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

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:** \[ 2x - 5y = -5 \quad \text{and} \quad 14x + 5y = -5 \]