Question

Solve the following system of equations using the elimination method.

2x−3y=14
2
𝑥
−
3
𝑦
=
14
, 4x+3y=10

Answer 1

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 \( (4, -2) \).