Problem 23 - A Graphical Approach to College Algebra

Question

Solve each system by substitution. $$\begin{aligned}&3 x-7 y=15\\\&3 x+7 y=15\end{aligned}$$

Step-by-Step Solution

Answer

The solution is \( x = 5 \), \( y = 0 \).

Step 1: Solve One Equation for a Variable

We start by solving the first equation for one of the variables. Let's solve for \( x \) from the first equation: \( 3x - 7y = 15 \). Add \( 7y \) to both sides: \( 3x = 7y + 15 \). Then divide everything by 3: \( x = \frac{7y + 15}{3} \).

Step 2: Substitute into the Second Equation

Take the expression we found for \( x \) in Step 1 and substitute it into the second equation \( 3x + 7y = 15 \). This gives us: \( 3\left(\frac{7y+15}{3}\right) + 7y = 15 \). Simplify: \( 7y + 15 + 7y = 15 \).

Step 3: Simplify and Solve for y

Combine like terms: \( 14y + 15 = 15 \). To find \( y \), subtract 15 from both sides: \( 14y = 0 \). Divide by 14: \( y = 0 \).

Step 4: Substitute y Back to Find x

Now that we have \( y = 0 \), substitute back into the expression for \( x \) from Step 1: \( x = \frac{7(0) + 15}{3} \). Simplify: \( x = \frac{15}{3} \), so \( x = 5 \).

Step 5: Verify Solution

Substitute \( x = 5 \) and \( y = 0 \) back into the original equations to verify: For \( 3x - 7y = 15 \): \( 3(5) - 7(0) = 15 \), which simplifies to \( 15 = 15 \). For \( 3x + 7y = 15 \): \( 3(5) + 7(0) = 15 \), which also simplifies to \( 15 = 15 \). Both equations are satisfied.

Key Concepts

Substitution Method

The substitution method is a popular technique for solving systems of equations, particularly effective for systems involving linear equations. It involves a series of straightforward steps: isolating a variable in one equation, substituting the resulting expression into another equation, and then solving this new equation.

Here's how it works in practice:

• First, choose one equation and solve for one of the variables. This step typically involves some algebraic manipulation to isolate the variable on one side of the equation.
• Next, substitute this expression into the other equation. By doing this, you eliminate one variable and are left with an equation in a single variable, making it easier to solve.
• After finding the value of this variable, substitute it back into the expression obtained from the first step to determine the value of the other variable.

Through substitution, we effectively reduce a system of two equations in two variables to two simpler problems: solve an equation for one variable and then substitute back to find the other. This method is systematic and allows for a straightforward verification of solutions.

Linear Equations

Linear equations are equations of the first order, and they represent straight lines when graphed on a coordinate plane. They are typically written in the standard form, which often looks like \(ax + by = c\).

Key characteristics of linear equations include:

• Each term is either a constant or the product of a constant and a single variable.
• The highest power of the variable is one, meaning these equations will never include terms like \(x^2\) or \(xy\).
• Their solution sets, when plotted, form a straight line, hence the name "linear."

In the context of solving systems of linear equations, we look for a set of variable values that satisfy all the equations in the system simultaneously. When using the substitution method, linear equations are particularly manageable due to their straightforward structure and the ease with which variables can be isolated and substituted.

Algebraic Manipulation

Algebraic manipulation is a fundamental skill in solving equations, particularly useful in the substitution method. It involves rearranging and simplifying equations to isolate variables, making them easier to solve.

When performing algebraic manipulation, consider:

• Addition and subtraction are used to move terms from one side of the equation to the other. This is useful for collecting like terms or isolating a specific variable.
• Multiplication or division can adjust coefficients, allowing for straightforward isolation of variables. For instance, you might divide all terms by a coefficient to simplify an equation.
• Understanding and applying properties of equality is crucial. Whatever operation you perform on one side of an equation, you must also perform on the other to keep the equation balanced.

Practicing these operations increases your proficiency in handling more complex systems of equations. It also builds a solid foundation for higher-level algebra and calculus courses.