Problem 43 - Algebra for College Students

Question

Solve each system by using the substitution method. \(\left(\begin{array}{rl}\frac{2}{3} x+\frac{1}{2} y & =\frac{1}{6} \\ 4 x+6 y & =-1\end{array}\right)\)

Step-by-Step Solution

Answer

The solution is \( x = \frac{3}{4} \) and \( y = -\frac{2}{3} \).

Step 1: Solve one equation for one variable

Let's take the first equation \( \frac{2}{3}x + \frac{1}{2}y = \frac{1}{6} \) and solve for \( y \). Subtract \( \frac{2}{3}x \) from both sides: \( \frac{1}{2}y = \frac{1}{6} - \frac{2}{3}x \). Multiply every term by 2 to solve for \( y \): \( y = \frac{2}{6} - \frac{4}{3}x \), which simplifies to \( y = \frac{1}{3} - \frac{4}{3}x \).

Step 2: Substitute into the other equation

Substitute \( y = \frac{1}{3} - \frac{4}{3}x \) into the second equation \( 4x + 6y = -1 \). Replace \( y \) in the equation: \( 4x + 6(\frac{1}{3} - \frac{4}{3}x) = -1 \).

Step 3: Simplify and Solve

Distribute 6 in the equation: \( 4x + 2 - 8x = -1 \). Combine like terms to get \( -4x + 2 = -1 \). Subtract 2 from both sides to isolate terms with \( x \): \( -4x = -3 \). Divide both sides by -4 to solve for \( x \): \( x = \frac{3}{4} \).

Step 4: Substitute back to find second variable

Using the value \( x = \frac{3}{4} \), substitute it back into \( y = \frac{1}{3} - \frac{4}{3}x \) to find \( y \). Calculate \( y = \frac{1}{3} - \frac{4}{3}(\frac{3}{4}) \). Perform multiplication: \( \frac{4}{3} \times \frac{3}{4} = 1 \), so \( y = \frac{1}{3} - 1 = -\frac{2}{3} \).

Step 5: Solution Check

Verify the solution \( x = \frac{3}{4} \) and \( y = -\frac{2}{3} \) in both original equations. For the first equation, \( \frac{2}{3}x + \frac{1}{2}y = \frac{1}{6} \). Substitute values to check: \( \frac{2}{3}(\frac{3}{4}) + \frac{1}{2}(-\frac{2}{3}) = \frac{1}{6} \). Calculate: \( \frac{1}{2} - \frac{1}{3} = \frac{1}{6} \). This holds true. For the second, \( 4x + 6y = -1 \). Substitute values to check: \( 4(\frac{3}{4}) + 6(-\frac{2}{3}) = -1 \). Calculate \( 3 - 4 = -1 \), which is correct.

Key Concepts

Solving Systems of Equations

Solving a system of equations involves finding the values of variables that satisfy all equations simultaneously. When using the substitution method, we aim to express one variable in terms of the other and substitute this expression into another equation. This reduces the system to a single equation with one unknown. In our example, we started by solving the first equation for \( y \). After isolating \( y \), we substituted its expression into the second equation. This allowed us to find the value of \( x \), which we then used to find \( y \). Always check that the solution satisfies both equations in the system to ensure accuracy.
This method is especially useful when one of the equations is simple to manipulate, making it easier to substitute.

Algebraic Manipulation

Algebraic manipulation is key in solving equations, allowing us to rearrange and simplify expressions. In this method:

• Move terms to isolate variables, like subtracting \( \frac{2}{3}x \) from both sides to solve for \( y \).
• Use multiplication or division to clear fractions or coefficients, as seen when multiplying every term by 2 to solve for \( y \).
• Combining like terms, such as \( 4x + 2 - 8x \), to simplify down to \( -4x + 2 \).

The goal is to make the equation more approachable and easier to solve, while ensuring balanced equations by performing the same operations on both sides.

Fraction Operations

Working with fractions is essential in substituting and simplifying equations within systems. During substitution, fractions often arise:

• When solving \( \frac{2}{3}x + \frac{1}{2}y = \frac{1}{6} \) for \( y \), fractions are handled by finding a common denominator.
• Fraction operations include addition, subtraction, multiplication, and division, which are performed carefully to maintain the balance of equations.
• For examples of simplification, multiplying all terms by 2 to avoid fractions or simplifying \( \frac{1}{3} - 1 \) to \( -\frac{2}{3} \).

Fraction operations often require clear and methodical steps to ensure errors are minimized, making it easier to handle complex equations.

System of Linear Equations

A system of linear equations consists of two or more linear equations with the same variables. Linear equations are equations of the first degree, which graph as straight lines in a coordinate plane. Solving such systems requires finding a point where these lines intersect.
The substitution method is particularly effective when one equation can easily be solved for one variable, simplifying the system.

• In our problem, we dealt with two equations: \( \frac{2}{3}x + \frac{1}{2}y = \frac{1}{6} \) and \( 4x + 6y = -1 \).
• We transformed the system into a single-variable equation that was then solved to find exact values of \( x \) and \( y \).
• Finally, checking the intersection confirms the accuracy of solutions, both algebraically and graphically, ensuring they satisfy the original equations.

Systems of equations are foundational in algebra, providing solutions that describe relationships or intersections in various contexts.