Let's find the weight of the circle labeled C. We have two scales. The first scale shows us that one square S S has a weight of three. The second scale shows one circle + 1 square balances with a weight of 7.
Since we know S is three, we can substitute that value. So C + 3 = 7.
This means C must be four.
Let's find the weight of a square.
The first scale shows one triangle has the same weight as two squares. We can write this as t = 2 s.
The second scale shows 1 square + 1 triangle = 12. We can substitute 2 s for t. That gives us s + 2 s = 12. Combining like terms, we get 3 s= 12. If three squares weigh 12, then each square must weigh four. This process is called substitution.
The first equation gave us the value of a triangle in terms of squares. We use that information to replace the triangle in the second equation.
Now that we know a square weighs four, let's find the weight of the triangle.
Looking at our first scale, we know that one triangle equals two squares. Since s= 4, we can calculate that t = 2 * 4, which is 8.
Let's look at a different problem. The first scale shows that two triangles equal 5 circles. So 2 t = 5 c.
The second scale shows one circle plus two triangles equals 24. Let's rewrite this second equation without using t. We can substitute 5 C in for 2T.
This gives us C + 5 C = 24. Combining the terms on the left side, we get 6 C.
Now we have a simple problem. Six circles equal a weight of 24.
To find the weight of one circle, we divide 24 by 6. That tells us C is 4.
Finally, let's find the weight of the triangle T. Knowing that C is 4, we can use our first equation 2T = 5 C. By substituting 4 for C, we get 2T = 5 * 4, which is 20.
If 2t= 20, then a single t must be 10.
This is the power of substitution. By replacing a variable with an equivalent value from another equation, we can create simpler problems that are easier to solve.
