Problem 42 - 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. $$\left\\{\begin{array}{rr}u-30 v= & -5 \\\\-3 u+80 v= & 5\end{array}\right.$$

Step-by-Step Solution

Answer

The solution is \((25, 1)\).

Step 1: Express One Variable in Terms of the Other

From the first equation, solve for \( u \) in terms of \( v \): \( u = 30v - 5 \). This expresses \( u \) in terms of \( v \).

Step 2: Substitute the Expression into the Second Equation

Substitute the expression for \( u \) from Step 1 into the second equation: \[-3(30v - 5) + 80v = 5\] Simplify the equation to solve for \( v \).

Step 3: Simplify and Solve for v

Simplify the substituted equation:\[-90v + 15 + 80v = 5\]Combine like terms:\[-10v + 15 = 5\]Then solve for \( v \):\[-10v = 5 - 15\]\[-10v = -10\]\( v = 1 \).

Step 4: Solve for u Using the Value of v

Substitute \( v = 1 \) back into the expression for \( u \) from Step 1:\[ u = 30(1) - 5 \]\[ u = 30 - 5 \]\( u = 25 \).

Step 5: Write the Solution as an Ordered Pair

The solution to the system with the values found is an ordered pair where \( u = 25 \) and \( v = 1 \), so the solution is \((25, 1)\).

Key Concepts

Solving Linear Equations

Linear equations are fundamental in mathematics. They describe lines through algebraic equations and can often be written in the form of a system of equations. When you have a system of linear equations, like in our example, you are trying to find values for the variables that satisfy all the equations at the same time.
This means you want to find where the lines, represented by each equation, intersect. Intersections are key because the point where they meet is the solution.

Here's a simple breakdown of the process:

• Identify the variables in the equations. In our case, these are \( u \) and \( v \).
• Manipulate the equations to isolate one variable, making it easier to solve for the others.
• Use any consistent method to solve, like substitution or elimination.

Finding the solution involves checking that the values you obtain satisfy all the original equations.

Substitution Method

The substitution method is one effective technique for solving a system of linear equations. The goal is to substitute one variable with an expression involving the other variable, making it simpler to solve.
This method works best when one of the variables is easy to isolate in one of the equations.

In the given exercise:

• Start by solving the first equation \( u - 30v = -5 \) for \( u \). You get \( u = 30v - 5 \). This is your first step.
• Next, substitute this expression for \( u \) in the second equation \(-3u + 80v = 5 \). This gives you a new equation only in terms of \( v \).
• Once you solve for \( v \), use its value in the expression for \( u \) to find \( u \) itself.

The substitution method is particularly handy when equations lend themselves to easy manipulation, avoiding extra complexity.

Ordered Pairs

An ordered pair, like \((25, 1)\), represents a solution to the system of equations. They show the specific values of the variables that satisfy both equations simultaneously.
Think of ordered pairs as coordinates on a graph where each pair corresponds to a unique point of intersection between lines represented by the equations.

In various contexts:

• Ordered pairs can be considered solutions in mathematics where the outcome is dependent on two variables.
• In our solved system, the ordered pair \((25, 1)\) means that when \( u \) is 25 and \( v \) is 1, both equations hold true.
• This is shown by substituting back and verifying that both sides of the equations equal each other according to the ordered pair.

Expressing solutions as ordered pairs is a clear way to communicate the precise values that solve the system.