Heart of Algebra — solving systems of two linear equations by substitution.

Pick the easier equation to solve for one variable. Here, equation 2 is simpler, so solve for x:
x - y = 1
x = y + 1

Now you've got x in terms of y. Substitute this expression for x into the first equation:
2(y + 1) + 3y = 11
2y + 2 + 3y = 11
5y + 2 = 11
5y = 9
y = 9/5

Finally, substitute y = 9/5 back into x = y + 1:
x = 9/5 + 1 = 9/5 + 5/5 = 14/5

Your solution is x = 14/5, y = 9/5.

Most students learn substitution with nice equations like $y = 2x + 1$. Real problems aren't that clean. You'll often need to solve for a variable first, dealing with equations like $3x - 4y = 7$.

When neither equation has an isolated variable, choose strategically. Look for the variable with coefficient 1 or -1, or the one that creates the least messy fractions. From $3x - 4y = 7$, solving for $x$ gives $x = \frac{7 + 4y}{3}$. That's messier than solving $2y - x = 5$ for $x$: $x = 2y - 5$.

Here's your roadmap every single time:

Step 1: Pick an equation and isolate one variable. Look for the easiest variable to isolate — usually one with a coefficient of 1 or -1. If you see something like $y = 3x + 5$, you're golden. If everything has messy coefficients, pick the simplest-looking option.

Step 2: Substitute that expression into the other equation. Take what you found in Step 1 and plug it in wherever you see that variable in the second equation. You'll end up with one equation containing only one variable.

If you have:
- 3x + y = 7
- 2x - 4y = 10

The first equation is perfect because you can quickly get y = 7 - 3x. Don't overthink this step—go with what looks cleanest.

Sometimes both equations need work to isolate a variable. Pick the one with the smallest coefficient (like 2x instead of 5x) or no fractions if possible.

Using our example:
3x + y = 7
Subtract 3x from both sides: y = 7 - 3x

The substitution method follows four clear steps. First, you pick one equation and solve for one variable in terms of the other. Second, you substitute that expression into the other equation. Third, you solve the resulting single-variable equation. Fourth, you plug your answer back into either original equation to find the second variable.

Let's say you have this system:

When you solve for a variable, you're essentially undoing everything that's been done to it. Look at 3x + 7 = 22. The variable x has been multiplied by 3, then had 7 added to it. To isolate x, you reverse these operations in opposite order:

First, undo the addition: 3x + 7 - 7 = 22 - 7, so 3x = 15
Then, undo the multiplication: 3x ÷ 3 = 15 ÷ 3, so x = 5

The key insight? You always work backwards through the order of operations. If the variable was multiplied then something was added, you subtract first, then divide.

The substitution method works by solving one equation for one variable, then substituting that expression into the other equation. Choose the equation and variable that'll give you the simplest algebra. Look for coefficients of 1 or -1, or variables that are already isolated.

For example, if you have:
- 3x + y = 11
- 2x - y = 4

The first equation makes it easy to solve for y: y = 11 - 3x. That's your substitution expression.

When you see an equation like 3x + 7 = 22, your job is to get x by itself. Think backwards through the order of operations. What happened to x? First it got multiplied by 3, then 7 was added. To undo this, you'll subtract 7 first, then divide by 3.

Start with: 3x + 7 = 22
Subtract 7 from both sides: 3x = 15
Divide both sides by 3: x = 5

The key principle: whatever operation is "touching" your variable, do the opposite operation to both sides. Addition becomes subtraction, multiplication becomes division, and so on.

Look for the equation where isolating a variable looks easiest. Here, the second equation has x with a coefficient of 1, so solving for x will be cleanest. You'd get x = 2y - 8 without fractions.

If both equations look equally messy, pick the one with the smallest coefficients. You're going to be substituting this expression into the other equation, so simpler is better.

Let's say you solved the first equation for y and got y = 11 - 3x. Now you substitute this whole thing for every y in the second equation:

Let's see this with 3x + 5 = 14:
- Subtract 5 from both sides: 3x = 9
- Divide both sides by 3: x = 3

Always check your answer by substituting back: 3(3) + 5 = 9 + 5 = 14 ✓

For equations where the variable appears on both sides like 5x - 3 = 2x + 9, collect all x terms on one side first:
- Subtract 2x from both sides: 3x - 3 = 9
- Add 3 to both sides: 3x = 12
- Divide by 3: x = 4