Solve the system of equations by substitution.

Question

left brace Start 2 By 1 Matrix 1st Row 1st Column x equals 5 y plus 10 2nd Row 1st Column 3 x plus 8 y equals negative 39 EndMatrix

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Part 1
The solution of the system is xequals
  
enter your response here and yequals
  
enter your response here.
​(Type integers or simplified​ fractions.)

Answer 1

To solve the system of equations by substitution, we'll start with the equations given:

1. \( x = 5y + 10 \) (Equation 1)
2. \( 3x + 8y = -39 \) (Equation 2)

We will substitute the expression for \( x \) from Equation 1 into Equation 2.

Step 1: Substitute \( x \) in Equation 2

Substituting \( x \) in Equation 2:

\[ 3(5y + 10) + 8y = -39 \]

Step 2: Distribute and simplify

Distributing the \( 3 \):

\[ 15y + 30 + 8y = -39 \]

Combine like terms:

\[ 23y + 30 = -39 \]

Step 3: Solve for \( y \)

To isolate \( y \), subtract \( 30 \) from both sides:

\[ 23y = -39 - 30 \]

\[ 23y = -69 \]

Now, divide both sides by \( 23 \):

\[ y = -3 \]

Step 4: Substitute \( y \) back to find \( x \)

Now that we have \( y \), we'll substitute it back into Equation 1 to find \( x \):

\[ x = 5(-3) + 10 \]

\[ x = -15 + 10 \]

\[ x = -5 \]

Conclusion

The solution to the system is:

\[ x = -5 \quad \text{and} \quad y = -3 \]

Let's enter the final answers:

• \( x = -5 \)
• \( y = -3 \)