Question

part 2: solving by substitution:
solve the following systems using

1. ( x + y = 4 )

( y = 3x )

1. ( 2x ) (partially obscured)
2. ( 2y + x = 4 )

( 5x = 30 )

1. ( x = 5y - 4 )

( y + x = 20 ) (note: original has a typo-like ≠ but likely =)

Explanation:

Problem 1:

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

We know that \( y \) is equal to \( 3x \), so we can replace \( y \) in the first equation with \( 3x \). This gives us \( x + 3x = 4 \).

Step2: Simplify and solve for \( x \)

Combining like terms, \( x + 3x = 4x \), so the equation becomes \( 4x = 4 \). Dividing both sides by 4, we get \( x = \frac{4}{4} = 1 \).

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

Now that we know \( x = 1 \), we can find \( y \) by plugging \( x = 1 \) into the equation \( y = 3x \). So \( y = 3(1) = 3 \).

Step1: Solve \( 5x = 30 \) for \( x \)

We can solve the second equation \( 5x = 30 \) for \( x \) by dividing both sides by 5. So \( x = \frac{30}{5} = 6 \).

Step2: Substitute \( x = 6 \) into \( 2y + x = 4 \)

Now that we know \( x = 6 \), we substitute this value into the first equation \( 2y + x = 4 \). This gives us \( 2y + 6 = 4 \).

Step3: Solve for \( y \)

Subtract 6 from both sides of the equation \( 2y + 6 = 4 \): \( 2y = 4 - 6 = -2 \). Then divide both sides by 2: \( y = \frac{-2}{2} = -1 \).

Step1: Substitute \( x = 5y - 4 \) into \( y + x = 20 \)

We know that \( x = 5y - 4 \), so we can replace \( x \) in the second equation \( y + x = 20 \) with \( 5y - 4 \). This gives us \( y + (5y - 4) = 20 \).

Step2: Simplify and solve for \( y \)

Combining like terms, \( y + 5y - 4 = 6y - 4 \), so the equation becomes \( 6y - 4 = 20 \). Add 4 to both sides: \( 6y = 20 + 4 = 24 \). Then divide both sides by 6: \( y = \frac{24}{6} = 4 \).

Step3: Substitute \( y = 4 \) back into \( x = 5y - 4 \)

Now that we know \( y = 4 \), we can find \( x \) by plugging \( y = 4 \) into the equation \( x = 5y - 4 \). So \( x = 5(4) - 4 = 20 - 4 = 16 \).

Answer:

The solution to the system is \( x = 1 \), \( y = 3 \) or the ordered pair \( (1, 3) \).

Problem 3: