Problem 48 - 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. $$\left\\{\begin{array}{rr}-\frac{1}{10} x+\frac{1}{2} y= & 4 \\\2 x-10 y= & -80\end{array}\right.$$

Step-by-Step Solution

Answer

The system has infinitely many solutions: \((x, y) = (5t - 40, t)\).

Step 1: Eliminate Fractions

To start, eliminate the fractions in the first equation. Multiply the entire equation by 10 to get integer coefficients: \(-\frac{1}{10}x + \frac{1}{2}y = 4\) becomes \[-x + 5y = 40\]

Step 2: Write the System of Equations

Now, the system of equations looks like this:\(\begin{array}{c}-x + 5y = 40 \2x - 10y = -80\end{array}\)

Step 3: Multiply to Align Equations for Elimination

Multiply the first equation by 2 to help eliminate the \(x\) terms:\(\begin{array}{c}-2x + 10y = 80 \2x - 10y = -80\end{array}\)

Step 4: Add Equations to Eliminate Variable

Add the two equations from Step 3:\[(-2x + 10y) + (2x - 10y) = 80 + (-80)\]This simplifies to \(0 = 0\), which indicates that the system has infinitely many solutions.

Step 5: Express Solutions in Parameter Form

Choose one of the original equations to express in terms of a parameter. For instance, from \[-x + 5y = 40\] let \(y = t\).Then, solve for \(x\):\[\begin{align*}-x + 5t &= 40 \x &= 5t - 40\end{align*}\]

Step 6: Write the Solution as Ordered Pairs

With \(x = 5t - 40\) and \(y = t\), the solutions can be expressed as ordered pairs \((x, y) = (5t - 40, t)\) where \(t\) is any real number.

Key Concepts

Infinitely Many Solutions

When dealing with systems of linear equations, they can have three types of solutions: a unique solution, no solution, or infinitely many solutions. Infinitely many solutions occur when the system's equations represent the same line, so every solution on one curve is also on the other.
In our exercise, after performing operations to simplify the given equations, we were led to the statement \(0 = 0\), a true statement showing there are infinitely many solutions. This means the system does not constrain the variables to a single set of values but rather allows a range of values, forming a line of solutions.
To express these solutions, we use a parameter. For instance, by setting \(y = t\), we derive that \(x = 5t - 40\), which gives us the ordered pairs of solutions as \((5t - 40, t)\) for any real number \(t\). This representation helps visualize all the possibilities that satisfy the system.

Elimination Method

The elimination method is a powerful technique for solving systems of linear equations. The goal is to remove one variable by adding or subtracting the equations, which simplifies solving for the others.
Starting with our system, \(-\frac{1}{10} x + \frac{1}{2} y = 4\) and \(2x - 10y = -80\), we first simplify the equations to integer coefficients. Then, by aligning equations through multiplication, similar terms can be eliminated. In this case, we multiplied the first equation by 2, giving us \(-2x + 10y = 80\), which can be directly added to the second equation, \(2x - 10y = -80\), to eliminate \(x\).
With elimination, the resulting zero equation \(0 = 0\) suggested infinitely many solutions, guiding us to parametrize the solution set. It's a structured method, enabling focus on finding how variables relate instead of guessing values.

Fractions Elimination

Dealing with fractions in equations can sometimes complicate calculations. Thus, a helpful strategy is to clear them out by using a common multiple that turns all coefficients into whole numbers, making the system easier to manage.
In the initial system we tackled, the fractions were present in \(-\frac{1}{10} x + \frac{1}{2} y = 4\). To eliminate them, we multiplied the entire equation by 10, the least common multiple of the denominators used, transforming it into \(-x + 5y = 40\).
This simplification makes the next steps—like aligning and eliminating variables—more straightforward, as working with integers can prevent errors and speed up processing. Fraction elimination is particularly beneficial in ensuring accuracy and clarity throughout solving systems of equations.