Problem 10 - Algebra A Combined Function

Question

Solve each system of equations by the addition method. If a system contains fractions or decimals, you may want to first clear each equation of fractions or decimals.$$ \left\\{\begin{array}{l} x+4 y=14 \\ 5 x+3 y=2 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \(x = -2\) and \(y = 4\).

Step 1: Make coefficients of one variable equal

We will aim to eliminate the variable \(x\). To do so, we need to make the coefficients of \(x\) in the two equations equal. Multiply the first equation by 5: \(5(x + 4y) = 5(14)\) becomes \(5x + 20y = 70\).

Step 2: Subtract equations to eliminate a variable

Now subtract the second equation \(5x + 3y = 2\) from the newly formed equation \(5x + 20y = 70\): \((5x + 20y) - (5x + 3y) = 70 - 2\). Simplifying gives us \(17y = 68\).

Step 3: Solve for one variable

Solve for \(y\) by dividing both sides of the equation \(17y = 68\) by 17: \(y = 4\).

Step 4: Substitute back to find the other variable

Substitute \(y = 4\) into the first original equation \(x + 4y = 14\): \(x + 4(4) = 14\), which simplifies to \(x + 16 = 14\). Solve for \(x\): \(x = 14 - 16\) so \(x = -2\).

Step 5: Verify the solution in the original equations

Substitute \(x = -2\) and \(y = 4\) into the second original equation \(5x + 3y = 2\): \(5(-2) + 3(4) = -10 + 12 = 2\). Both equations are satisfied, confirming the solution is correct.

Key Concepts

Solving Systems of Equations

When dealing with systems of equations, your goal is to find values of the variables that simultaneously satisfy all the equations in the system. Typically, these equations involve two variables, like in our original problem with variables \(x\) and \(y\).
Several methods can be utilized to solve these systems, each depending on the specific scenario and type of equations you are provided with. Understanding these methods will allow you to approach any system of linear equations with confidence and flexibility.

• Graphical Method: Plotting each equation on a graph to find the point of intersection.
• Substitution Method: Solving one equation for a single variable and substituting it into the other equation.
• Elimination Method or Addition Method: Manipulating the equations to cancel out one of the variables.

In linear systems like \(x + 4y = 14\) and \(5x + 3y = 2\), finding a solution means identifying the values of \(x\) and \(y\) that satisfy both equations, thus effectively reducing the problem to solving linear equations one at a time.

Elimination Method

The elimination method, often termed the addition method, is a technique used to solve systems of linear equations. By adding or subtracting equations, you aim to eliminate one variable, making it easier to solve for the other.
In our example, to eliminate \(x\), you matched the coefficients of \(x\) by multiplying the first equation by 5:
\[5(x + 4y) = 5(14)\]
This gives:
\[5x + 20y = 70\]
Subtracting the original second equation \(5x + 3y = 2\) from this adjusted equation \(5x + 20y = 70\) results in a new equation \(17y = 68\).

• This step removes the \(x\) variable entirely from the equation.
• Now, you can focus solving for \(y\).

Once \(y\) is found, you can substitute it back into either of the original equations to solve for \(x\), completing the procedure. This method is particularly handy for removing fractions or decimals from the equation.

Fractions and Decimals in Equations

Fractions and decimals can make equations look more intimidating, but they are manageable with a few adjustments. When faced with these, your first step in the elimination method is to clear fractions or decimals to simplify calculation.
Here’s how:

• Multiply every term by the least common denominator (LCD) to clear fractions.
• Multiply each term by 10, 100, etc. to shift decimal points.

This transformation turns equations into simpler forms that are easier to handle, resulting in whole number coefficients and constants. While our example didn’t specifically use fractions or decimals, understanding this process conserves clarity and simplicity when they do appear. Clarifying equations acts as a helpful foundation for any manipulation, including employing the elimination method.

Substitution Method

The substitution method is another powerful tool for solving systems of equations. It can be particularly useful when one of the equations is already solved for a variable.
The process involves a few steps:

• Solve one of the equations for one variable.
• Substitute this expression into the other equation.
• Solve the new equation for the variable still present.
• Back-substitute the found value into one of the original equations to find the remaining variable.

In the context of our example, once \(y\) was found using the elimination method, you substituted \(y = 4\) into the first equation \(x + 4y = 14\) to find \(x = -2\). Substitution is particularly useful when the coefficient of one variable is 1, simplifying the task of isolating that variable. Both elimination and substitution are valuable methods in your algebra toolkit.