Solve the following system of equations using the elimination method.

2x−3y=14
2
𝑥

3
𝑦
=
14
, 4x+3y=10

1 answer

To solve the system of equations using the elimination method, we have the following equations:

  1. \( 2x - 3y = 14 \) (Equation 1)
  2. \( 4x + 3y = 10 \) (Equation 2)

Step 1: Align the equations

We have: \[ 2x - 3y = 14 \quad (1) \] \[ 4x + 3y = 10 \quad (2) \]

Step 2: Align the equations for elimination

We can eliminate \(y\) by adding the equations, but we first notice that we can modify Equation 1 to make the coefficients of \(y\) opposite.

We can multiply Equation 1 by 1 (to keep it the same for easy addition): \[ 2x - 3y = 14 \quad (1) \] Equation 2 is kept as is: \[ 4x + 3y = 10 \quad (2) \]

Step 3: Add the equations

Now we'll add the two equations to eliminate \(y\): \[ (2x - 3y) + (4x + 3y) = 14 + 10 \] This simplifies to: \[ 6x = 24 \]

Step 4: Solve for \(x\)

Now divide by 6: \[ x = 4 \]

Step 5: Substitute \(x\) back into one of the original equations

We can substitute \(x = 4\) back into Equation 1: \[ 2(4) - 3y = 14 \] This simplifies to: \[ 8 - 3y = 14 \] Now we isolate \(y\): \[ -3y = 14 - 8 \] \[ -3y = 6 \] \[ y = -2 \]

Solution

The solution to the system of equations is: \[ x = 4, \quad y = -2 \]

Thus, the solution is \( (4, -2) \).