4x+6y=16

−7x+2y=−3

1 answer

To solve the system of equations:

  1. \( 4x + 6y = 16 \) (Equation 1)
  2. \( -7x + 2y = -3 \) (Equation 2)

We can use the method of substitution or elimination. Here, we'll use the elimination method.

First, we will manipulate the equations to align their coefficients.

Step 1: Align the coefficients

To eliminate \( y \), we can manipulate the equations such that the coefficients of \( y \) will cancel each other out. We can do this by making the coefficients of \( y \) in both equations the same.

Multiply Equation 1 by 1 and Equation 2 by 3:

  • \( 4x + 6y = 16 \) (Equation 1)
  • \( -21x + 6y = -9 \) (Equation 2 multiplied by 3)

Step 2: Subtract Equation 1 from Equation 2

Now we subtract Equation 1 from the modified Equation 2:

\[ (-21x + 6y) - (4x + 6y) = -9 - 16 \]

This simplifies to:

\[ -25x = -25 \]

Step 3: Solve for \( x \)

Divide both sides by -25:

\[ x = 1 \]

Step 4: Substitute \( x \) back into one of the original equations

Now we substitute \( x = 1 \) back into Equation 1 to find \( y \):

\[ 4(1) + 6y = 16 \] \[ 4 + 6y = 16 \] \[ 6y = 16 - 4 \] \[ 6y = 12 \] \[ y = 2 \]

Final Solution

The solution to the system of equations is: \[ x = 1, \quad y = 2 \]

Thus, the solution is \( (1, 2) \).