Problem 14 - A Graphical Approach to Precalculus with Limits

Question

Solve each system by substitution. $$\begin{aligned}&7 x-y=-10\\\&3 y-x=10\end{aligned}$$

Step-by-Step Solution

Answer

The solution to the system is \( x = -1 \) and \( y = 3 \).

Step 1: Solve for y in the First Equation

Start by solving the first equation for \( y \). The first equation is \( 7x - y = -10 \). Add \( y \) to both sides to get \( 7x = y - 10 \). Then add 10 to both sides to isolate \( y \): \( y = 7x + 10 \).

Step 2: Substitute the Expression for y into the Second Equation

Now, substitute \( y = 7x + 10 \) into the second equation \( 3y - x = 10 \). This gives you \( 3(7x + 10) - x = 10 \).

Step 3: Simplify the Equation

Simplify the equation \( 3(7x + 10) - x = 10 \). First, distribute the 3: \( 21x + 30 - x = 10 \). Combine like terms to get \( 20x + 30 = 10 \).

Step 4: Solve for x

Subtract 30 from both sides to isolate terms with \( x \): \( 20x = 10 - 30 \). Simplify to get \( 20x = -20 \). Divide both sides by 20 to solve for \( x \): \( x = -1 \).

Step 5: Substitute x Back into Expression for y

Substitute \( x = -1 \) back into the equation \( y = 7x + 10 \). This yields \( y = 7(-1) + 10 \). Simplify to find \( y = -7 + 10 \), which simplifies to \( y = 3 \).

Step 6: Verify the Solution

Plug \( x = -1 \) and \( y = 3 \) back into the original equations to ensure they satisfy both equations. For the first equation, \( 7(-1) - 3 = -7 - 3 = -10 \). For the second equation, \( 3(3) - (-1) = 9 + 1 = 10 \). Both equations are satisfied, confirming the solution is correct.

Key Concepts

Substitution Method

The substitution method is a common technique used to solve systems of equations when trying to find where two equations intersect or share a solution point. Essentially, this approach entails solving one of the equations for a single variable.

• First, isolate a variable in one of the equations. This can either be done for \(x\) or \(y\), depending on what seems easier.
• Once a variable is isolated, substitute its expression into the other equation. This substitution means replacing the variable with an equivalent expression from the first equation.
• Solving the resulting equation gives you the value of one variable.
• Lastly, substitute back into the isolated expression to find the other variable.

This method is particularly helpful when the equations are not immediately simple to graph, or do not readily display a common intersection upon inspection.

Linear Equations

Linear equations are mathematical expressions involving variables that are raised to the power of one. These equations are essential in many fields of study as they represent relationships with constant rates of change.Here are key characteristics of linear equations:

• They appear in the format \( ax + by = c \), where \( a \), \( b \), and \( c \) are constants.
• The graph of a linear equation is a straight line, hence the term "linear."
• These equations can be combined into systems to find common solutions, where they intersect on a graph.

In our case, the system of equations provided consisted of two linear equations. Utilizing linear equations in a system allows for the determination of shared solutions through various methods like substitution.

Solution Verification

Solution verification ensures that the solution obtained actually satisfies the original system of equations. This step is crucial because it confirms the accuracy of your work.For verification:

• Substitute the found values back into both original equations. In this exercise, substituting \( x = -1 \) and \( y = 3 \) into both equations confirmed that they both hold true.
• Check each equation independently to ensure consistency. Both must equal the constant terms on the right side of the equation.

Rechecking your solutions not only confirms correctness but also boosts confidence in your computational skills. Missing or incorrect verification steps can lead to errors that might seem unnoticeable without careful review.