Problem 26 - Algebra 1

Question

Choose a method to solve the linear system. Explain your choice, and then solve the system. $$ \begin{aligned} &8 x+y=15\\\ &9=2 y+2 x \end{aligned} $$

Step-by-Step Solution

Answer

The solution of the system is \(x = 1\) and \(y = 7\).

Step 1: Multiply the Equations

First, let's multiply the first equation by 2 and the second equation by 4: \[16x + 2y = 30\] and \[8x + 4y = 36\]

Step 2: Subtract the Equations

Now, subtract the first equation from the second one: \(8x + 4y - 16x - 2y = 36 - 30\). After simplifying we get: \[-8x + 2y = 6\]. Division by -2 leads to: \(4x - y = -3\). Thus, from this equation we find that \(y = 4x + 3\).

Step 3: Substitute y into the first Equation

The first equation \[8x + y = 15\] becomes: \[8x + 4x + 3 = 15\]. So, solving for x gives \(x = 1\).

Step 4: Substitute x into the equation for y

Substitute \(x = 1\) into the equation \(y = 4x + 3\): \(y=4*1+3\), so \(y = 7\).

Key Concepts

Linear Equations

Linear equations play an essential role in algebra and are the basis for solving linear systems. They are equations of the first order, meaning the highest exponent of the variable is one. A linear equation can often be expressed in the standard form: \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants. The beauty of linear equations is their simplicity and predictability.
Linear equations can represent real-life situations, like determining costs, calculating distance, or predicting profits.
For example:

• If a car rental costs \(20 a day plus a one-time fee of \)30, the total cost \(C\) for \(d\) days can be represented by the equation: \(C = 20d + 30\).
• Such equations allow us to estimate or calculate necessary variables under certain conditions.

Understanding linear equations is fundamental to mastering other math concepts and solving more complex problems involving multiple variables.

Substitution Method

The substitution method is a technique used to solve a system of equations. It involves solving one of the equations for one variable and then substituting this expression into the other equation. This makes it possible to find the value of one variable in terms of the other, thereby reducing the problem to one equation with one variable.

Here’s how the substitution method works in practice:

• Start with two linear equations.
• Solve one equation for one of the variables (e.g., solve for \(y\) in terms of \(x\)).
• Substitute this expression into the other equation.
• Solve the resulting equation to find the value of one variable.
• Plug this value back into the expression obtained in step 2, solving for the other variable.

This method is particularly useful when one of the equations has a simple variable that can be easily isolated. However, it might become cumbersome when dealing with more complex equations. In the given exercise, the substitution method proved effective as once we rearranged the equation to find \(y\), substituting back into the first equation allowed us to solve for \(x\) quickly.

Elimination Method

The elimination method is another powerful tool for solving systems of linear equations. Instead of solving for one variable at a time like the substitution method, it works by eliminating one variable through addition or subtraction.
To use the elimination method effectively:

• Align your equations vertically so that like terms are in the same columns.
• If necessary, multiply one or both equations by a constant to obtain coefficients for one of the variables that are opposites.
• Add or subtract the equations to eliminate one variable.
• Solve for the remaining variable.
• Substitute this value back into one of the original equations to find the other variable.

The elimination method is often preferred when the coefficients of a variable are easily manipulated to be equal or opposite, as seen in the given exercise. By multiplying the equations and subtracting, we effectively eliminated \(y\), allowing us to solve for \(x\) more straightforwardly.