Problem 28 - Algebra and Trigonometry

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 3. $$\left\\{\begin{aligned} u-30 v &=-5 \\\\-3 u+80 v &=5 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The system has a unique solution: \((25, 1)\).

Step 1: Identify the System of Equations

Identify the system of linear equations from the problem statement. The given system is: \( u - 30v = -5 \) and \( -3u + 80v = 5 \).

Step 2: Choose a Variable to Eliminate

To solve the system, choose one variable to eliminate by addition or subtraction. We can eliminate \(u\) by multiplying the first equation by 3, which will make the coefficients of \(u\) in both equations equal but opposite.

Step 3: Multiply to Prepare for Elimination

Multiply the first equation by 3:\[3(u - 30v) = 3(-5) \3u - 90v = -15\]Now the system is:\[3u - 90v = -15 \-3u + 80v = 5\]

Step 4: Add Equations to Eliminate \(u\)

Add the two equations together to eliminate \(u\):\[(3u - 90v) + (-3u + 80v) = -15 + 5 \0u - 10v = -10\] This simplifies to \(-10v = -10\).

Step 5: Solve for \(v\)

Solve the equation \(-10v = -10\) for \(v\) by dividing both sides by -10:\[v = 1\]

Step 6: Substitute Back to Find \(u\)

Substitute \(v = 1\) back into one of the original equations to find \(u\). Using the first equation \(u - 30v = -5\):\[u - 30(1) = -5 \u - 30 = -5 \u = 25\]

Step 7: Write the Solution as an Ordered Pair

The solution to the system of equations is \((u, v) = (25, 1)\). No other solutions exist, so the solution is unique.

Key Concepts

Elimination Method

The elimination method helps to solve a system of equations by removing one variable, simplifying the process. It's a powerful technique for efficiently handling linear equations. To use this method, follow these primary steps:

• Select one variable to eliminate in both equations. This could be achieved by addition or subtraction.
• Modify one or both equations by multiplying them with suitable numbers, so that when you add or subtract the equations, one variable cancels out.

In our example problem, we want to eliminate the variable \(u\). By multiplying the first equation by 3, the coefficients of \(u\) become opposites (3 and -3) in the two equations. Adding them cancels \(u\) out, making it easier to solve for the remaining variable \(v\).

The elimination method is helpful because it often reduces a complex problem into a simpler one, making the solution straightforward and clear.

Linear Equations

Linear equations are equations of the first degree, which means each variable is raised to the power of one. They typically appear in a simple format like \(ax + by = c\). These equations represent straight lines when graphed on a coordinate plane. Here are some features of linear equations:

• They are characterized by constant coefficients, like the -30 and 80 in the example equations.
• These equations often combine to form a larger system that can be solved using different techniques, such as the elimination method.

In the case of our exercise, we have two linear equations involving the variables \(u\) and \(v\). By working through the problem using the elimination method, we can solve these linear equations for unique solutions. Linear equations form the backbone of the system, with each line intersecting at a particular point, representing the solution to the system.

Solving Systems of Equations

Solving a system of equations means finding values for the variables that satisfy all the involved equations simultaneously. Systems of equations can vary in complexity depending on the number and type of equations involved. Here are the main steps and considerations when solving a system of linear equations like the one in our exercise:

• Identify whether there's one solution (unique), no solutions, or infinitely many solutions.
• Utilize methods such as substitution, elimination, or graphing. The elimination method is typically efficient for linear equations.
• Always verify the solution by substituting the values back into the original equations.

In our exercise, after we applied elimination, the resulting equations were easy to solve. We found \(v = 1\), and by substituting back, \(u = 25\). Thus, the solution is \((u, v) = (25, 1)\), confirming the system has a unique solution. Solving systems effectively allows you to understand the relationship between variables in different equations.