Solve the system of linear equations by the method of elimination, find (x,y). $\left\{ \begin{array} { l }
- 0.06 x + 0.01 y = 0.03 \
- 0.09 x - 0.04 y = 0.09
\end{array} \right.$
A) $\left( - \frac { 11 } { 7 } , - \frac { 11 } { 9 } \right)$
B) $\left( - \frac { 14 } { 11 } , - \frac { 9 } { 11 } \right)$
C) $\left( - \frac { 7 } { 11 } , - \frac { 27 } { 11 } \right)$
D)No solution
E) $\left( - \frac { 7 } { 11 } , - \frac { 9 } { 11 } \right)$

Question

Solve the system of linear equations by the method of elimination, find (x,y).   $\left\{ \begin{array} { l } 
- 0.06 x + 0.01 y = 0.03 \
- 0.09 x - 0.04 y = 0.09
\end{array} \right.$  
A)  $\left( - \frac { 11 } { 7 } , - \frac { 11 } { 9 } \right)$
B) $\left( - \frac { 14 } { 11 } , - \frac { 9 } { 11 } \right)$
C)  $\left( - \frac { 7 } { 11 } , - \frac { 27 } { 11 } \right)$
D)No solution
E)  $\left( - \frac { 7 } { 11 } , - \frac { 9 } { 11 } \right)$

Quizplus · Accepted Answer

To solve the system of equations by elimination, we first multiply each equation by a number that will make the coefficients of either x or y the same in both equations. Multiplying the first equation by 4 and the second by 1 gives us:$$ -0.24x + 0.04y = 0.12 $$$$ -0.09x - 0.04y = 0.09 $$Adding these equations eliminates y, leading to:$$ -0.33x = 0.21 $$Solving for x gives $ x = -\frac{0.21}{0.33} = -\frac{7}{11} $.Substitute $ x = -\frac{7}{11} $ into one of the original equations to solve for y. Using the first equation:$$ -0.06(-\frac{7}{11}) + 0.01y = 0.03 $$$$ \frac{7}{110} + 0.01y = 0.03 $$$$ 0.01y = 0.03 - \frac{7}{110} $$$$ 0.01y = \frac{33}{110} - \frac{7}{110} $$$$ 0.01y = \frac{26}{110} $$$$ y = \frac{26}{110} \div 0.01 $$$$ y = \frac{26}{110} \times 100 $$$$ y = \frac{26}{1.1} $$$$ y = -\frac{9}{11} $$Thus, the solution is $ x = -\frac{7}{11}, y = -\frac{9}{11} $, which corresponds to choice E.

Solve the System of Linear Equations by the Method of Elimination