Problem 13 - Intermediate Algebra

Question

For Problems \(1-26\), solve each system by using the substitution method. (Objective 1) $$ \left(\begin{array}{c} 7 x-3 y=-2 \\ x=\frac{3}{4} y+1 \end{array}\right) $$

Step-by-Step Solution

Answer

The solution is \(x = -2\) and \(y = -4\).

Step 1: Understand the Substitution Method

The substitution method involves solving one of the equations for one variable and then substituting that expression into the other equation. This allows us to solve for one variable and then back-substitute to find the other.

Step 2: Solve the Second Equation for x

The given system includes the equation \(x = \frac{3}{4}y + 1\). This equation is already solved for \(x\), which makes it the best candidate for substitution.

Step 3: Substitute the Expression for x into the First Equation

Take the expression \(x = \frac{3}{4}y + 1\) and substitute it into the first equation \(7x - 3y = -2\). This becomes: \(7\left(\frac{3}{4}y + 1\right) - 3y = -2\).

Step 4: Simplify and Solve for y

Distribute the 7 in the substituted equation: \(7 \times \frac{3}{4}y + 7 \times 1 - 3y = -2\). This simplifies to \(\frac{21}{4}y + 7 - 3y = -2\). Combine like terms: \(\frac{21}{4}y - \frac{12}{4}y = -2 - 7\). This becomes: \(\frac{9}{4}y = -9\). Now solve for \(y\) by multiplying both sides by \(\frac{4}{9}\): \(y = -4\).

Step 5: Substitute Back to Find x

Now that we have \(y = -4\), substitute it back into the expression for \(x\): \(x = \frac{3}{4}(-4) + 1\). Simplify this to get \(x = -3 + 1 = -2\).

Step 6: Write the Solution

The solution to the system of equations is \(x = -2\) and \(y = -4\). Verify by substituting these values into the original equations to ensure they hold true.

Key Concepts

Systems of Equations

A system of equations is a collection of two or more equations with the same set of variables. In algebra, solving such systems means finding the values of the variables that satisfy each equation in the system. These equations can be linear, making them appear as straight lines when graphed. There are different methods for solving systems of equations, including substitution, elimination, and graphing.

• Solving a system of equations involves finding a common solution or point of intersection for all equations.
• Each method has its own strategic advantages, depending on the form and complexity of the equations.
• In this exercise, we focus particularly on the substitution method, which can be very effective for certain types of systems.

Understanding systems of equations is crucial because they are used across various fields, from engineering to economics, to describe phenomena where multiple conditions must be met simultaneously.

Solving Linear Equations

Linear equations are expressions in which each term is either a constant or the product of a constant and a single variable. When solving linear equations, the goal is to isolate the variable on one side of the equation to find its value. Linear equations can be simple with a single variable or part of a system with multiple variables.

• To solve a linear equation, perform operations that simplify the equation, allowing the unknown to be isolated.
• Operations may include addition, subtraction, multiplication, and division, used to eliminate constants or coefficients.
• It's important to follow the order of operations and apply the same operation to both sides of the equation.

In the substitution method, solving linear equations accurately and systematically is key to finding the correct values for the variables involved. In our provided solution, we solved for one variable first, which helped simplify the second equation considerably.

Algebraic Methods

Algebraic methods refer to the various techniques used to manipulate and solve equations. The substitution method, used in this exercise, is a potent algebraic tool, especially when one of the equations is easily solvable for one of the variables. The goal is to reduce a system of equations into one equation with one variable, making it more straightforward to solve.

• The substitution method starts by identifying an equation that can be isolated for one variable.
• The isolated expression is then substituted into the other equation, turning it into a single variable equation.
• This new equation can then be solved using basic algebraic steps, and the found value is substituted back to find the other variable.

This approach not only provides clarity but often simplifies what seems like a complex problem into smaller, more manageable parts. Understanding this method gives students a solid foundation for tackling more complex algebraic problems in the future.