Question

solve the following systems using the substitution method:

1. $x + y = 10$

$y = 2x + 1$

1. $x = 3y$

$x + 4y = 14$

1. $y = 3x + 1$

$x + 2y = 9$

1. $x = 2y$

$y = x + 4$

Explanation:

Problem 1:

Step1: Substitute \( y = 2x + 1 \) into \( x + y = 10 \)

We get \( x + (2x + 1) = 10 \)

Step2: Simplify and solve for \( x \)

\( x + 2x + 1 = 10 \)
\( 3x + 1 = 10 \)
\( 3x = 10 - 1 \)
\( 3x = 9 \)
\( x = \frac{9}{3} = 3 \)

Step3: Substitute \( x = 3 \) into \( y = 2x + 1 \)

\( y = 2(3) + 1 = 6 + 1 = 7 \)

Step1: Substitute \( x = 3y \) into \( x + 4y = 14 \)

We get \( 3y + 4y = 14 \)

Step2: Simplify and solve for \( y \)

\( 7y = 14 \)
\( y = \frac{14}{7} = 2 \)

Step3: Substitute \( y = 2 \) into \( x = 3y \)

\( x = 3(2) = 6 \)

Step1: Substitute \( y = 3x + 1 \) into \( x + 2y = 9 \)

We get \( x + 2(3x + 1) = 9 \)

Step2: Simplify and solve for \( x \)

\( x + 6x + 2 = 9 \)
\( 7x + 2 = 9 \)
\( 7x = 9 - 2 \)
\( 7x = 7 \)
\( x = \frac{7}{7} = 1 \)

Step3: Substitute \( x = 1 \) into \( y = 3x + 1 \)

\( y = 3(1) + 1 = 3 + 1 = 4 \)

Answer:

\( x = 3 \), \( y = 7 \)

Problem 2: