Problem 32 - 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. $$\left\\{\begin{array}{rr}0.2 x-0.2 y= & -1.8 \\\\-0.3 x+0.5 y= & 3.3\end{array}\right.$$

Step-by-Step Solution

Answer

The system has one solution: \((-6, 3)\).

Step 1: Eliminate Decimals

To make calculations easier, we first eliminate decimals by multiplying every term in the equations by 10. This gives us the new system: \[\begin{aligned}2x - 2y &= -18, \-3x + 5y &= 33.\end{aligned}\]

Step 2: Use the Elimination Method

We add the two equations to eliminate one of the variables. The first equation is \(2x - 2y = -18\) and the second is \(-3x + 5y = 33\). If we add 3 times the first equation to 2 times the second equation:\[\begin{aligned}(3)(2x - 2y) + (2)(-3x + 5y) &= (3)(-18) + (2)(33),\6x - 6y - 6x + 10y &= -54 + 66,\4y &= 12.\end{aligned}\]Solving for \(y\), we get \(y = 3\).

Step 3: Substitute Back to Find x

Substitute \(y = 3\) back into one of the original equations to find \(x\). Let's use the first equation \(2x - 2y = -18\):\[\begin{aligned}2x - 2(3) &= -18,\2x - 6 &= -18,\2x &= -12,\x &= -6.\end{aligned}\]

Step 4: Write the Solution in Ordered Pair Form

The solution to the system is \((x, y) = (-6, 3)\). This indicates that the only solution to the system is this ordered pair.

Key Concepts

Elimination Method

When solving a system of equations, one powerful technique is the elimination method. This method involves manipulating the equations to cancel one of the variables, making it easier to solve for the remaining variable.
Here's a simple guide:

• First, set up your equations so that they can be easily added or subtracted. This may involve multiplying the entire equation by a number to align the coefficients of a variable.
• Combine (add or subtract) the equations to eliminate one of the variables completely.
• Solve for the remaining variable.

In our original problem, we multiplied the equations to eliminate decimals and then added them strategically to get rid of one variable (in this case, by making sure the coefficients allowed a variable to vanish when combined). This step is crucial as it simplifies a two-variable system to a single variable equation, which is much easier to handle.

Ordered Pairs

An ordered pair is a way of denoting a particular solution to a system of equations and is typically written in the form \((x, y)\).
In the context of our exercise, after solving for values of `x` and `y`, we present them as an ordered pair which indicates the point of intersection of the given equations on a coordinate plane.
Here are some key points to understand ordered pairs:

• The first element of the pair corresponds to the `x` value, and the second element corresponds to the `y` value.
• It signifies a specific location in the Cartesian plane where both equations intersect.
• An ordered pair being the solution of a system confirms that it satisfies all the given equations.

In our scenario, the solution (-6, 3) is the ordered pair indicating that at `x = -6`, and `y = 3`, both equations are true. It is crucial as it not only solves the equations but represents a real-world coordinate that both equations agree upon.

Substitution Method

The substitution method is another technique to solve systems of equations. It involves solving one equation for a single variable and then substituting this expression into the other equation. Although it wasn't explicitly needed in the original solution due to the efficiency of the elimination method, understanding substitution adds another tool to your problem-solving toolkit.

• Solve one of the equations for one variable in terms of the other.
• Substitute this expression into the second equation. This yields an equation with only one variable.
• Solve this equation to find the value of one variable.
• Finally, substitute this value back into the rearranged original equation to solve for the other variable.

This method is particularly useful when one of the equations is easily solvable for a variable, or when working with non-linear systems. In exercises like this, the substitution method can provide clarity and a different approach to cross-check solutions obtained through elimination, ensuring accuracy and completeness in understanding systems of equations.