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You have two simultaneous equations:

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

### Step 1: Solve one equation for one variable
Let’s solve the second equation for \( y \):

\[
2x + y = -3 \quad \Rightarrow \quad y = -3 - 2x
\]

### Step 2: Substitute into the first equation
Now substitute \( y = -3 - 2x \) into the first equation:

\[
4x - 3(-3 - 2x) = 14
\]

Simplify:

\[
4x + 9 + 6x = 14
\]

\[
10x + 9 = 14
\]

\[
10x = 5
\]

\[
x = \frac{5}{10} = \frac{1}{2}
\]

### Step 3: Substitute \( x = \frac{1}{2} \) into the equation for \( y \)
Substitute \( x = \frac{1}{2} \) into \( y = -3 - 2x \):

\[
y = -3 - 2\left(\frac{1}{2}\right) = -3 - 1 = -4
\]

### Final Solution:
The solution to the system is:

\[
x = \frac{1}{2}, \quad y = -4
\]

Would you like any further details or explanations?

Here are 5 related questions to expand your understanding:

1. How do you graphically interpret the solution to a system of linear equations?
2. What methods other than substitution can be used to solve simultaneous equations?
3. What happens if the system of equations has no solutions or infinitely many solutions?
4. Can this system be solved using matrices or determinants?
5. How would the solution change if one of the equations was nonlinear?

**Tip:** Always check your solution by plugging the values of \( x \) and \( y \) back into both original equations.

Solve the simultaneous equations: 4x - 3y = 14 and 2x + y = -3

Learn how to solve a system of simultaneous linear equations: 4x - 3y = 14 and 2x + y = -3. This step-by-step solution walks through solving using the substitution method. Suitable for students in Grades 8-10.

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