Sure! Solving a system of equations using substitution involves solving one equation for one variable and then substituting that expression into the other equation. Let's go through a step-by-step process with an example.

Example System of Equations:

$1) \, 2x + y = 10$ $2) \, x - y = 3$

Step 1: Solve one equation for one variable.

Let's solve equation $(2)$ for $x$:

$x = y + 3$

Step 2: Substitute this expression into the other equation.

Now, substitute $x = y + 3$ into equation $(1)$:

$2(y + 3) + y = 10$

Step 3: Simplify and solve for $y$.

Distribute and combine like terms:

$2y + 6 + y = 10$ $3y + 6 = 10$ $3y = 10 - 6$ $3y = 4$ $y = \frac{4}{3}$

Step 4: Substitute $y = \frac{4}{3}$ back into the expression for $x$.

Now that we have $y$, substitute it back into $x = y + 3$:

$x = \frac{4}{3} + 3$ $x = \frac{4}{3} + \frac{9}{3}$ $x = \frac{13}{3}$

Final Solution:

The solution to the system of equations is: $x = \frac{13}{3}, \, y = \frac{4}{3}$

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Would you like further details or additional examples? Let me know if you have any questions!

Here are some related questions you might find useful:

1. How can you solve a system of equations by substitution when the coefficients are not integers?
2. What happens if you have more than two equations in the system?
3. How does substitution compare to the elimination method for solving systems?
4. Can substitution be used for non-linear systems (e.g., quadratics)?
5. What to do when the substitution results in an identity (like $0 = 0$)?

Tip: Always try to isolate a variable in a simpler equation for substitution, ideally the one that has the smallest coefficients!