Problem 35 - Precalculus

Question

Solve each system of linear equations by elimination. $$\begin{aligned} &4 x-5 y=22\\\ &3 x+4 y=1 \end{aligned}$$

Step-by-Step Solution

Answer

The solution is \(x = 3\), \(y = -2\).

Step 1: Align the Equations

We start by aligning the given equations:\[ \begin{aligned} 4x - 5y &= 22 \ 3x + 4y &= 1 \end{aligned} \] Our goal is to eliminate one of the variables by adding or subtracting the equations.

Step 2: Equalize the Coefficients of y

To eliminate \(y\), let’s make the coefficients of \(y\) in both equations equal. The coefficients are \(-5\) and \(4\). We will multiply the first equation by \(4\) and the second equation by \(5\) to have equal coefficients for \(y\).\[ \begin{aligned} (4)(4x - 5y) &= (4)(22) \ (5)(3x + 4y) &= (5)(1) \end{aligned} \] which gives us: \[ \begin{aligned} 16x - 20y &= 88 \ 15x + 20y &= 5 \end{aligned} \]

Step 3: Add the Equations

Add the two new equations to eliminate \(y\):\[ \begin{aligned} 16x - 20y + 15x + 20y &= 88 + 5 \end{aligned} \]The \(y\) terms cancel, resulting in:\[ 31x = 93 \]

Step 4: Solve for x

Divide both sides of the equation by \(31\) to solve for \(x\):\[ x = \frac{93}{31} = 3 \]

Step 5: Substitute to Find y

Now that we have \(x = 3\), substitute this value back into one of the original equations to solve for \(y\). Using the equation \(3x + 4y = 1\):\[ 3(3) + 4y = 1 \]\[ 9 + 4y = 1 \]\[ 4y = 1 - 9 \]\[ 4y = -8 \]Finally, divide by \(4\):\[ y = \frac{-8}{4} = -2 \]

Step 6: Verification

Verify the solution \((x, y) = (3, -2)\) by substituting into the other original equation:\[ 4x - 5y = 22 \]Substitute \(x = 3\), \(y = -2\):\[ 4(3) - 5(-2) = 12 + 10 = 22 \]Both sides are equal, so the solution is verified.

Key Concepts

Systems of Linear Equations

Systems of linear equations are sets of equations where each involves the same set of variables. In the given exercise, there are two equations:

• Equation 1: \( 4x - 5y = 22 \)
• Equation 2: \( 3x + 4y = 1 \)

These two equations form a system because they share the same variables, \(x\) and \(y\). The goal is to find the values of these variables that satisfy both equations simultaneously.
To solve such systems, various methods like substitution, elimination, or graphing can be used. In this case, we use the elimination method. This method involves combining the equations to "eliminate" one of the variables, reducing the system to a single equation with one unknown. This is particularly useful when trying to solve systems with a reasonable level of complexity.
Once one variable is eliminated, we solve for the remaining variable and then back-substitute to find the other. Understanding and solving systems of linear equations is a fundamental part of algebra that forms the base for more advanced topics like linear algebra.

Solution Verification

After finding the solution to a system of linear equations, it is crucial to verify it. Verification ensures that the calculated values for the variables are correct and satisfy all given equations.To verify a solution, plug the values of \(x\) and \(y\) back into both of the original equations.

• Using \( (x, y) = (3, -2) \), substitute into the first equation: \[ 4(3) - 5(-2) = 12 + 10 = 22 \]Both sides equal, so it's correct.
• Next, check using the second equation: \[ 3(3) + 4(-2) = 9 - 8 = 1 \]Again, both sides equal, confirming the solution.

By confirming the solution satisfies both equations, one ensures the accuracy and reliability of the solution process. Verification is essential because even small calculation errors can lead to incorrect conclusions, hence it solidifies our confidence in the results.

Coefficients

Coefficients are the numerical or constant parts of a term in an equation that multiply the variables. In our system of linear equations:

• For the term involving \(x\) in the first equation \(4x - 5y = 22\), 4 is the coefficient of \(x\).
• For the term involving \(y\), \(-5\) is the coefficient of \(y\).
• Similarly, in the second equation \(3x + 4y = 1\), 3 is the coefficient of \(x\) and 4 is the coefficient of \(y\).

During the elimination process, we manipulate these coefficients to make one of the variables' coefficients identical or additive inverses in both equations, enabling the subtraction or addition of the equations to eliminate that variable.
A strong grasp of coefficients is vital as they play a key role in the elimination method. Being able to manipulate and understand coefficients helps in simplifying equations and finding solutions efficiently.