Problem 31 - Thinking Mathematically

Question

Solve each system by the addition method. Be sure to check all proposed solutions. \(\left\\{\begin{array}{l}4 x+3 y=15 \\ 2 x-5 y=1\end{array}\right.\)

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = 33/14\) and \(y = 1\).

Step 1: Multiply both equations by necessary multiples to prepare for elimination

In order to eliminate one of the variables by adding the two equations, they need to be multiplied by suitable numbers. If we multiply the first equation by 2 and the second by 3 we have the following: \(8x + 6y = 30\) and \(6x - 15y = 3\).

Step 2: Add the two equations to eliminate one variable

Add these two equations to eliminate variable 'y'. This results in \(14x = 33\)

Step 3: Solve for x

Solving for 'x' in the resulting equation provides \(x = 33/14\)

Step 4: Substitute x into original equation

Substitute \(x = 33/14\) into the first original equation: \(4*(33/14) + 3y = 15\). Simplicity to find 'y'. This results in \(y = (15 - 12)/3 = 1\)

Step 5: Check the proposed solutions

Check the calculated values of 'x' and 'y' by substituting them back into both original equations. For the first equation, \(4*(33/14) + 3*1 = 15\) which holds true. For the second equation \(2*(33/14) - 5*1 = 1\) which also holds true.

Key Concepts

System of Equations

A system of equations is essentially a group of two or more equations that are interconnected and must be solved simultaneously. These equations share common variables.
For example, in our given problem, we have two equations:

• \(4x + 3y = 15\)
• \(2x - 5y = 1\)

Here, \(x\) and \(y\) are the shared variables.
The primary goal is to find a set of values for these variables that satisfy all equations in the system at the same time.
These systems can either have one solution, infinitely many solutions, or no solution at all, depending on the nature and arrangement of the equations.

Addition Method

The addition method, also called the elimination method, is a strategy used to solve systems of linear equations.
The basic idea is to manipulate the equations such that adding them together eliminates one of the variables.
In our given problem, we prepared the equations for elimination by multiplying them by specific factors.

• The first equation \(4x + 3y = 15\) is multiplied by 2, giving us \(8x + 6y = 30\).
• The second equation \(2x - 5y = 1\) is multiplied by 3, resulting in \(6x - 15y = 3\).

Now, you add these transformed equations, aiming to cancel out one of the variables. In this problem, our intention was to eliminate the \(y\) variable, making it easier to solve for \(x\).

Variable Elimination

Variable elimination refers to the process of removing one of the variables to simplify solving a system of equations.
In this method, we managed to remove the variable \(y\) by adding the two modified equations from the earlier step:
\(8x + 6y = 30\) and \(6x - 15y = 3\).
By adding, we get:

• \(8x + 6y + 6x - 15y = 30 + 3\)
• \(14x - 9y = 33\)
• \(14x = 33\) [since the \(y\) terms cancel each other out]

Now, solving this simplified equation allows us to find \(x\) easily. This step drastically simplifies the system, focusing the solution process on a single variable equation.

Solution Verification

Solution verification is a crucial step that involves ensuring the derived values of variables truly satisfy the original system of equations.
In our exercise, we found \(x = \frac{33}{14}\) and \(y = 1\).
To verify these values, substitute them back into the initial equations:

• First equation: \(4\left(\frac{33}{14}\right) + 3 \times 1 = 15\).
• Second equation: \(2\left(\frac{33}{14}\right) - 5 \times 1 = 1\).

Both equations hold true with these values, confirming the solution is accurate.
Solution verification not only confirms accuracy but also deepens one’s understanding of linear equations and their reliability.