Problem 47 - Intermediate Algebra

Question

For Problems 19-48, solve each system by using either the substitution or the elimination-by-addition method, whichever seems more appropriate. (Objective 2) $$ \left(\begin{array}{l} \frac{3 x+y}{2}+\frac{x-2 y}{5}=8 \\ \frac{x-y}{3}-\frac{x+y}{6}=\frac{10}{3} \end{array}\right) $$

Step-by-Step Solution

Answer

The solution is \( x = 5 \), \( y = -5 \).

Step 1: Simplify the Fractions and Equations

First, we'll rewrite both equations by clearing the fractions. Multiply the first equation by 10 (LCM of 2 and 5) to eliminate the denominators:\[ 10 \left( \frac{3x + y}{2} \right) + 10 \left( \frac{x - 2y}{5} \right) = 10 \times 8 \]This gives:\[ 5(3x + y) + 2(x - 2y) = 80 \]Simplifying yields:\[ 15x + 5y + 2x - 4y = 80 \]\[ 17x + y = 80 \] For the second equation, multiply by 6 (LCM of 3 and 6) to clear fractions:\[ 6 \left( \frac{x - y}{3} \right) - 6 \left( \frac{x + y}{6} \right) = 6 \times \frac{10}{3} \]This results in:\[ 2(x - y) - (x + y) = 20 \]Simplifying gives:\[ 2x - 2y - x - y = 20 \]\[ x - 3y = 20 \]

Step 2: Use Substitution Method

Now, we'll use the substitution method. From the equation:\[ x - 3y = 20 \]Solve for \( x \):\[ x = 3y + 20 \]Substitute \( x = 3y + 20 \) into the first equation (\( 17x + y = 80 \)):\[ 17(3y + 20) + y = 80 \]Simplify and solve for \( y \):\[ 51y + 340 + y = 80 \]Combine like terms:\[ 52y + 340 = 80 \]Subtract 340 from both sides:\[ 52y = -260 \]Solve for \( y \):\[ y = -5 \]

Step 3: Solve for x

With \( y = -5 \), substitute back into the expression for \( x \):\[ x = 3(-5) + 20 \]Calculate \( x \):\[ x = -15 + 20 \]\[ x = 5 \]

Step 4: Verify the Solution

Check the solution \( x = 5 \) and \( y = -5 \) in both original equations:For the first equation:\[ \frac{3(5) + (-5)}{2} + \frac{5 - 2(-5)}{5} \]Simplifies to:\[ \frac{15 - 5}{2} + \frac{5 + 10}{5} = \frac{10}{2} + 3 = 5 + 3 = 8 \]This checks out.For the second equation:\[ \frac{5 - (-5)}{3} - \frac{5 + (-5)}{6} \]Simplifies to:\[ \frac{10}{3} - 0 = \frac{10}{3} \]Both check out, confirming the solution is correct.

Key Concepts

Substitution Method

The substitution method is a strategic approach in solving systems of equations. It's perfect when one equation is simple enough to easily solve for one variable.
The essence of this method is to isolate a variable in one equation and substitute this value into the other equation.

• For example, in the step-by-step solution, the equation \( x - 3y = 20 \) is solved for \( x \), giving \( x = 3y + 20 \).
• This calculated expression for \( x \) is then substituted into the equation \( 17x + y = 80 \).

This substitution reduces the system from two equations to one equation with one unknown, which simplifies solving the system. When you work through substituting back, it is essential to verify your solutions by plugging them back into the original equations to ensure they satisfy both.

Elimination Method

The elimination method is another powerful technique to solve systems of equations. The goal here is to eliminate one of the variables by adding or subtracting equations, thus simplifying the system.

• The method involves aligning equations and finding the least common multiple (LCM) to create equivalent terms that can be canceled out.
• For example, if the coefficients of \( y \) in two equations are opposites, adding them will eliminate \( y \). If not, multiply one or both equations to set this up.

In the original problem, finding a common denominator cleared the fractions, simplifying the process. This method aims at making variables disappear, streamlining the way to a solution. After the variables are eliminated, solve the remaining equation for the remaining variable, backtrack to find the other variable, and verify the solution with original equations.

Fractions in Equations

Handling fractions in equations can sometimes seem daunting, but the process becomes more straightforward with careful steps.

• To eliminate fractions, identify the least common multiple (LCM) of the denominators across the equations.
• Multiply every term by this LCM to transform the equation into one with integers only, easing the solving process.

In the exercise, you see both equations having fractional coefficients. To tackle this, the first equation was multiplied by 10 (LCM of 2 and 5), while the second was multiplied by 6 (LCM of 3 and 6). This approach simplifies complex fractions to more manageable forms.
By clearing the fractions, numerical calculations become easier, reducing room for error and making the equation more approachable.