Problem 19 - Algebra and Trigonometry Real Mathematics, Real People

Question

Solve the system by the method of elimination and check any solutions algebraically. $$\left\\{\begin{array}{l} 5 u+6 v=24 \\ 3 u+5 v=18 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution of the system of equations is \( u = 12 / 7 \) and \( v = 54 / 35 \).

Step 1 : Multiply both equations by necessary multiples

We aim to eliminate one of the variables. Let's eliminate \( v \). Multiply the first equation by 5 and the second by 6. \n 5*(5u + 6v) = 5*24 and 6*(3u + 5v) = 6*18 getting us \n 25u + 30v = 120 and 18u + 30v = 108

Step 2 : Subtract the second equation from the first

Subtract the second resulting equation from the first to eliminate \( v \). \n (25u + 30v) - (18u + 30v) = 120 - 108 \n this yields \n 7u = 12

Step 3: Solve for the first variable

To find the value of \( u \), divide both sides of the equation from Step 2 by 7 \n u = 12 / 7 \n So, u = 12 / 7.

Step 4: Substitute u in one of the original equations

Now, substituting \( u = 12/7 \) in the second original equation to find \( v \): \n 3*(12/7) + 5v = 18 \n which simplifies to 5v = 18 - 36/7 = 90/7 - 36/7 = 54/7, \n and solving for \( v \), we have \( v = (54/7) / 5 = 54/35 \n thus, v = 54/35.

Step 5: Check the solutions

Substitute \( u = 12 /7 \) and \( v = 54 / 35 \) into original equations and check if both left sides equals to right sides. The check confirms that the solutions are correct.

Key Concepts

Understanding Systems of Equations

A system of equations is a collection of two or more equations with a set of unknowns. In mathematics, solving these equations requires finding values for the unknowns that satisfy all the equations at the same time.
In the exercise given, we encountered two equations:

• Equation 1: \(5u + 6v = 24\)
• Equation 2: \(3u + 5v = 18\)

These equations form a system because they share the same variables, \(u\) and \(v\). A solution to this system is a pair of numbers that make both equations true. Finding such a solution involves using various techniques like substitution or elimination.

The Concept of Variable Elimination

Variable elimination is a strategic approach to solving systems of equations. The goal is to make one of the variables cancel out, allowing us to solve for the remaining variable more easily.
In our exercise, we chose to eliminate the variable \(v\) by manipulating the equations. We achieved this by:

• Multiplying the first equation by 5, making it \(25u + 30v = 120\)
• Multiplying the second by 6, resulting in \(18u + 30v = 108\)

This method, known as the elimination method, relies on the principle of adding or subtracting equations after aligning coefficients of one of the variables, leading to its cancellation. In this instance, subtracting the modified second equation from the first leaves us with one equation in \(u\).

Solving Algebraically

After successfully eliminating a variable using the elimination method, the next step is to solve the remaining equation algebraically.
In our given problem, by cancelling out \(v\), we derived the equation \(7u = 12\). Solving this is straightforward:

• Divide both sides by 7, giving \(u = \frac{12}{7}\).

Now, with \(u\) known, we substitute it back into one of the original equations to find \(v\). Utilizing Equation 2: \(3u + 5v = 18\), plugging in \(u = \frac{12}{7}\) leads us to a new expression to solve for \(v\):

• Solve for \(v\) to find \(v = \frac{54}{35}\).

Thus, we have our values for both unknowns.

Checking Solutions Algebraically

Once solutions are found, it’s crucial to verify they are correct, especially in exams or real-world applications. This process, known as checking algebraically, confirms the solutions satisfy all given equations.
Let's substitute \(u = \frac{12}{7}\) and \(v = \frac{54}{35}\) back into the original equations:

• For Equation 1: \(5u + 6v = 24\), substitute: \(5(\frac{12}{7}) + 6(\frac{54}{35})\) should equal 24.
• For Equation 2: \(3u + 5v = 18\), substitute: \(3(\frac{12}{7}) + 5(\frac{54}{35})\) should equal 18.

In both cases, simplifying the left sides should verify that each equals the right side of the equation. This confirmation process validates that the solutions are not only mathematically correct but also applicable to the original equations, ensuring they work in context.