Problem 17 - Intermediate Algebra

Question

For Problems \(1-26\), solve each system by using the substitution method. (Objective 1) $$ \left(\begin{array}{l} 4 x+3 y=-40 \\ 5 x-y=-12 \end{array}\right) $$

Step-by-Step Solution

Answer

The solution is \(x = -4\) and \(y = -8\).

Step 1: Solve one equation for one variable

Start by solving the second equation for one of the variables to use in substitution. The second equation is \(5x - y = -12\). Solve for \(y\):\[ y = 5x + 12 \]

Step 2: Substitute the expression into the other equation

Substitute \(y = 5x + 12\) back into the first equation \(4x + 3y = -40\):\[ 4x + 3(5x + 12) = -40 \]

Step 3: Solve for x

Distribute and simplify the expression from substitution:\[ 4x + 15x + 36 = -40 \]Combine like terms:\[ 19x + 36 = -40 \]Subtract 36 from both sides:\[ 19x = -76 \]Divide by 19:\[ x = -4 \]

Step 4: Substitute back to find y

Use the value of \(x\) to find \(y\) using the expression found in Step 1, \(y = 5x + 12\):\[ y = 5(-4) + 12 = -20 + 12 = -8 \]

Step 5: Verify the solution

Substitute \(x = -4\) and \(y = -8\) into the original equations to verify the solution:Equation 1: \(4x + 3y = -40\)\[4(-4) + 3(-8) = -16 - 24 = -40\] which holds true.Equation 2: \(5x - y = -12\)\[5(-4) - (-8) = -20 + 8 = -12\] which also holds true.

Key Concepts

Solving Systems of Equations

A system of equations consists of two or more equations with the same set of variables. To find solutions, we need values that make all equations true simultaneously. Often, these solutions represent intersecting points on graphs of the equations.

The substitution method is a common technique used to solve systems of equations. This involves solving one equation for a variable and then substituting this expression into another equation. For example:

• Solve one equation for a single variable (e.g., solve for \(y\)).
• Substitute this expression into another equation. This helps eliminate one variable, enabling you to solve for the other.
• Use the solution to find the value of the eliminated variable by back-substitution.
• Verify the solutions by plugging them back into the original equations.

This method is particularly useful when dealing with linear systems and can often be less cumbersome than graphing if the system involves complex coefficients.

Linear Equations

Linear equations are equations where the highest power of the variable is one. They form straight lines when graphed on a coordinate plane.

In our system, equations are of the form \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants. Both equations in the original problem :

• \(4x + 3y = -40\)
• \(5x - y = -12\)

are linear equations. To solve these using substitution, the key is to isolate one variable in terms of the other.

For the given exercise, we isolated \(y\) in the second equation, making it easier to substitute it into the first equation. Recognizing each equation's linearity allows us to approach them systematically, simplifying expressions and ensuring all solutions satisfy the combined system.

Intermediate Algebra

Intermediate algebra involves working with more complex equations and functions compared to basic algebra. A core component of this is solving systems of equations effectively. In intermediate algebra, one needs a solid understanding of rearranging equations, a process pivotal for utilizing the substitution method.

The process can include:

• Rearranging formulas: Adjusting equations to make a specific variable the subject.
• Combining and simplifying expressions: Efficiently handling terms and coefficients to streamline calculations.
• Verification: Ensuring that every solution satisfies all given equations.

By practicing these principles, students develop a more profound grasp of mathematical concepts, enabling them to tackle more sophisticated mathematical challenges in courses that follow intermediate algebra.