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Solve the systems of equations by any convenient​ method, if a solution exists.

left brace Start 2 By 1 Matrix 1st Row 1st Column 3 x equals 7 y minus 1 2nd Row 1st Column 7 x equals 19 minus 5 y EndMatrix

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Part 1
What is the solution to the system of​ equations? Select the correct answer below​ and, if​ necessary, fill in any answer boxes to complete your choice.

1 answer

To solve the system of equations given in the matrix format:

  1. \( 3x = 7y - 1 \)
  2. \( 7x = 19 - 5y \)

Let's rearrange both equations in a standard form (Ax + By = C).

For the first equation:

\( 3x - 7y = -1 \)

For the second equation:

\( 7x + 5y = 19 \)

Now we have the following system:

  1. \( 3x - 7y = -1 \) (Equation 1)
  2. \( 7x + 5y = 19 \) (Equation 2)

Next, we can use either the substitution method or the elimination method to solve the system. We will use the elimination method here.

To eliminate \(x\), we will multiply Equation 1 by \(7\) and Equation 2 by \(3\):

  1. Multiply Equation 1 by \(7\): \[ 21x - 49y = -7 \]

  2. Multiply Equation 2 by \(3\): \[ 21x + 15y = 57 \]

Now we can write the new system of equations:

  1. \( 21x - 49y = -7 \)
  2. \( 21x + 15y = 57 \)

Next, subtract the first equation from the second equation to eliminate \(x\): \[ (21x + 15y) - (21x - 49y) = 57 - (-7) \] This simplifies to: \[ 64y = 64 \] Divide both sides by \(64\): \[ y = 1 \]

Now that we have \(y = 1\), we can substitute \(y\) back into one of the original equations to solve for \(x\). We'll use Equation 1: \[ 3x - 7(1) = -1 \] This simplifies to: \[ 3x - 7 = -1 \] \[ 3x = 6 \] \[ x = 2 \]

Final Solution

The solution to the system of equations is: \[ \boxed{(2, 1)} \]