## Elementary Linear Algebra 7th Edition

b.) The system is consistent according to the graph. c.) The solution is approximately $x = \frac{1}{2}, y = -\frac{1}{4}$. d.) By adding -1/4 times Equation 1 to Equation 2, we can then solve that $y = -\frac{1}{4}$. Through back substitution, we can solve that $x = \frac{1}{2}$. e.) Both solutions are identical, we can conclude that the system is consistent and the solution is $x = \frac{1}{2}, y = -\frac{1}{4}$.
b.) By graphing the equations, we can see that they intersect and are therefore consistent. d.) By adding $-\frac{1}{4}$ times Equation 1 to Equation 2, we can then eliminate the $x$ variable in Equation 2. $2x -8y = 3$ $0x + 3y = -\frac{3}{4}$ We can then conclude from the modified Equation 2, that $y = -\frac{1}{4}$. This value can then be substituted into Equation 1 to then solve for $x$. $2x-8*(-\frac{1}{4}) = 3$ $2x = 1$ $x = \frac{1}{2}$