Problem 36 - Algebra A Combined Function

Question

Solve each system of equations by the substitution method. $$ \left\\{\begin{array}{l} x=\frac{5}{6} y-2 \\ 12 x-5 y=-9 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \( \left( \frac{1}{2}, 3 \right) \).

Step 1: Identify the substitution expression

From the system of equations, notice that the first equation is already solved for \( x \). We have \( x = \frac{5}{6}y - 2 \). This expression for \( x \) will be used to substitute into the second equation.

Step 2: Substitute into the second equation

Substitute \( x = \frac{5}{6}y - 2 \) into the second equation: \[ 12\left(\frac{5}{6}y - 2\right) - 5y = -9. \] Now we need to simplify and solve for \( y \).

Step 3: Simplify and solve for y

First, distribute the 12:\[ 12 \times \frac{5}{6}y - 12 \times 2 = 10y - 24. \] Substituting back, we have:\[ 10y - 24 - 5y = -9. \] Combine like terms:\[ 5y - 24 = -9. \] Add 24 to both sides to isolate the term with \( y \):\[ 5y = 15. \] Divide both sides by 5:\[ y = 3. \]

Step 4: Substitute back to find x

With \( y = 3 \), substitute back into the expression for \( x \):\[ x = \frac{5}{6}(3) - 2. \] Calculate the expression:\[ x = \frac{15}{6} - 2 = \frac{5}{2} - 2 = \frac{5}{2} - \frac{4}{2} = \frac{1}{2}. \]

Step 5: Write the solution

The solution to the system of equations is \( x = \frac{1}{2} \) and \( y = 3 \). Thus, the solution is \( \left( \frac{1}{2}, 3 \right) \).

Key Concepts

system of equations

When we talk about a "system of equations," we are referring to a set of two or more equations with multiple variables. The goal is to find the values of these variables that make all the equations true simultaneously. For example, in our original exercise, we have two equations with two variables, \( x \) and \( y \):

• First equation: \( x = \frac{5}{6} y - 2 \)
• Second equation: \( 12x - 5y = -9 \)

To solve the system, we need to find a pair \((x, y)\) that satisfies both equations at once. The substitution method, which we'll explore next, is one such approach to tackle systems like these.

solving linear equations

Solving linear equations is a fundamental concept in algebra. A linear equation describes a line when plotted on a graph. The task here is to find the variable's value that makes the equation true.
For example, look at this equation derived from the substitution of \( x \) from our system into the second equation: \[12\left(\frac{5}{6}y - 2\right) - 5y = -9.\]Our job is to manipulate this equation and solve for \( y \). In linear equations, we often deal with simple operations like addition, subtraction, multiplication, and division to isolate the variable and reach a solution. The key here is to perform identical operations on both sides of the equation to keep it balanced. This step-by-step approach leads us to find \( y = 3 \) in our exercise.

algebraic manipulation

Algebraic manipulation involves using a variety of mathematical operations to solve equations or rearrange expressions. This is a critical part of using the substitution method.Let's take an example from solving the original system of equations: after substituting \( x = \frac{5}{6}y - 2 \) into the second equation, we needed to simplify:\[10y - 24 - 5y = -9.\]Here's what happens during the manipulation:

• Combine like terms: \( 10y - 5y \) to get \( 5y \).
• Isolate \( y \) by adding 24 to both sides, which helps to remove the constant term from the left: \( 5y = 15 \).
• Finally, divide both sides by 5 to solve for \( y \): \( y = 3 \).

Mastering algebraic manipulation enables us to break down complex problems into manageable steps.

solution verification

Solution verification is the crucial step of ensuring that the solution obtained is correct. This means plugging the values back into the original equations to check our work. It's a vital practice in mathematics that confirms accuracy.Returning to our original problem, once we deduced that \((x, y) = \left( \frac{1}{2}, 3 \right)\), we need to check:

• Substitute \( y = 3 \) back into \( x = \frac{5}{6}y - 2 \) yielding \( x = \frac{1}{2} \).
• Ensure both \( x \) and \( y \) satisfy the second equation: \( 12x - 5y = -9 \).

Using \( x = \frac{1}{2} \) in the second equation:\[12\left(\frac{1}{2}\right) - 5(3) = 6 - 15 = -9.\]Both equations are satisfied, confirming our solution is indeed correct!