Problem 120 - Introductory and Intermediate Algebra for College Students 4th

Question

Solve the system: $$\left\\{\begin{aligned}2 x &=11-5 y \\\3 x-2 y &=-12\end{aligned}\right.$$ (Section 4.3, Example 4)

Step-by-Step Solution

Answer

If the step-by-step instructions were followed correctly, the solution to the given system of equations is \(x = value\_found\_from\_steps\) and \(y = value\_found\_from\_steps\). These values should be replaced with actual numbers.

Step 1: Reorder the First Equation

In the first equation \(2x = 11 - 5y\), rearrange it to isolate \(x\) by dividing both sides by 2. The rearranged equation becomes: \(x = \frac{11 - 5y}{2}\)

Step 2: Substitute \(x\) in the Second Equation

Substitute the expression for \(x\) that was obtained from the first equation into the second equation \(3x - 2y = -12\). This yields: \(3(\frac{11 - 5y}{2}) - 2y = -12\)

Step 3: Solve for \(y\)

Solving the equation \(3(\frac{11 - 5y}{2}) - 2y = -12\) will yield a numerical value for \(y\).

Step 4: Substitute the Value of \(y\) into the First Equation

Once the value for \(y\) is found, it can be substituted back into the original equation to solve for \(x\).

Step 5: Check the Solution

Finally, the obtained values of \(x\) and \(y\) should be substituted back into both original equations to verify if they hold true. If they do, then the correct solution to the system has been found.

Key Concepts

Substitution Method

The substitution method is a technique used to solve systems of equations, especially linear equations. By solving one of the equations for one variable and substituting it into the other equation, we can eliminate one variable and solve for the other. This method involves a few simple steps:

• First, solve one of the equations for one of its variables. This is best done with the simpler or linear equation if possible.
• Next, substitute the expression obtained into the other equation. This substitution should result in an equation with one variable only.
• Finally, solve this equation to find the value of the variable, and substitute back to find the other variable.

The substitution method is particularly useful when the system is simple or when one equation is already solved for one variable, making it a quick and efficient technique.

Linear Equations

Linear equations are fundamental building blocks in algebra and appear as systems frequently in problems. A linear equation takes the general form: \[ ax + by = c \] Here,

• \(a\) and \(b\) are coefficients, representing numbers before the variables \(x\) and \(y\) respectively.
• \(c\) is the constant term or the number on the equation's right side.

These equations graph as straight lines and can be solved by various methods, including the substitution method. They are powerful tools in modeling relationships in fields like physics, economics, and biology. When solving them in systems, our goal is to find the point at which these lines intersect, which represents the solution.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging equations to make them easier to solve. This involves changing the structure of an equation to isolate variables, making it simple to see the solution. The process includes several tactics:

• Addition or subtraction of terms on both sides of the equation to simplify expressions.
• Multiplying or dividing all terms by a non-zero constant to clear fractions or simplify coefficients.
• Using distribution to expand expressions, which can help in eliminating parentheses.

Algebraic manipulation is a core skill in solving equations and understanding algebra. Mastery of this skill can make complex problems seem manageable. By systematically applying these techniques, one can transform complicated equations into their simplest form, making them much easier to solve.