Problem 26 - Beginning and Intermediate Algebra

Question

Solve each system by substitution. $$\begin{aligned}&12 x-9 y=-8\\\&-6 x+5 y=5\end{aligned}$$

Step-by-Step Solution

Answer

The solution to the system of equations \(\begin{aligned}&12 x-9 y=-8\\&-6 x+5 y=5\end{aligned}\) is \(x = \frac{5}{6}\) and \(y = 2\).

Step 1: Solve one of the equations for one of the variables

We will solve the second equation for x in terms of y: \(-6x + 5y = 5\) Add 6x to both sides: \(5y = 6x + 5\) Divide both sides by 6: \(\frac{5y - 5}{6} = x\)

Step 2: Substitute the expression from step 1 into the other equation

Now, substitute the expression for x (\(\frac{5y - 5}{6}\)) into the first equation: \(12\left(\frac{5y - 5}{6}\right) - 9y = -8\)

Step 3: Solve the resulting equation in one variable

First, simplify the equation: \(12\left(\frac{5y - 5}{6}\right) - 9y = 2(5y - 5) - 9y = 10y - 10 - 9y = y - 10\) Now, set the simplified equation equal to -8: \(y - 10 = -8\) Add 10 to both sides: \(y = 2\)

Step 4: Substitute the solution from step 3 back into the expression from step 1 to find the value of the other variable

Now, substitute y = 2 into the expression for x from step 1: \(\frac{5y - 5}{6} = \frac{5(2) - 5}{6} = \frac{5}{6}\) So, x = \(\frac{5}{6}\)

Step 5: Check the solutions

Finally, we will check our solutions by substituting x = \(\frac{5}{6}\) and y = 2 back into both equations: For the first equation: \(12x - 9y = 12\left(\frac{5}{6}\right) - 9(2) = 10 - 18 = -8\) For the second equation: \(-6x + 5y = -6\left(\frac{5}{6}\right) + 5(2) = -5 + 10 = 5\) Both of the original equations are satisfied by the given values x = \(\frac{5}{6}\) and y = 2, so these are the correct solutions.

Key Concepts

Solving Systems of Equations

A system of equations consists of two or more equations that share two or more variables. The goal in solving these systems is to find the values of the variables that make all the equations true at the same time. In the problem we're looking at, it's a system with two linear equations involving variables x and y. There are different methods to solve systems of equations, such as graphing, elimination, and substitution.

The substitution method is particularly useful when one of the equations can be easily rearranged to solve for one variable in terms of the other. This solution is then substituted into the other equation to find the second variable. The steps usually involve:

• Rearranging one equation to express one variable in terms of the other.
• Substitution of this expression into the other equation.
• Simplifying the resulting equation to find the value of one variable.
• Substituting back to find the other variable.

By doing this, we systematically reduce the problem of solving two equations to solving just one equation.

Linear Equations

Linear equations are algebraic expressions of a straight line and are usually in the form: \(ax + by = c\). The equations in this exercise, \(12x - 9y = -8\) and \(-6x + 5y = 5\), represent linear relationships between x and y. In such equations:

• a and b are the coefficients of x and y, respectively.
• c is the constant term.
• The solutions to these equations can be graphically represented as points of intersection on a coordinate plane.

Linear equations are foundational in algebra because they establish a basic understanding of how variables can be balanced to maintain equality. By solving them, we can uncover the relationships between involved quantities and predict or determine unknown values.

Algebraic Substitution

Algebraic substitution is a crucial technique used in solving systems of equations, particularly when approaching them via the substitution method. This technique involves replacing one variable with an expression derived from another equation. Here's how it works:

• Choose one equation from the system to solve for a particular variable.
• Express this variable in terms of the other variable(s) appearing in the equation.
• Substitute this expression in place of the variable in the other equation(s).

This blend of solving and substituting works seamlessly because it changes the system from two equations with two unknowns to a single equation with one unknown, making it simpler to solve. In the given problem, solving for \( x \) from the second equation and then replacing \( x \) in the first, simplifies the task of finding \( y \), after which \( x \) can be easily determined. This method is not just applicable for linear equations—it can also be adapted for more complex non-linear systems as well.