Problem 25 - Intermediate Algebra

Question

Solve each system by elimination. See Examples 3 and 4 $$ \left\\{\begin{array}{l} 5 x+2 y=11 \\ 7 x+6 y=9 \end{array}\right. $$

Step-by-Step Solution

Answer

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

Step 1: Multiply First Equation

Multiply the entire first equation by 3 in order to eliminate the variable \( y \). This gives us: \[ 3(5x + 2y) = 3(11) \] Simplifying, we get:\[ 15x + 6y = 33 \].

Step 2: Subtract Equations

Subtract the second equation from the modified first equation to eliminate \( y \):\[ (15x + 6y) - (7x + 6y) = 33 - 9 \]Simplify the resulting equation:\[ 8x = 24 \].

Step 3: Solve for x

Divide both sides of the equation \( 8x = 24 \) by 8 to solve for \( x \):\[ x = \frac{24}{8} \]This simplifies to:\[ x = 3 \].

Step 4: Substitute x back into the First Equation

Now substitute \( x = 3 \) back into the original first equation: \[ 5x + 2y = 11 \] Substitute \( x = 3 \):\[ 5(3) + 2y = 11 \].

Step 5: Solve for y

Simplify and solve for \( y \):\[ 15 + 2y = 11 \]Subtract 15 from both sides:\[ 2y = 11 - 15 \] \[ 2y = -4 \]. Now, divide by 2:\[ y = \frac{-4}{2} = -2 \].

Step 6: Solution Verification

Check the values \( x = 3 \) and \( y = -2 \) in both original equations to ensure correctness:For the first equation:\[ 5x + 2y = 11 \]:\[ 5(3) + 2(-2) = 15 - 4 = 11 \].Correct!For the second equation:\[ 7x + 6y = 9 \]:\[ 7(3) + 6(-2) = 21 - 12 = 9 \].Correct!Values are verified.

Key Concepts

Elimination Method

The elimination method is a powerful technique to solve a system of linear equations. It involves combining two equations to eliminate one of the variables, making it easier to solve for the remaining variable. Here's the basic idea:

• Choose a variable to eliminate from one of the equations.
• Alter one or both equations so that the coefficients of the chosen variable are equal in magnitude but opposite in sign.
• Add or subtract the equations to eliminate the variable.
• Solve the resulting equation for the one remaining variable.

In the given problem, the system of equations is \( 5x + 2y = 11 \) and \( 7x + 6y = 9 \). We chose to eliminate \( y \) by multiplying the entire first equation by 3 so that the terms containing \( y \) are equal (\( 6y \) in both equations). The equations become \( 15x + 6y = 33 \) and \( 7x + 6y = 9 \). By subtracting the second equation from the first, the \( y \) terms cancel out, making it possible to solve directly for \( x \) in this simplified problem.
Elimination is useful when straightforward multiplication or addition can allow for quick cancellation, and it often simplifies to dealing with one variable at a time.

Solving Linear Equations

Solving linear equations involves finding the values that satisfy each equation simultaneously. Linear equations are algebraic expressions where each term is either a constant or the product of a constant and a single variable. Linear equations take the form \( ax + by = c \), and can appear in systems as in the current exercise.
The main goal when solving a system of linear equations is to find the intersection of the lines represented by the equations. This intersection represents the values of \( x \) and \( y \) that solve both equations at once. Techniques like the elimination method, substitution, and graphing can be used. Choosing the most efficient method depends on the specific problem.

• The Elimination Method simplifies by removing one variable through addition or subtraction.
• The Substitution Method replaces one variable with an expression from another equation.
• Graphing can help visualize solutions, but is often less precise when reactions are complex or values need exact solutions.

Understanding each equation's layout and managing arithmetic operations is crucial. For our system, once \( y \) was eliminated, we derived and solved a single equation for \( x = 3 \). With the value of \( x \) known, we back-calculated to find \( y = -2 \), ensuring that both solutions satisfy each equation.

Substitution Method

The substitution method is an alternative technique to solve systems of equations. It's particularly useful when one equation is easily rearranged to express a variable in terms of the other. Here's how it works:

• Solve one of the equations for one variable in terms of the other variable.
• Substitute this expression into the other equation.
• Solve the resulting single-variable equation.
• Once a value for one variable is found, substitute it back into one of the original equations to get the value of the other variable.

For our given system, substitution could be done by first solving the first equation for \( y \), giving \( y = \frac{11 - 5x}{2} \). Then, substitute this expression into the second equation, \( 7x + 6\left(\frac{11 - 5x}{2}\right) = 9 \). This equation only has one unknown - \( x \), which can be solved directly. After finding \( x \), substitute this value back into the expression for \( y \) to find its value.
Substitution is often best when one variable is already isolated or can be easily manipulated. While in this specific problem elimination was useful because both coefficients of \( y \) lined up smoothly, understanding substitution is key for a complete toolkit in solving linear equations. This method emphasizes understanding how solutions develop through manipulation and replacement within equations.