Which system of equations has a solution of approximately $$ (-0.6,0.8) $$ ? $$ 2 x-5 y=-5 $$ and $$ 14 x+5 y=-5 $$ $$ 2 x-5 y=-5 $$ and $$ 11 x-5 y=15 $$ $$ 11 x-5 y=15 $$ and $$ 14 x+5 y=-5 $$ $$ 11 x-5 y=15 $$ and $$ 9 x+5 y=5 $$ Mark this and return

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Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}11x-5y=15\14x+5y=-5\end{array}\right.$$
- step1: Solve the equation:
  $$\left\{ \begin{array}{l}x=\frac{15+5y}{11}\14x+5y=-5\end{array}\right.$$
- step2: Substitute the value of $$x:$$
  $$14\times \frac{15+5y}{11}+5y=-5$$
- step3: Simplify:
  $$\frac{14\left(15+5y\right)}{11}+5y=-5$$
- step4: Multiply both sides of the equation by LCD:
  $$\left(\frac{14\left(15+5y\right)}{11}+5y\right)\times 11=-5\times 11$$
- step5: Simplify the equation:
  $$210+125y=-55$$
- step6: Move the constant to the right side:
  $$125y=-55-210$$
- step7: Subtract the numbers:
  $$125y=-265$$
- step8: Divide both sides:
  $$\frac{125y}{125}=\frac{-265}{125}$$
- step9: Divide the numbers:
  $$y=-\frac{53}{25}$$
- step10: Substitute the value of $$y:$$
  $$x=\frac{15+5\left(-\frac{53}{25}\right)}{11}$$
- step11: Calculate:
  $$x=\frac{2}{5}$$
- step12: Calculate:
  $$\left\{ \begin{array}{l}x=\frac{2}{5}\y=-\frac{53}{25}\end{array}\right.$$
- step13: Check the solution:
  $$\left\{ \begin{array}{l}x=\frac{2}{5}\y=-\frac{53}{25}\end{array}\right.$$
- step14: Rewrite:
  $$\left(x,y\right) = \left(\frac{2}{5},-\frac{53}{25}\right)$$
Solve the system of equations $$ 2 x-5 y=-5; 11 x-5 y=15 $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}2x-5y=-5\11x-5y=15\end{array}\right.$$
- step1: Solve the equation:
  $$\left\{ \begin{array}{l}x=\frac{-5+5y}{2}\11x-5y=15\end{array}\right.$$
- step2: Substitute the value of $$x:$$
  $$11\times \frac{-5+5y}{2}-5y=15$$
- step3: Simplify:
  $$\frac{11\left(-5+5y\right)}{2}-5y=15$$
- step4: Multiply both sides of the equation by LCD:
  $$\left(\frac{11\left(-5+5y\right)}{2}-5y\right)\times 2=15\times 2$$
- step5: Simplify the equation:
  $$-55+45y=30$$
- step6: Move the constant to the right side:
  $$45y=30+55$$
- step7: Add the numbers:
  $$45y=85$$
- step8: Divide both sides:
  $$\frac{45y}{45}=\frac{85}{45}$$
- step9: Divide the numbers:
  $$y=\frac{17}{9}$$
- step10: Substitute the value of $$y:$$
  $$x=\frac{-5+5\times \frac{17}{9}}{2}$$
- step11: Calculate:
  $$x=\frac{20}{9}$$
- step12: Calculate:
  $$\left\{ \begin{array}{l}x=\frac{20}{9}\y=\frac{17}{9}\end{array}\right.$$
- step13: Check the solution:
  $$\left\{ \begin{array}{l}x=\frac{20}{9}\y=\frac{17}{9}\end{array}\right.$$
- step14: Rewrite:
  $$\left(x,y\right) = \left(\frac{20}{9},\frac{17}{9}\right)$$
Solve the system of equations $$ 11 x-5 y=15; 9 x+5 y=5 $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}11x-5y=15\9x+5y=5\end{array}\right.$$
- step1: Solve the equation:
  $$\left\{ \begin{array}{l}x=\frac{15+5y}{11}\9x+5y=5\end{array}\right.$$
- step2: Substitute the value of $$x:$$
  $$9\times \frac{15+5y}{11}+5y=5$$
- step3: Simplify:
  $$\frac{9\left(15+5y\right)}{11}+5y=5$$
- step4: Multiply both sides of the equation by LCD:
  $$\left(\frac{9\left(15+5y\right)}{11}+5y\right)\times 11=5\times 11$$
- step5: Simplify the equation:
  $$135+100y=55$$
- step6: Move the constant to the right side:
  $$100y=55-135$$
- step7: Subtract the numbers:
  $$100y=-80$$
- step8: Divide both sides:
  $$\frac{100y}{100}=\frac{-80}{100}$$
- step9: Divide the numbers:
  $$y=-\frac{4}{5}$$
- step10: Substitute the value of $$y:$$
  $$x=\frac{15+5\left(-\frac{4}{5}\right)}{11}$$
- step11: Calculate:
  $$x=1$$
- step12: Calculate:
  $$\left\{ \begin{array}{l}x=1\y=-\frac{4}{5}\end{array}\right.$$
- step13: Check the solution:
  $$\left\{ \begin{array}{l}x=1\y=-\frac{4}{5}\end{array}\right.$$
- step14: Rewrite:
  $$\left(x,y\right) = \left(1,-\frac{4}{5}\right)$$
Solve the system of equations $$ 2 x-5 y=-5; 14 x+5 y=-5 $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}2x-5y=-5\14x+5y=-5\end{array}\right.$$
- step1: Solve the equation:
  $$\left\{ \begin{array}{l}x=\frac{-5+5y}{2}\14x+5y=-5\end{array}\right.$$
- step2: Substitute the value of $$x:$$
  $$14\times \frac{-5+5y}{2}+5y=-5$$
- step3: Simplify:
  $$-35+40y=-5$$
- step4: Move the constant to the right side:
  $$40y=-5+35$$
- step5: Add the numbers:
  $$40y=30$$
- step6: Divide both sides:
  $$\frac{40y}{40}=\frac{30}{40}$$
- step7: Divide the numbers:
  $$y=\frac{3}{4}$$
- step8: Substitute the value of $$y:$$
  $$x=\frac{-5+5\times \frac{3}{4}}{2}$$
- step9: Calculate:
  $$x=-\frac{5}{8}$$
- step10: Calculate:
  $$\left\{ \begin{array}{l}x=-\frac{5}{8}\y=\frac{3}{4}\end{array}\right.$$
- step11: Check the solution:
  $$\left\{ \begin{array}{l}x=-\frac{5}{8}\y=\frac{3}{4}\end{array}\right.$$
- step12: Rewrite:
  $$\left(x,y\right) = \left(-\frac{5}{8},\frac{3}{4}\right)$$
Let's analyze the solutions for each system of equations:

