Problem 50 - Precalculus: Mathematics for Calculus

Question

Solve the system, or show that it has no solution. If the system has infinitely many solutions, express them in the ordered-pair form given in Example 6. $$\left\\{\begin{aligned} -\frac{1}{10} x+\frac{1}{2} y &=4 \\ 2 x-10 y &=-80 \end{aligned}\right.$$

Step-by-Step Solution

Answer

Infinitely many solutions: \((x, y) = (5y - 40, y)\).

Step 1: Clear the Fractions

To begin solving the system, let's get rid of the fractions in the first equation. Multiply every term in the first equation by 10 (the least common denominator of the fractions) to eliminate the fractions: \[-\frac{1}{10}x + \frac{1}{2}y = 4\] becomes \[-x + 5y = 40.\] So the system is now:\[\begin{aligned} -x + 5y &= 40 \ 2x - 10y &= -80 \end{aligned}\]

Step 2: Solve One Equation for x

Let's solve the first equation for \(x\). Rearrange the equation \(-x + 5y = 40\) to express \(x\) in terms of \(y\):\[-x = -5y + 40.\] Thus, \[x = 5y - 40.\]

Step 3: Substitute for x in the Second Equation

Using the expression we found for \(x\) from Step 2, substitute \(x = 5y - 40\) into the second equation \(2x - 10y = -80\):\[2(5y - 40) - 10y = -80.\]This becomes \[10y - 80 - 10y = -80.\]

Step 4: Simplify and Check

Simplify the expression from Step 3:\[10y - 10y - 80 = -80.\]This reduces to \[-80 = -80,\] which is always true. This means the two equations are consistent and dependent.

Step 5: Express the Solution Set

Since the system has infinitely many solutions, we will express them in terms of the variable \(y\). Using \(x = 5y - 40\), the solutions can be written as ordered pairs:\[ (x, y) = (5y - 40, y). \] That is, for every value of \(y\), there is a corresponding value of \(x\) given by the expression \(x = 5y - 40\).

Key Concepts

Infinitely Many Solutions

When you solve a system of equations and end up with an identity, like \(-80 = -80\), it means the system has infinitely many solutions. This happens because the two equations represent the same line. As a result, every point on this line is a solution.

In practical terms, if your final equation simplifies to a true statement without variables, your equations are consistent. You can express the solution set in terms of one of the variables. For instance, in our exercise, we choose \(y\) and find \(x\) in terms of \(y\): \[ (x, y) = (5y - 40, y). \] This notation tells us that for any value of \(y\), there's a corresponding \(x\).

• The form "\(x = 5y - 40\)" gives you flexibility to find any solution along the line described by the system.
• To find a specific solution, pick a value for \(y\) and use the equation to find \(x\).

Eliminate Fractions

Fractions can make equations harder to work with, so eliminating them makes solving easier. Start by identifying a common denominator for all fractional terms in an equation. Multiply each term by this denominator to remove fractions.

In our exercise, the first equation has fractions \(-\frac{1}{10}x\) and \(\frac{1}{2}y\). The common denominator is 10. By multiplying the entire equation by 10, we convert fractions to whole numbers:

• \(-\frac{1}{10}x \times 10\) becomes \(-x\).
• \(\frac{1}{2}y \times 10\) becomes \(5y\).

This results in the simplified equation: \(-x + 5y = 40\).
Eliminating fractions not only simplifies calculations but also minimizes errors that often occur with fractions.

Substitution Method

The substitution method involves expressing one variable in terms of another, then substituting this expression into the other equation. This method is particularly useful when one equation is already solved for a variable.

In our problem, after simplifying, we have \(-x + 5y = 40\). We solve it for \(x\) and get: \[ x = 5y - 40. \]This expression is then substituted into the second equation to eliminate one of the variables.

Substituting \(x = 5y - 40\) into \(2x - 10y = -80\) allows us to work with a single variable, \(y\):
After substitution and simplification, we encounter \[-80 = -80,\] which confirms that the equations are consistent. Remember, when using substitution, always simplify the resulting equation fully and check the logic behind your substitution.

Dependent Equations

Dependent equations occur when two or more equations describe the same line in a coordinate plane. They are 'dependent' because one equation can be derived from the other by simple algebraic manipulation.

In the given problem, after eliminating fractions and performing substitution, we realize that any solution from one equation satisfies the other. The simplification leading to a true identity, such as \(-80 = -80\), is a strong indicator of dependent equations.
To identify dependent equations:

• Check if one equation is a scalar multiple of the other.
• After substitutions and simplifications, look for true equalities that hold for any values of the variables.

Such systems have infinitely many solutions because both equations are essentially the same line repeated, and every point on that line satisfies both equations.