Problem 40 - Intermediate Algebra

Question

For Problems 19-48, solve each system by using either the substitution or the elimination-by-addition method, whichever seems more appropriate. (Objective 2) $$ \left(\begin{array}{rr} -2(x+2)+4(y-3) & =-34 \\ 3(x+4)-5(y+2) & =23 \end{array}\right) $$

Step-by-Step Solution

Answer

The solution is \(x = -3, y = -6\).

Step 1: Simplify the Equations

Let's begin by expanding both equations. For the first equation:\[-2(x + 2) + 4(y - 3) = -34\]Expanding yields:\[-2x - 4 + 4y - 12 = -34\]Simplifying gives:\[-2x + 4y - 16 = -34\]Adding 16 to both sides results in:\[-2x + 4y = -18\]For the second equation:\[3(x + 4) - 5(y + 2) = 23\]Expanding gives:\[3x + 12 - 5y - 10 = 23\]Simplifying results in:\[3x - 5y + 2 = 23\]Subtracting 2 from both sides gives:\[3x - 5y = 21\]

Step 2: Decide on the Method

The simplified system of equations is:\[-2x + 4y = -18\]\[3x - 5y = 21\]Given the structure, using the elimination method seems appropriate since the coefficients of \(x\) and \(y\) terms can be easily manipulated to cancel one variable.

Step 3: Eliminate a Variable

To eliminate \(x\), multiply the first equation by 3 and the second equation by 2:1. \[3(-2x + 4y) = 3(-18)\] yields \[-6x + 12y = -54\]2. \[2(3x - 5y) = 2(21)\] yields \[6x - 10y = 42\]Add these equations:\[-6x + 12y + 6x - 10y = -54 + 42\]Simplifying, the \(x\) terms cancel each other:\[2y = -12\]

Step 4: Solve for \(y\)

Divide both sides by 2:\[y = -6\]

Step 5: Substitute and Solve for \(x\)

Substitute \(y = -6\) back into one of the original equations, such as \(-2x + 4y = -18\):\[-2x + 4(-6) = -18\]Simplify:\[-2x - 24 = -18\]Add 24 to both sides:\[-2x = 6\]Divide both sides by -2:\[x = -3\]

Step 6: Verify the Solution

Substitute \(x = -3\) and \(y = -6\) into the second original equation \(3x - 5y = 21\):\[3(-3) - 5(-6) = 21\]Calculate:\[-9 + 30 = 21\]Both sides equal, confirming our solution is correct.

Key Concepts

Substitution Method

The substitution method is a very handy tool when solving systems of equations. It involves solving one of the equations for one variable, and then using that expression to replace the variable in the other equation. This reduces the system to a single equation with one variable, which is typically easier to solve. Here's how it usually goes:

• Solve one of the equations for one of the variables. For example, if you have two equations with variables \(x\) and \(y\), you might solve one equation to express \(x\) in terms of \(y\).
• Substitute the expression for that variable into the other equation. This will give you an equation with only one variable.
• Solve the resulting equation to find the value of the remaining variable.
• Substitute the value back into the expression you found in the first step to solve for the other variable.

The substitution method is particularly effective when one of the equations is easily solved for one of the variables, or when the coefficients are simple integers. This makes the calculations easier and less prone to error.However, in complex systems with difficult coefficients, like the one in our original exercise, the elimination method might be more efficient, as you can directly reduce the potential for complex fractions or bulky calculations.

Elimination Method

The elimination method is another popular way to solve systems of equations and it involves combining the equations to eliminate one of the variables. It's like artfully plotting to "cancel out" one of the variables and make your system manageably simpler. In practice:

• Align the equations so that similar terms are in columns. This could be the coefficients of \(x\), \(y\), and the constants.
• Adjust the equations by multiplication so that the coefficients of one of the variables are opposites. This means, if one equation has a term \(2x\), the other should be \(-2x\) if you want to eliminate \(x\).
• Add or subtract the equations to eliminate the chosen variable.
• The result should be a simple one-variable equation that you can solve easily.
• Use the found value to solve for the other variable by substituting it back into one of the original equations.

This method works excellently in scenarios, like our exercise, where manipulating coefficients leads neatly to a cancellation of variables. Its efficiency grows when handling systems where coefficients are non-trivial, allowing quick progress to a solution.

Simplifying Equations

Simplifying equations is a crucial step in solving any equation system. It involves making the expressions as straightforward as possible, which often means fewer terms or smaller coefficients. This makes equations easier to interpret and handle in things like mathematical operations. Typical actions in simplifying equations include:

• Expanding parentheses to remove brackets.
• Combining like terms, for instance, grouping all \(x\) terms together.
• Balancing the equation by ensuring both sides are equivalent and aligned to necessary forms.
• Ensuring that variables are easily isolated for either substitution or elimination.

In the original problem, simplifying both equations by expanding and rearranging the terms set the foundation for using the elimination method successfully. Simplification is not only about making the math neat, but also getting closer to the solution by revealing the system's underlying structure that can be exploited in subsequent steps. Always remember, a well-simplified equation can almost solve itself!