1. **System 1:**
   $$
   2x - 5y = -5 \
   14x + 5y = -5
   $$
   The solution is approximately $$ (x,y) = (-0.625, 0.75) $$.

2. **System 2:**
   $$
   2x - 5y = -5 \
   11x - 5y = 15
   $$
   The solution is approximately $$ (x,y) = \left(\frac{20}{9}, \frac{17}{9}\right) $$ or $$ (2.22, 1.89) $$.

3. **System 3:**
   $$
   11x - 5y = 15 \
   14x + 5y = -5
   $$
   The solution is approximately $$ (x,y) = \left(\frac{2}{5}, -\frac{53}{25}\right) $$ or $$ (0.4, -2.12) $$.

4. **System 4:**
   $$
   11x - 5y = 15 \
   9x + 5y = 5
   $$
   The solution is approximately $$ (x,y) = (1, -0.8) $$.

Now, comparing these results with the target solution of approximately $$ (-0.6, 0.8) $$:

- The first system gives $$ (-0.625, 0.75) $$, which is close to $$ (-0.6, 0.8) $$.
- The other systems do not yield a solution close to $$ (-0.6, 0.8) $$.

Thus, the system of equations that has a solution closest to $$ (-0.6, 0.8) $$ is:

**System 1:**
$$
2x - 5y = -5 \quad \text{and} \quad 14x + 5y = -5
$$
The system of equations $$ 2x - 5y = -5 $$ and $$ 14x + 5y = -5 $$ has a solution closest to $$ (-0.6, 0.8) $$.

Which system of equations has a solution of approximately \( (-0.6,0.8) \) ? \( 2 x-5 y=-5 \) and \( 14 x+5 y=-5 \) \( 2 x-5 y=-5 \) and \( 11 x-5 y=15 \) \( 11 x-5 y=15 \) and \( 14 x+5 y=-5 \) \( 11 x-5 y=15 \) and \( 9 x+5 y=5 \) Mark this and return

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