Problem 65 - Introductory Algebra for College Students

Question

Solve: $$\left\\{\begin{array}{l}3 x+2 y=6 \\\8 x-3 y=1\end{array}\right.$$

Step-by-Step Solution

Answer

The solution of the simultaneous equations is \(x = 0.8\) and \(y = 1.8\)

Step 1: Use The Substitution Method

The first equation can be rewritten to express y in terms of x: \(y = \frac{6 - 3x}{2}\).

Step 2: Substitute y in Second Equation

Substitute \(y\) from the first equation into the second equation: \(8x - 3 (\frac{6 - 3x}{2}) = 1\) and simplify to get \(8x - 9 + \frac{9x}{2} = 1\). By multiplying every term by 2 to avoid fractions, it becomes \(16x - 18 + 9x = 2\), which simplifies to \(25x = 20\). From this, we can solve for x, getting \(x = \frac{20}{25} = 0.8\)

Step 3: Substitute x in First Equation

Now that we have x, we can substitute it into the first equation \(3*0.8 + 2y = 6\), which simplifies to \(2.4 + 2y = 6\) and further simplifies to \(2y = 6 - 2.4 = 3.6\). Solving for y gives \(y = \frac{3.6}{2} = 1.8\)

Key Concepts

Substitution Method

The substitution method is a helpful technique used to solve simultaneous linear equations. In this method, you solve one equation for one variable, and then substitute this expression into the other equation. This allows you to reduce a system of equations to a single equation with one variable. Here’s a quick run through the substitution method:

• Choose one of the equations to solve for one of the variables. For simplicity, try to select an equation and a variable that leads to easy algebraic manipulation.
• Once you have solved for the variable, take the expression you found and substitute it into the other equation. This will usually result in a single equation with a single variable.
• Finally, solve this equation to find the value of the variable. Once one variable is known, substitute it back into any of the original equations to find the other variable’s value.

In our example problem, the first step involved solving the first equation for \(y\) by rewriting it as \(y = \frac{6 - 3x}{2}\). This expression was then substituted into the second equation.

Equation Solving

Equation solving involves finding values for variables that make an equation true. For linear equations, the solution involves finding where lines intersect. In a system of two equations, solving these equations will find the values of the variables where both equations hold true simultaneously.
To effectively solve these equations, especially when using substitution, careful attention to detail is necessary.

• When you substitute an expression into another equation, simplifying the resulting equation is crucial.
• It is important to perform operations accurately to keep track of coefficients and constants correctly.
• Once the equation is simplified, solving it typically involves straightforward algebraic steps like isolating the variable.

In the provided example, substituting \(y\) into the second equation, obtained a new equation \(8x - 3 (\frac{6 - 3x}{2}) = 1\), which required simplifying and solving for \(x\). This ultimately led to \(25x = 20\), resulting in \(x = 0.8\).

Algebraic Manipulation

Algebraic manipulation refers to rearranging and simplifying mathematical expressions in order to solve equations and systems of equations efficiently. Mastery of algebraic manipulation is essential for using methods like substitution.

• Start by clearly organizing terms to simplify expressions while following operations like addition, subtraction, multiplication, and division.
• Moving terms across the equals sign requires changing their signs, often leading to more manageable expressions.
• Multiplying through by a common factor can eliminate fractions from equations, as seen here to simplify calculations.

In the solution of our simultaneous equations, multiplication was used to clear fractions by multiplying all terms by 2. This transformed the equation into \(16x - 18 + 9x = 2\). By solving further algebraic manipulations like collecting \(x\) terms and constants separately, it simplified to \(25x = 20\), where \(x\) could be easily isolated.