Problem 8 - Intermediate Algebra

Question

For Problems \(1-26\), solve each system by using the substitution method. (Objective 1) $$ \left(\begin{array}{l} 3 x-4 y=-14 \\ 4 x+3 y=23 \end{array}\right) $$

Step-by-Step Solution

Answer

The solution is \((2, 5)\).

Step 1: Solve one equation for one variable

Choose one of the equations to solve for a variable. Here, we can solve the first equation for \( x \). The equation is \( 3x - 4y = -14 \). Solving for \( x \), we get: \[ 3x = 4y - 14 \] \[ x = \frac{4y - 14}{3} \].

Step 2: Substitute into the other equation

Substitute the expression for \( x \) from Step 1 into the second equation. The second equation is \( 4x + 3y = 23 \). Substitute to get: \[ 4\left(\frac{4y - 14}{3}\right) + 3y = 23 \].

Step 3: Simplify and solve for the remaining variable

Distribute and simplify the equation from Step 2: \[ \frac{16y - 56}{3} + 3y = 23 \]. Multiply through by 3 to clear the fraction: \[ 16y - 56 + 9y = 69 \]. Simplify to find \( y \): \[ 25y - 56 = 69 \]. Add 56 to both sides: \[ 25y = 125 \]. Divide by 25: \( y = 5 \).

Step 4: Substitute back to find the other variable

Take the value of \( y = 5 \) and substitute it back into the expression for \( x \) from Step 1: \[ x = \frac{4(5) - 14}{3} \]. Calculate: \[ x = \frac{20 - 14}{3} \]. Simplify: \[ x = \frac{6}{3} \]. Thus, \( x = 2 \).

Step 5: Write the solution as an ordered pair

The solution to the system of equations is the ordered pair where both equations intersect. Based on the values found, the solution is \( (x, y) = (2, 5) \).

Key Concepts

Systems of Equations

A system of equations is a set of two or more equations that involve the same set of variables. In the context of algebra, these systems usually include linear equations. Solving a system of equations means finding a set of values for the variables that satisfy all equations in the system simultaneously. For instance, if you're dealing with two variables, you will have two equations.
When you solve these systems, you're essentially looking for the point(s) at which the equations intersect on a graph. Graphical intersection is one approach, but algebraic methods like substitution or elimination are often more precise, especially for complex systems. The substitution method is particularly useful for systems where one variable can be easily isolated.

• Linear equations in a system can either have one solution (intersect at a single point), infinitely many solutions (coincide as one line), or no solution (parallel and never meet).
• Solving systems by substitution involves replacing one variable in an equation with an expression derived from another equation.

By systematically using these methods, you can ensure both accuracy and efficiency.

Solution of Equations

Finding the solution to a system of equations essentially involves determining the values of the variables that make all equations true simultaneously. In our original exercise, we've used the substitution method because it offers a structured path when one of the variables is easily isolated.
The steps of the substitution method are straightforward:

• Isolate a Variable: Choose one equation, and manipulate it to express one variable in terms of the other.
• Substitute and Solve: Take this expression and substitute it into the other equation. This step helps in telling us exactly what the value of one unknown is.
• Back-Substitution: With one value known, substitute back into the expression from the first step to find the other variable.

This method is effective in systems where at least one equation is easily solvable for one of the variables. Once solved, the solution is typically presented as an ordered pair (for two variables) that satisfies both equations.

Algebraic Manipulation

Algebraic manipulation involves rearranging equations and expressions to make them easier to work with. This often involves using basic algebraic operations such as addition, subtraction, multiplication, and division. In solving a system by the substitution method, algebraic manipulation plays a key role in each step.
There are some essential manipulations in the given exercise:

• Rearranging Equations: Start by rearranging terms to isolate a variable. For instance, to express one variable in terms of another.
• Substitution: After rearranging, substituting one equation into another to eliminate a variable.
• Simplification: Simplifying the equations to remove fractions or combine like terms. Multiplying through by a common denominator often clears fractions for easier handling.

Being proficient at algebraic manipulation is crucial as it allows for more complex systems to be tackled. This skill not only boosts accuracy but also enhances problem-solving efficiency across various mathematical problems.