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# More than Two Variables

So far, our visual and algebraic solving of balance puzzles has primarily involved two types of shapes, or variables. Over the next several questions, we will extend our thinking to three or more quantities.

If a circle has a weight of 4, what is the weight of a triangle?

## More than Two Variables

### Balancing Scales

# More than Two Variables

As in previous balance puzzles, we want to know how to convert a set of a balance scales into a series of algebraic equations.

Let's revisit the last problem, and write equations to represent the balance scales.

If squares are $$x,$$ triangles are $$y,$$ and circles are $$z,$$ then we get this system of equations: \begin{align} x + y &= z \\ 2y &= 2x + z \\3x &= y. \end{align}

## More than Two Variables

### Balancing Scales

# More than Two Variables

What is the weight of a circle?

## More than Two Variables

### Balancing Scales

# More than Two Variables

What is the weight of one triangle?

## More than Two Variables

### Balancing Scales

# More than Two Variables

Given the equations below, what is the value of $$z\,?$$ \begin{align} x&=3 \\ 3x + y &= 10 \\ 2x + 2y + z &= 12 \end{align}

## More than Two Variables

### Balancing Scales

# More than Two Variables

Which variable has the largest value, $$x, y,$$ or $$z\,?$$ \begin{align} 2x + y + z &= 11 \\ x + 3y + 2z &= 17 \\ 2x + 2y + 3z &= 17 \end{align}

## More than Two Variables

### Balancing Scales

# More than Two Variables

Given the system of equations below, how many of the variables have a negative value? \begin{align}a + b + c + d &= 8 \\ 2a + 3b + 2c + d &= 15 \\ 3a + 2b - c + d &= 20 \\ 5a - 4b + 3c + d &= 12 \end{align}

## More than Two Variables

### Balancing Scales

# More than Two Variables

If squares are $$x,$$ triangles are $$y,$$ and circles are $$z,$$ then we get this system of equations: \begin{align} x + y &= z \\ 2y &= 2x + z \\3x &= y. \end{align} Is it possible to determine the weight of one circle?

## More than Two Variables

### Balancing Scales

# More than Two Variables

As we saw in the previous problem, sometimes it is not possible to actually solve for unknown values in balance puzzles or equations. In these cases, we are able to see relationships between variables, such as the fact that two circles are equivalent to a square, but not actually determine the weight, or value, of each.

The previous problem is shown below. Notice that we do not know any weights for various shapes or combinations of shapes. Therefore, we are only able to determine relationships between shapes, not actual weights of individual shapes.

Here is another example of a system of equations that we cannot solve because we do not have enough information: \begin{align} 2x + y + z &= 8 \\ y - z &= 2. \end{align} We can only eliminate either $$y$$ or $$z$$ from the system, leaving us with many options for the values of the two remaining variables.

## More than Two Variables

### Balancing Scales

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