For 5-8, Solve each System of Equations using the method (Substitution Method or Elimination Method)of your choice. 5. ) x=2-3y x+2y=0 6. 3x+6y=15 and 2x+4y=3

Let's solve each system of equations using the appropriate method.<br /><br />### Problem 5<br />Given the system:<br />\[ \left\{\begin{array}{l}x = 2 - 3y \\ x + 2y = 0\end{array}\right. \]<br /><br />We will use the **Substitution Method** for this problem.<br /><br />1. From the first equation, we already have \( x \) expressed in terms of \( y \):<br /> \[ x = 2 - 3y \]<br /><br />2. Substitute \( x = 2 - 3y \) into the second equation:<br /> \[ (2 - 3y) + 2y = 0 \]<br /><br />3. Simplify and solve for \( y \):<br /> \[ 2 - 3y + 2y = 0 \]<br /> \[ 2 - y = 0 \]<br /> \[ y = 2 \]<br /><br />4. Substitute \( y = 2 \) back into the first equation to find \( x \):<br /> \[ x = 2 - 3(2) \]<br /> \[ x = 2 - 6 \]<br /> \[ x = -4 \]<br /><br />So, the solution to the system is:<br />\[ (x, y) = (-4, 2) \]<br /><br />### Problem 6<br />Given the system:<br />\[ \left\{\begin{array}{l}3x + 6y = 15 \\ 2x + 4y = 3\end{array}\right. \]<br /><br />We will use the **Elimination Method** for this problem.<br /><br />1. First, let's multiply the second equation by 1.5 to make the coefficients of \( x \) in both equations equal:<br /> \[ 1.5(2x + 4y) = 1.5(3) \]<br /> \[ 3x + 6y = 4.5 \]<br /><br />2. Now we have the system:<br /> \[ \left\{\begin{array}{l}3x + 6y = 15 \\ 3x + 6y = 4.5\end{array}\right. \]<br /><br />3. Subtract the second equation from the first equation:<br /> \[ (3x + 6y) - (3x + 6y) = 15 - 4.5 \]<br /> \[ 0 = 10.5 \]<br /><br />This results in a contradiction, which means that there is no solution to this system of equations. The lines represented by these equations are parallel and do not intersect.<br /><br />So, the system has **no solution**.