Problem 34 - Algebra 2

Question

Solve each system of equations by using either substitution or elimination. \(3 u+5 v=6\) \(2 u-4 v=-7\)

Step-by-Step Solution

Answer

The solution is \(u = -\frac{1}{2}\), \(v = \frac{3}{2}\).

Step 1: Choose a Method

We need to solve the system of equations: \[3u + 5v = 6\] \[2u - 4v = -7\]We will use the elimination method here, as it seems straightforward to eliminate one variable by aligning terms properly.

Step 2: Align Coefficients for Elimination

To use the elimination method, align one of the variables by obtaining equal coefficients. Let's eliminate \(v\). Multiply the first equation by 4, and the second equation by 5:\[4(3u + 5v) = 4(6)\]\[5(2u - 4v) = 5(-7)\]This simplifies to:\[12u + 20v = 24\]\[10u - 20v = -35\]

Step 3: Add and Eliminate One Variable

Add the two modified equations to eliminate \(v\):\[(12u + 20v) + (10u - 20v) = 24 + (-35)\]This simplifies to:\[22u = -11\]

Step 4: Solve for Remaining Variable

Divide both sides by 22 to solve for \(u\):\[u = \frac{-11}{22} = -\frac{1}{2}\]

Step 5: Substitute Back to Find Second Variable

Substitute \(u = -\frac{1}{2}\) back into one of the original equations to find \(v\). Let's use the first equation:\[3(-\frac{1}{2}) + 5v = 6\]\[-\frac{3}{2} + 5v = 6\]Add \(\frac{3}{2}\) to both sides:\[5v = 6 + \frac{3}{2}\]Convert 6 to a fraction with denominator 2:\[5v = \frac{12}{2} + \frac{3}{2} = \frac{15}{2}\]

Step 6: Solve for Second Variable

Divide both sides by 5 to solve for \(v\):\[v = \frac{15}{2} \times \frac{1}{5}\]\[v = \frac{15}{10} = \frac{3}{2}\]

Step 7: Verify the Solutions

Substitute \(u = -\frac{1}{2}\) and \(v = \frac{3}{2}\) back into both original equations to ensure they satisfy both.

Key Concepts

Elimination Method

The elimination method is a handy technique for solving systems of linear equations. This method involves eliminating one of the variables by adding or subtracting the equations. This is particularly useful when the coefficients of one variable are identical or can easily be made identical.
To apply the elimination method, follow these simple steps:

• Analyze the system of linear equations to decide which variable to eliminate.
• Multiply one or both equations by suitable numbers to align the coefficients of one variable.
• Add or subtract the equations to eliminate one variable, thereby reducing the system to a single equation.
• Solve the resulting equation for the remaining variable.

Finally, substitute the found value back into one of the original equations to find the value of the other variable. Verify the solution by plugging back both values into the original system to ensure consistency.

Substitution Method

While the elimination method is highly effective, the substitution method is another equally valuable technique for solving linear equations. This method is perfect when it is easy to isolate one variable in one of the equations.
The substitution method involves these steps:

• Solve one of the equations for one variable in terms of the other variable.
• Substitute the expression into the other equation to eliminate one variable.
• Simplify the equation and solve for the single remaining variable.

Following these steps will give you the value of one variable. You can then substitute this value back into one of the original equations to find the other variable's value. Remember to check your solution by inserting the values back into the original equations.

Linear Equations

Linear equations are fundamental in mathematics. They describe relationships in which the variables are all on the first power or degree, which ensures they graph as straight lines.
A linear equation typically looks like this: \[ ax + by = c \]where \(a\), \(b\), and \(c\) are constants.A system of linear equations refers to multiple linear equations that you work with simultaneously. Solving these systems means finding the values of variables that satisfy all equations at once.Characteristics of linear equations include:

• They result in straight lines on a graph.
• They have a constant rate of change.
• The solutions can be found using algebraic methods like elimination or substitution.

Understanding linear equations is essential as they form the basis for many more complex mathematical and real-world problems. They help in modeling relationships and predicting outcomes, making them widely applicable in various fields.