Problem 36 - Elementary Algebra

Question

Solve each system by using either the substitution or the elimination-by- addition method, whichever seems more appropriate. $$\left(\begin{array}{l}-2(x+2)+4(y-3)=-34 \\\ 3(x+4)-5(y+2)=23\end{array}\right)$$

Step-by-Step Solution

Answer

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

Step 1: Convert Equation Into Standard Form

The given system of equations is:\[-2(x+2) + 4(y-3) = -34\]\[3(x+4) - 5(y+2) = 23\]First, distribute the terms in both equations to simplify them.\[-2x - 4 + 4y - 12 = -34\]Simplify to get:\[-2x + 4y = -18\]For the second equation, distribute:\[3x + 12 - 5y - 10 = 23\]Simplify to get:\[3x - 5y = 21\] Hence, the system of equations in standard form is: \[-2x + 4y = -18\] and \[3x - 5y = 21\].

Step 2: Use Elimination Method to Eliminate x

We have the system:\[-2x + 4y = -18\]\[3x - 5y = 21\]To eliminate \(x\), we will find a common multiple of the coefficients of \(x\). Multiply the first equation by 3 and the second by 2:\[3(-2x + 4y) = 3(-18)\] resulting in \[-6x + 12y = -54\]\[2(3x - 5y) = 2(21)\] resulting in \[6x - 10y = 42\]Add these equations to eliminate \(x\):\[-6x + 12y = -54\]\[+6x - 10y = 42\]\[2y = -12\]Solve for \(y\): \(y = -6\).

Step 3: Substitute y to Find x

Now substitute \(y = -6\) back into one of the original equations to solve for \(x\). Use the first equation \(-2x + 4y = -18\):\[-2x + 4(-6) = -18\]\[-2x - 24 = -18\]Add 24 to both sides:\[-2x = 6\]Divide both sides by -2 to solve for \(x\):\[x = -3\].

Step 4: Verify the Solution

To ensure the solution \((x, y) = (-3, -6)\) is correct, substitute both values back into the second original equation to verify:\[3(-3) - 5(-6) = 21\]\[-9 + 30 = 21\]This simplifies to \[21 = 21\], confirming that the solution satisfies both equations.

Key Concepts

Elimination Method

The elimination method is a powerful technique for solving systems of equations. It involves manipulating the equations in such a way that one of the variables is eliminated, allowing the other to be solved more easily. This method often requires both equations to be carefully aligned so that adding or subtracting them will cancel out one variable.

Here's how the elimination method works:

• Align the equations: Begin by writing both equations in standard form, aligning like terms.
• Find common coefficients: Seek a common multiple for the coefficients of one variable across both equations. For instance, if you aim to eliminate the variable \(x\), adjust the coefficients so they cancel when the equations are added or subtracted.
• Add or subtract the equations: This will eliminate one variable, leaving an equation in one variable.
• Solve for the remaining variable: Use basic algebra to find the value of this variable.
• Back-substitute: Once you have one variable, substitute it back into one of the original equations to find the other variable.

This systematic approach is useful when equations are easily manipulated to eliminate one variable and is often preferred for systems where variables have coefficients that are multiples of each other.

Substitution Method

Another common method for solving systems of equations is the substitution method. This technique is particularly useful when one equation is already solved for one variable or can be easily rearranged to do so.

In the substitution method, you follow these steps:

• Solve one equation for one variable: Choose the equation that is easiest to manipulate and express one variable in terms of the other variable.
• Substitute into the second equation: Take the expression obtained and substitute it for the corresponding variable in the other equation, resulting in an equation with a single variable.
• Solve for the single variable: Simplify and solve this equation to find the value of one variable.
• Substitute back to find the other variable: Use the value found to solve for the remaining variable back in the expression initially derived.
• Verify the solution: It is always a good practice to substitute the solution back into the original equations to ensure their accuracy.

The substitution method is intuitive and works well when the equations are readily set up for substitution. It's especially helpful when the coefficients are fractions or decimals, simplifying calculations considerably.

Algebraic Problem-Solving

Algebraic problem-solving involves a range of mathematical concepts and techniques used to find unknown variables. When dealing with systems of equations, algebraic solutions often involve either graphical methods or algebraic methods like substitution and elimination.

Here's a general approach for tackling algebraic problems:

• Understand the problem: Carefully read and break down the problem to determine what is being asked and identify the variables involved.
• Set up equations: Translate the problem statement into mathematical equations.
• Choose a method: Depending on the problem's complexity, decide whether the substitution method, elimination method, or another technique is most appropriate.
• Manipulate the equations: Use algebraic operations like addition, subtraction, multiplication, and division to simplify and solve the equations.
• Interpret the results: Ensure the solution makes sense in the context of the original problem and verify the accuracy.

Effective algebraic problem-solving requires practice, attention to detail, and a structured approach to achieve consistent results. Mastery of basic algebraic techniques allows for more creative solutions to complex problems, making difficult tasks more manageable.