\begin{tabular}{l} 19. Solve using substitution. \\ \( \qquad \begin{array}{l}x+2 y=6 \\ x-3 y=-19\end{array} \) \\ \hline\end{tabular}
Real Tutor Solution
Answer
Solution
Sign in to Unlock Answers for Free!
A Learning Platform Trusted by Millions of Real Students and Teachers.
Reviewed and approved by the UpStudy tutoring team
Mind Expander
To solve the system of equations using substitution, let's start with the first equation: 1. \( x + 2y = 6 \) We can express \( x \) in terms of \( y \): \( x = 6 - 2y \). Now we substitute this expression for \( x \) into the second equation: 2. \( x - 3y = -19 \) Substitute \( x \): \( (6 - 2y) - 3y = -19 \). Simplifying gives: \( 6 - 5y = -19 \). Rearranging yields: \( -5y = -19 - 6 \) \( -5y = -25 \) So, \( y = 5 \). Next, we'll substitute \( y \) back to find \( x \): \( x = 6 - 2(5) = 6 - 10 = -4 \). Hence, the solution to the system of equations is \( x = -4 \) and \( y = 5 \). So the final answer is \( (x, y) = (-4, 5) \).