Problem 56 - Intermediate Algebra

Question

Solve each system by any method, if possible. If a system is inconsistent or if the equations are dependent, state this. $$ \left\\{\begin{array}{l} 4 y+5 x-7=0 \\ \frac{10}{7} x-\frac{4}{9} y=\frac{17}{21} \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \(x = \frac{4}{5}\) and \(y = \frac{3}{4}\).

Step 1: Write the System of Equations

The given system of equations is \[\begin{align*}4y + 5x &= 7 \, (1)\\frac{10}{7}x - \frac{4}{9}y &= \frac{17}{21} \, (2)\end{align*}\] We'll solve this system using substitution or elimination.

Step 2: Express one variable from Equation (1)

From Equation \((1)\), solve for \(y\):\[4y = 7 - 5x\]\[y = \frac{7 - 5x}{4}\]

Step 3: Substitute into Equation (2)

Substitute \(y = \frac{7 - 5x}{4}\) into Equation \((2)\):\[\frac{10}{7}x - \frac{4}{9}\left(\frac{7 - 5x}{4}\right) = \frac{17}{21}\]Simplify as follows:\[\frac{10}{7}x - \frac{1}{9}(7 - 5x) = \frac{17}{21}\]

Step 4: Simplify and solve for x

Distribute and simplify:\[\frac{10}{7}x - \frac{7}{9} + \frac{5x}{9} = \frac{17}{21} \]Combine like terms:\[\left(\frac{10}{7} + \frac{5}{9}\right)x = \frac{17}{21} + \frac{7}{9}\]To add the fractions with \(x\), find a common denominator and solve for \(x\), which will require clearing denominations.

Step 5: Calculate LCD and solve numerator equation for x

Find the least common multiple (LCM) for denominators, which is 63:\[\frac{90}{63}x + \frac{35}{63}x = \frac{51}{63} + \frac{49}{63}\]Combine terms:\[\frac{125}{63}x = \frac{100}{63}\]This gives:\[x = \frac{100}{125} = \frac{4}{5}\]

Step 6: Substitute x back to find y

Substitute \(x = \frac{4}{5}\) into \(y = \frac{7 - 5x}{4}\):\[y = \frac{7 - 5 \left(\frac{4}{5}\right)}{4} = \frac{7 - 4}{4} = \frac{3}{4}\]

Step 7: Verify the solution

Substitute both \(x = \frac{4}{5}\) and \(y = \frac{3}{4}\) back into the original equations to verify they satisfy both equations, confirming our solution.

Key Concepts

Substitution Method

The substitution method is one of the techniques used to solve systems of equations. It's particularly useful when you can express one variable in terms of the other from one of the equations. In our case, we started with the system

• 4y + 5x = 7
• \( \frac{10}{7}x - \frac{4}{9}y = \frac{17}{21} \)

To use substitution, we first solved the first equation for \( y \). By rearranging, we found: \[ y = \frac{7 - 5x}{4} \]This expression for \( y \) is then substituted into the second equation, replacing \( y \), thus only \( x \) remains as the variable to solve. This method reduces the number of variables in an equation, making it simpler to solve for the remaining variable. Once \( x \) is found, it is substituted back into the expression for \( y \) to find its corresponding value. The substitution method is particularly advantageous in cases where an equation is easily solvable for one variable.

Elimination Method

The elimination method might not have been the primary method chosen in this scenario, but it's a powerful tool for solving systems of equations. The idea is to eliminate one of the variables by adding or subtracting equations. This is achieved by aligning the equations so that adding or subtracting them cancels out one of the variables entirely.Consider our initial system:

• 4y + 5x = 7
• \( \frac{10}{7}x - \frac{4}{9}y = \frac{17}{21} \)

To use elimination effectively, you often need to manipulate the equations to have common coefficients for either \( x \) or \( y \). This sometimes involves multiplying entire equations by necessary values. Once variables are canceled, you solve for the remaining variable, substitute back to find the other variable, and thus find the solution for the system.Although elimination wasn’t chosen in this example due to the complexity introduced by fractions, it's often a go-to method for simpler coefficient systems.

Linear Equations

Linear equations are fundamental in algebra, characterized by variables raised only to the first power, without any exponents above one. These equations graph as straight lines and are represented in the form:\[ ax + by = c \]The system from our example contains two linear equations involving \( x \) and \( y \).

• The first equation: \( 4y + 5x = 7 \)
• The second equation: \( \frac{10}{7}x - \frac{4}{9}y = \frac{17}{21} \)

We solve these equations to find the exact point where both lines intersect, which represents the solution to the system. If the lines were parallel, there would be no intersection, and thus no solution, indicating an inconsistent system. Alternatively, if the lines overlap completely, they are dependent equations, leading to infinitely many solutions.Understanding that we are dealing with linear equations helps in determining the appropriate method to use and predicting the possible outcomes regarding the nature of the system's solutions.

Fractions in Algebra

Fractions often complicate algebraic equations, but understanding them is crucial for solving systems efficiently. In the example system given, fractions appear in the coefficients:

• \( \frac{10}{7}x - \frac{4}{9}y = \frac{17}{21} \)

Dealing with fractions generally requires finding a common denominator to simplify operations, such as addition or subtraction. In our solution, we looked for the least common multiple (LCM) of fractions' denominators at certain steps to combine and solve the expressions efficiently. Multiplying through by the LCM can clear fractions to simplify the equation. When working with fractions, be sure to perform operations methodically to ensure all steps lead to a valid and solvable equation.By effectively managing the fractions, we found a cleaner path to solve for \( x \) and \( y \). It's essential to be precise when handling fractions to avoid errors and ensure the solution is exact.