+20y=30 \end{matrix}\right.\\ 23x=66\\ x=\frac{66}{23}\\ 2\cdot\frac{66}{23}-5y=9\\ 5y=\frac{132}{23}-9\\ 5y=-\frac{75}{23}\\ y=-\frac{15}{23}\\(\frac{66}{23},-\frac{15}{23})$

$2. \\ 3x+4y=3\\ 4x-5y=5\\ \left \{\begin{matrix} 5(3x+4y)=5\cdot3 \\ 4(4x-5y)=4\cdot5 \end{matrix}\right.\\ \left \{\begin{matrix} 15x+20y=15 \\ 16x-20y=20 \end{matrix}\right.\\ 31x=35\\ x=\frac{35}{31}\\ 3\cdot\frac{35}{31}+4y=3\\ 4y=3-\frac{105}{31}\\ 4y=-\frac{12}{31}\\ y=-\frac{3}{31}\\(\frac{35}{31},-\frac{3}{31})$