Solving Systems by Substitution

Free sample questions, a clear explanation, and 5 practice skills with an AI tutor that guides without giving the answer away.

Concept Review

Solving Systems by Substitution: The Mathematical Switch

Imagine you're at a restaurant where the menu only shows combo meals, but you want to know the individual price of a burger and fries. If the server tells you "a burger costs $3 more than fries" and "a combo meal costs $9," how would you figure out each price? This is exactly what solving systems by substitution helps us do.

A system of equations is like having two different clues about the same mystery. When one equation is already solved for a variable (like y = 2x + 1), we can use substitution—essentially making a "switch" by replacing that variable in the other equation.

The Substitution Switch in Action

Let's solve a concrete example:

System:

y = 2x + 1 (Equation 1 - already solved for y)

3x + y = 11 (Equation 2)

Step 1:

Since y = 2x + 1, substitute this expression for y in equation 2:

3x + (2x + 1) = 11

Step 2:

Solve for x: 5x + 1 = 11 → 5x = 10 → x = 2

Step 3:

Substitute x = 2 back into y = 2x + 1:

y = 2(2) + 1 = 5

🔑 The "Already Solved" Advantage

Here's the counterintuitive part: when one equation is already solved for a variable, the hard work is already done for you. Many students think this makes the problem harder, but it actually makes it easier!

Think of it like having a translator ready—you don't need to isolate the variable because someone already did that step.

Back to the Restaurant

Remember our burger and fries problem? We can write it as:

• b = f + 3 (burger costs $3 more than fries)

• b + f = 9 (combo costs $9)

Substituting: (f + 3) + f = 9 → 2f + 3 = 9 → f = $3, so b = $6

🎯 Key Takeaway

Just like figuring out individual menu prices from combo deals, substitution lets us solve for unknown values when we have the right clues. When one equation is already solved for a variable, we have our "translator" ready—we just need to make the switch and solve.

Sample questions

1. Solve the system: y = 2x + 3 and y = 3x + 1

✓ (2, 7)

○ (1, 5)

○ (0, 3)

○ (3, 9)

Answer: (2, 7) — Set 2x + 3 = 3x + 1 → 3 - 1 = 3x - 2x → 2 = x, then y = 2(2)+3 = 7.

2. Solve: y = 4x - 2 and y = x + 7

○ (2, 6)

○ (4, 14)

✓ (3, 10)

○ (1, 2)

Answer: (3, 10) — 4x - 2 = x + 7 → 3x = 9 → x = 3, y = 3 + 7 = 10.

3. A system is given as x = 2y + 1 and 3x + y = 10. What is the solution?

○ (1, 0)

○ (5, 2)

○ (7, 3)

✓ (3, 1)

Answer: (3, 1) — Substitute x: 3(2y+1) + y = 10 → 6y+3+y=10 → 7y=7 → y=1, x=2(1)+1=3.

Skills in this topic

Solve a system of equations where one equation is already solved for a variable
Isolate a variable in one equation to solve the system by substitution
Solve a system of equations by substitution where the solution contains fractions or decimals
Identify no solution or infinite solutions algebraically during the substitution process
Check the algebraic solution to a system by substituting the values into both original equations

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