Problem 39 - Precalculus

Question

Solve each system of linear equations by elimination. $$\begin{array}{r} 3.4 x+1.7 y=8.33 \\ -2.7 x-7.8 y=15.96 \end{array}$$

Step-by-Step Solution

Answer

The solution is \(x = 4.20\), \(y = -3.50\).

Step 1: Align equations for elimination

The given system of equations is:\[ \begin{array}{r} 3.4x + 1.7y = 8.33 \ -2.7x - 7.8y = 15.96 \end{array} \]We will aim to eliminate one of the variables by making the coefficients of either \(x\) or \(y\) equal in both equations.

Step 2: Equalize coefficients of y by multiplication

Let's eliminate \(y\) by first making the coefficients of \(y\) in both equations equal in magnitude:- Multiply the first equation by 7.8, and - The second equation is multiplied by 1.7 to align the coefficients:\[ \begin{align*}3.4x \cdot 7.8 + 1.7y \cdot 7.8 &= 8.33 \cdot 7.8 \-2.7x \cdot 1.7 - 7.8y \cdot 1.7 &= 15.96 \cdot 1.7 \\end{align*} \]

Step 3: Simplify the equations

Perform the multiplications:- First Equation: \(26.52x + 13.26y = 65.034\)- Second Equation: \(-4.59x - 13.26y = 27.132\)

Step 4: Add the equations to eliminate y

Now add the two equations to eliminate \(y\):\[ \begin{align*}(26.52x + 13.26y) + (-4.59x - 13.26y) &= 65.034 + 27.132 \21.93x &= 92.166 \x &= \frac{92.166}{21.93}\end{align*} \]Calculate the value of \(x\).

Step 5: Solve for x

Compute the value of \(x\):\[ x = \frac{92.166}{21.93} \approx 4.20 \]

Step 6: Substitute x back to find y

Use the value of \(x\) to find \(y\). Substitute \(x = 4.20\) back into the first equation:\[ 3.4(4.20) + 1.7y = 8.33 \]

Step 7: Solve for y

Calculate the expression to find \(y\):\[ 14.28 + 1.7y = 8.33 \1.7y = 8.33 - 14.28 \1.7y = -5.95 \y = \frac{-5.95}{1.7} \]Compute the value of \(y\).

Step 8: Simplify y

Complete the division to find \(y\):\[ y = -3.50 \]

Key Concepts

Elimination Method

The elimination method is a technique used to solve systems of linear equations. It involves manipulating the equations to cancel out one of the variables, making it easier to solve for the remaining variable.
The basic steps in the elimination method include:

• Choosing which variable to eliminate.
• Aligning the coefficients of that variable in both equations to make them equal in magnitude.
• Adding or subtracting the equations to cancel the chosen variable out.

You often need to multiply one or both equations by certain numbers to achieve this alignment. In our problem, we eliminated the variable \(y\) by aligning their coefficients through multiplication, simplifying the process by leaving us with a single equation to solve for \(x\). After finding \(x\), we substitute it back into one of the original equations to find \(y\).
Using the elimination method effectively requires careful calculation, but it's a powerful technique for breaking down complex systems into simpler equations.

Linear Equations

Linear equations are mathematical expressions that describe straight lines. They typically have two variables and are in the form \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants.
In the given problem, we have the system of linear equations:

• \(3.4x + 1.7y = 8.33\)
• \(-2.7x - 7.8y = 15.96\)

Each equation represents a line on a graph. Solving the system means finding the point where these two lines intersect, which corresponds to the values of \(x\) and \(y\) that satisfy both equations simultaneously.
Linear equations can be solved using various methods, such as graphing, substitution, or elimination. The elimination method is particularly useful when the coefficients can be easily manipulated to cancel out one of the variables, as seen in this exercise.

Solving Equations

Solving equations, particularly in the context of linear systems, involves finding the values of the unknown variables that make all equations true. The solution should satisfy every equation in the system.
In our exercise:

• We used elimination to first solve for \(x\).
• Once we found \(x\), we substituted it back to find \(y\).
• The resulting values \(x = 4.20\) and \(y = -3.50\) were tested back in the original equations for verification.

When you solve these equations, double-check by plugging the solution back into the original equations to ensure they hold true. This method helps confirm that the point \((x, y)\) found is indeed the intersection of the two lines, thus solving the system correctly. Understanding the logic and mechanics of solving equations is crucial for successfully handling linear systems, whether you choose the elimination method or another solving strategy.