Question

A) To solve the system of linear equations by substitution, we can solve one equation for one variable and substitute it into the other equation.

From the first equation, we have:
\[ x = -11 - 3y \]
Substituting this value of \( x \) into the second equation:
\[ 3(-11 - 3y) + 2y = 30 \]
Simplifying:
\[ -33 - 9y + 2y = 30 \]
\[ -7y = 63 \]
\[ y = -9 \]
Substituting this value of \( y \) back into the first equation:
\[ x + 3(-9) = -11 \]
\[ x - 27 = -11 \]
\[ x = 16 \]
Therefore, the solution to the system of equations is \( x = 16 \) and \( y = -9 \).
b) To solve the second system of linear equations:
\[ \begin{array}{l} 2x - 3y = -4 \\ 4x + 5y = 18 \end{array} \]
We can use the method of elimination to solve this system.
Multiplying the first equation by 2, we get:
\[ 4x - 6y = -8 \]
Adding this equation to the second equation:
\[ (4x + 5y) + (4x - 6y) = 18 + (-8) \]
\[ 8x - y = 10 \]
Simplifying:
\[ y = 8x - 10 \]
Substituting this value of \( y \) into the first equation:
\[ 2x - 3(8x - 10) = -4 \]
\[ 2x - 24x + 30 = -4 \]
\[ -22x = -34 \]
\[ x = \frac{17}{11} \]
Substituting this value of \( x \) back into the equation for \( y \):
\[ y = 8\left(\frac{17}{11}\right) - 10 \]
\[ y = \frac{34}{11} - 10 \]
\[ y = -\frac{96}{11} \]
Therefore, the solution to the second system of equations is \( x = \frac{17}{11} \) and \( y = -\frac{96}{11} \).

Answer 1

The solution to the first system of equations is x = 16 and y = -9. The solution to the second system of equations is x = 17/11 and y = -96/11.

Step-by-step explanation:

a) To solve the system of linear equations by substitution, we can solve one equation for one variable and substitute it into the other equation. Solving the equations for x and y, we find that the solution is x = 16 and y = -9.

b) To solve the second system of equations using elimination, we can add or subtract the equations to eliminate one variable. By doing this, we find that the solution is x = 17/11 and y = -96/11.

Learn more about Solving Systems of Linear Equations