Solving Systems of Equations with Substitution

Let's figure out the unknown weights on this pair of scales. Each scale is perfectly balanced, meaning the total weight on the left side equals the total on the right. We have two unknown weights, a green square S, and a blue circle C. Our goal is to find the exact weight of each.

We can turn the top balance into an equation. The left side has two s squares and one one weight or 2 s + one.

The right side has a nine weight.

Because the scale is balanced, the two sides are equal. Our equation is 2 s + 1 = 9. Now let's do the same for the second scale. The left side has three C circles and a seven weight for a total of 3 C + 7. The right side has four s squares or 4 s. This gives us our second equation. 3 c + 7 = 4 s. We now have a system of equations. Since we have two different equations relating the same two unknowns, we can solve for both s and c. Let's solve the first equation for s. In 2 + 1 = 9, we need to isolate s. Subtract one from both sides. This leaves us with 2 S = 8. If 2 S squares weigh 8, then 1 S must weigh 4. Now that we know S= 4, we can use this in our second equation. We'll replace S in the second equation with its value 4. The right side is 4 S. So, we calculate 4 * 4, which is 16. The equation becomes 3 C + 7 = 16. Next, we solve for C. To isolate C, subtract 7 from both sides.

This leaves 3 C equals 9. If three C circles weigh 9, then each C must weigh 3. We found that S= 4 and C= 3. When you have two independent equations for two unknowns, you can find both values by solving one equation, then substituting that result into the other.