Solving Systems of Equations

Learn to spot the two equations hiding inside any real-world setup.

Every system gives you two facts about two unknowns. Write those facts as equations, then solve. The Regents loves coin problems, pricing problems, and age problems — they look different but use the exact same method.

The trick is about setting each unknown as one half of an equation, then using one of the two methods below to find the answer of one equation by using the other!

Click here for some tips on systems of equations

Here are the two ways to solve a system of equations

Substitution: solve one equation for a variable, substitute into the other.

Elimination: add or subtract equations to cancel a variable. Multiply one first if needed.

Use whichever feels more natural — the Regents doesn't care which method you use.

⚠Label before you solve. Write "let n = nickels, let q = quarters" before writing any equations. Students who skip this often mix up variables midway through and lose points on an otherwise correct solution.

Worked example

Problem: Solve the system of equations:

y = 2x

x + y = 12

Substitution Method

Goal: When you have two equations with two unknowns, you need to use one equation to eliminate a variable from the other — leaving you with one equation and one unknown, which you already know how to solve.

1. Y is already isolated in equation (1), so we know y = 2x. Anywhere we see y in equation (2), we can replace it with 2x — now equation (2) only has x in it, which we can solve.

2. Now that we know x = 4, we can substitute it back into either equation. We now have only one unknown left (y), so we can isolate it and solve.

   
   
   
   

OR

Elimination Method

Goal: Find a variable that appears with the same coefficient in both equations. Subtracting one equation from the other will then cancel that variable out — leaving you with one equation and one unknown.

1. Both equations have y with a coefficient of 1, so we can subtract the top equation from the bottom equation directly.

2. The y terms cancel, leaving only x. Solve for x as normal.

3. Now that we know x = 4, substitute it back into either equation to find y.

Try it yourself! Below is a real Regents problem from January 2024

Problem: Jim had a bag of coins. The number of nickels, n, and quarters, q, totaled 28 coins. Their combined value was $4.00. Write a system of equations that models this situation. Algebraically determine both n and q.

Let's get started.

1. Type your next step below the problem, or try clicking a number and dragging it to the other side of the "=".

2. Press enter to add a new line of math 

3. Our AI checks your work as you go 🟢means you're on the right track, ❌means something's off.

4. Made a mistake? Use Ctl+Z to undo a line. Right click to see other shortcuts.

Click here to see the answer

STEP-BY-STEP SOLUTION

1. Write the system:

   n + q = 28

   
   

   0.05n + 0.25q = 4.00

2. Multiply the first equation by 0.05:

   0.05n + 0.05q = 1.40

3. Subtract from the second equation to eliminate n:

   0.20q = 2.60 → q = 13

4. Substitute back:

   n + 13 = 28 → n = 15

5. Check: 15(0.05) + 13(0.25) = 0.75 + 3.25 = $4.00 ✓

15 nickels and 13 quarters. Always verify — one check line can catch a careless error.

✗ COMMON TRAP

Writing 5n + 25q = 4 — mixing cents and dollars. If you use cents on the left, the right side must be 400, not 4.

✓ THE FIX

Choose one unit and stick to it. All dollars: 0.05n + 0.25q = 4. All cents: 5n + 25q = 400. Both work.

For more practice with systems of equations, check out these resources:

https://momentofmath.com/mom.html? (look under "systems of linear equations")