\( \left. \begin{array} { c } { a + b = 0,45 \ldots 41 } \\ { - 3 a + b = - 0,15 \ldots 181 } \\ { a = ? } \\ { b = ? } \end{array} \right. \)

Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}a+b=0.4541\\-3a+b=-0.15181\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}a=0.4541-b\\-3a+b=-0.15181\end{array}\right.\) - step2: Substitute the value of \(a:\) \(-3\left(0.4541-b\right)+b=-0.15181\) - step3: Simplify: \(-1.3623+4b=-0.15181\) - step4: Move the constant to the right side: \(4b=-0.15181+1.3623\) - step5: Add the numbers: \(4b=1.21049\) - step6: Divide both sides: \(\frac{4b}{4}=\frac{1.21049}{4}\) - step7: Divide the numbers: \(b=\frac{121049}{400000}\) - step8: Substitute the value of \(b:\) \(a=0.4541-\frac{121049}{400000}\) - step9: Simplify: \(a=\frac{60591}{400000}\) - step10: Calculate: \(\left\{ \begin{array}{l}a=\frac{60591}{400000}\\b=\frac{121049}{400000}\end{array}\right.\) - step11: Check the solution: \(\left\{ \begin{array}{l}a=\frac{60591}{400000}\\b=\frac{121049}{400000}\end{array}\right.\) - step12: Rewrite: \(\left(a,b\right) = \left(\frac{60591}{400000},\frac{121049}{400000}\right)\) To solve the system of equations: 1. **Equations**: - \( a + b = 0.4541 \) - \( -3a + b = -0.15181 \) 2. **Using the results**: - The solutions for \( a \) and \( b \) are: - \( a = 0.1514775 \) - \( b = 0.3026225 \) Thus, the final answers are: - \( a \approx 0.1515 \) - \( b \approx 0.3026 \)