Problem 50 - Intermediate Algebra

Question

Solve each system by any method, if possible. If a system is inconsistent or if the equations are dependent, state this. $$ \left\\{\begin{array}{l} \frac{2}{3} x-\frac{1}{4} y=-8 \\ \frac{1}{2} x-\frac{3}{8} y=-9 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution to the system is \(x = -6\) and \(y = 16\).

Step 1: Clear Fractions

To eliminate the fractions in the equations, multiply each equation by the Least Common Multiple (LCM) of the denominators. The LCM of the denominators 3 and 4 in the first equation is 12. The LCM of the denominators 2 and 8 in the second equation is 8. So, multiply the first equation by 12 and the second by 8.

Step 2: Multiply the Equations

Multiply the first equation by 12: \[12 \left(\frac{2}{3} x\right) - 12\left(\frac{1}{4} y\right) = 12(-8) \]This simplifies to:\[8x - 3y = -96\]Multiply the second equation by 8:\[8 \left(\frac{1}{2} x\right) - 8 \left(\frac{3}{8} y\right) = 8(-9)\]This simplifies to:\[4x - 3y = -72\]

Step 3: Subtract the Equations

Subtract the second simplified equation from the first one to eliminate y:\[(8x - 3y) - (4x - 3y) = -96 - (-72)\]This simplifies to:\[4x = -24\]

Step 4: Solve for x

Solve the equation from Step 3 for \(x\):\[4x = -24\]Divide both sides by 4:\[x = -6\]

Step 5: Substitute and Solve for y

Substitute \(x = -6\) into one of the original equations to solve for \(y\). We'll use the first equation for convenience:\[\frac{2}{3}(-6) - \frac{1}{4}y = -8\]Simplify:\[-4 - \frac{1}{4}y = -8\]Add 4 to both sides:\[-\frac{1}{4}y = -4\]Multiply both sides by -4 to solve for \(y\):\[y = 16\]

Step 6: Verify the Solution

Substitute \(x = -6\) and \(y = 16\) back into the second original equation to verify:\[\frac{1}{2}(-6) - \frac{3}{8}(16) = -9\]Simplify:\[-3 - 6 = -9\]This verification confirms our solution.

Key Concepts

Solving Systems of Equations

Systems of equations involve multiple equations that you solve together. Each equation in the system contains the same set of variables, and the solution is the point(s) where the equations intersect, representing the values of the variables that satisfy all equations simultaneously. When solving systems of equations, we aim to find a common solution for the variables involved.

There are several powerful methods to solve systems of equations:

• Substitution: Solve one equation for one variable and then substitute this into the other equations. This simplifies the system step by step to find a solution sequentially for each variable.
• Elimination: Add or subtract equations to eliminate a variable, allowing you to solve for the remaining variables. This often involves multiplying one or both equations by a number to align coefficients for subtraction.
• Graphical: Plot each equation on a graph to visually determine the point of intersection, which represents the solution.

In our exercise, we used elimination after manipulation to solve the system, ensuring to clear fractions first. Systems can be consistent, with at least one solution, or inconsistent, where no solution exists. Dependent systems have infinitely many solutions and typically result from identical equations. In our solution, the system was consistent with a unique solution.

Linear Algebra

Linear algebra is a branch of mathematics focused on vectors, matrices, and linear transformations. It provides the tools to study systems of linear equations and much more.

The essence of linear algebra is captured in the ability to manipulate and solve linear systems—our task involves finding solutions where lines or planes intersect.

Key concepts include:

• Vectors: Basic elements that can represent points or directions in space.
• Matrices: Arrays of numbers that can simplify and speed up calculations involving linear relations.
• Row operations: Similar to the arithmetic manipulations we apply to equations, they aid in solving systems by bringing them into a form that is simpler to interpret.

Through linear algebra techniques, we can condense complex systems into manageable forms and gain insights that are not just algebraic but geometric, considering dimensions and orientations of solutions in space. In practical terms, this allowed us to simplify and neatly solve the linear system provided in our problem.

Algebraic Manipulation

Algebraic manipulation is crucial when dealing with systems of equations. It involves changing the form of equations without changing their solutions. This allows us to simplify problems and make solutions more reachable.

Key techniques of algebraic manipulation include:

• Rearranging terms: Moving terms around using arithmetic operations to frame the equation in a useful form.
• Scaling equations: Multiplying or dividing through by constants to align coefficients or cancel terms.
• Substitution and Elimination: Used to reduce the number of variables by isolating one at a time.

In our example, algebraic manipulation enabled us to eliminate the fractions by scaling, which cleared the path to using the elimination method effectively. By maintaining balance—what you do to one side, do to the other—you ensure that the manipulations remain legitimate.

Fractions in Equations

Fractions often complicate equations, so simplifying them is a helpful strategy when solving systems. Dealing with fractions generally involves finding a way to remove them to make calculations easier to handle.

Tips for handling fractions in equations:

• Least Common Multiple (LCM): Use the LCM of denominators to clear fractions. Multiply the entire equation by the LCM to convert fractions into whole numbers.
• Reciprocal multiplication: Consider multiplying by the reciprocal to clear terms had you isolated fractions.
• Simplify first: Spend time ensuring the equation starts in its simplest form, reducing calculation errors.

In our problem, clearing fractions was the initial step that paved the way for clean algebraic manipulation. By converting all terms into integer forms, each subsequent step became simpler, ensuring accuracy and efficiency in solving the system.