Problem 28 - Intermediate Algebra

Question

Solve each system by elimination. See Examples 3 and 4 $$ \left\\{\begin{array}{l} 2 x+3 y=31 \\ 3 x+2 y=39 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution to the system is \(x = 11, y = 3\).

Step 1: Multiply the Equations

To eliminate one of the variables, we first want the coefficients of either \(x\) or \(y\) to be the same. Let's eliminate \(y\). Multiply the first equation by 2 and the second equation by 3 to make the coefficients of \(y\) equal:\[\begin{align*}2(2x + 3y) &= 2(31) \3x + 2y &= 39 \3(3x + 2y) &= 3(39)\end{align*}\]

Step 2: Write the New System of Equations

Now rewrite the system with the new coefficients:\[\begin{array}{l}4x + 6y = 62 \9x + 6y = 117\end{array}\]

Step 3: Subtract the Equations

Subtract the first equation from the second to eliminate \(y\):\[(9x + 6y) - (4x + 6y) = 117 - 62\]This results in:\[5x = 55\]

Step 4: Solve for x

Divide both sides by 5 to solve for \(x\):\[x = \frac{55}{5} = 11\]

Step 5: Substitute to Find y

Substitute \(x = 11\) back into the first original equation to solve for \(y\):\[2(11) + 3y = 31\]\[22 + 3y = 31\]Subtract 22 from both sides:\[3y = 9\]

Step 6: Solve for y

Divide both sides by 3 to solve for \(y\):\[y = \frac{9}{3} = 3\]

Step 7: Solution Check

Substitute \(x = 11\) and \(y = 3\) back into the second original equation to ensure correctness:\[3(11) + 2(3) = 39\]\[33 + 6 = 39\]Since both sides of the equation equal, the solution \((x, y) = (11, 3)\) is verified.

Key Concepts

Elimination Method

The elimination method is a powerful technique for solving systems of equations. It's particularly useful when you need to solve for two variables, like in the system of equations:

• \(2x + 3y = 31\)
• \(3x + 2y = 39\)

The main goal with elimination is to cancel out one of the variables by making the coefficients of that variable the same in both equations.
To achieve this, you often multiply each equation by a certain number. For instance, in this problem, we multiplied the first equation by 2 and the second by 3 so that the coefficients of \(y\) become 6 in both equations.
This step allows us to eliminate \(y\) completely by subtracting one equation from the other. This is where the elimination method shines, as it reduces a system of equations into a single linear equation, making it much easier to solve.

Solving Linear Equations

Once you eliminate a variable using the elimination method, you're left with a simple linear equation. For example, after eliminating \(y\) in the previous step, you end up with:

• \(5x = 55\)

Solving a linear equation like this involves straightforward algebraic manipulation.
Here, you divide both sides by the coefficient of \(x\) (which is 5) to find the solution for \(x\).
In this instance, division gives you \(x = 11\). With \(x\) determined, you can then substitute this value back into one of the original equations to solve for \(y\).
This substitution helps complete the solution of the system by finding the values of both variables, \(x\) and \(y\). This process is essential in solving any system of linear equations.

Mathematical Verification

Mathematical verification is a crucial step to ensure your solution is accurate. After finding the values of \(x\) and \(y\), it's important to double-check by substituting them back into the original equations.
In our exercise, substituting \(x = 11\) and \(y = 3\) into the second equation gives:

• \(3(11) + 2(3) = 39\)

Simplifying this:

• \(33 + 6 = 39\)

The left side equals the right side, confirming that our solution \((11, 3)\) satisfies the original system.
Verification ensures there are no mistakes in arithmetic or algebraic manipulation.
It's a simple yet powerful tool for instilling confidence in your solution and demonstrating a solid understanding of solving systems of equations using elimination.