Problem 9 - College Algebra Essentials

Question

In Exercises \(5-18,\) solve each system by the substitution method. $$ \left\\{\begin{array}{l} x=4 y-2 \\ x=6 y+8 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution to the system of equations is \( x=-22, y=-5 \)

Step 1: Identify the substitution

Since both equations are solved for 'x', we can equate both the expressions for 'x' to each other. This gives us \[4y-2=6y+8\].

Step 2: Simplify the equation

Solving this equation for 'y' involves first moving terms involving 'y' to one side and constant terms to the other side. After rearranging, we get \[2y=-10\].

Step 3: Solve for 'y'

To find the value of 'y', divide both sides of the equation by coefficient of 'y'. Hence, we get \[y=-5\].

Step 4: Solve for 'x'

Now that we know 'y', substitute \[-5\] back into the first equation for 'y'. After simplifying, we get \[x=4*(-5)-2 =- 22\].

Key Concepts

Systems of Equations

When working with a system of equations, you are dealing with two or more equations that have common variables. The goal is to find the values of these variables that will satisfy all equations in the system simultaneously. In our example, we have two equations with two variables, 'x' and 'y'. The equations given to us form a system:

• Equation 1: \(x = 4y - 2\)
• Equation 2: \(x = 6y + 8\)

The substitution method is particularly useful when one variable is isolated on one side of the equation, as it is in both our equations. By substituting, we connect both equations, reducing the system to a single-variable equation, which is much simpler to solve.

Solving Linear Systems

Solving linear systems like the one in our exercise involves finding the point of intersection between the two lines represented by the equations. As you noticed, in our solution process, we started by equating the expressions for 'x' from both equations, turning the system of equations into one equation with one variable. After rearrangement, we end up with the simpler single-variable equation \(2y = -10\), which leads us to the value of 'y'. This process elegantly shows how substituting one equation into another transforms a system of equations into something more manageable.

Algebraic Equations

Algebraic equations are mathematical statements that express the equality of two algebraic expressions. They can contain constants, variables, and various arithmetic operations. Through the substitution method, we use algebraic manipulation to simplify equations and solve for the unknowns. For instance, in our exercise, once we isolated 'y', we found it to be \(y = -5\).
We then substitute the value of 'y' back into one of our original algebraic equations to find 'x' which in this case is \(x = -22\). This step-by-step approach is crucial for solving algebraic equations accurately. One important tip to remember is to always double-check your substitution to avoid any algebraic errors that might lead to incorrect solutions.