Problem 69 - Algebra 2

Question

Use the following system of equations \(\left\\{\begin{array}{l}{5 x-3 y=11} \\\ {-x+12 y=3.5}\end{array}\right.\) What is the value of \(x\) in the solution? Enter your answer as a decimal.

Step-by-Step Solution

Answer

The value of \(x\) is \(2.5\).

Step 1: Express one variable in terms of the other (if necessary)

From the second equation \( -x+12y=3.5 \), express \( x \) in terms of \( y \): \[-x=3.5-12y \Rightarrow x=-3.5+12y\].

Step 2: Substitute the expression into the other equation

Substitute \( x=-3.5+12y \) into the first equation \(( 5x-3y=11 \)): \[ 5(-3.5+12y)-3y=11. \]

Step 3: Solve for \( y \) after the substitution

Expanding and simplifying the substituted equation: \[-17.5+60y-3y=11 \Rightarrow 57y=11+17.5 \Rightarrow 57y=28.5 \Rightarrow y=0.5.\]

Step 4: Find \( x \) using the value of \( y \) obtained in step 3

Substitute \( y=0.5 \) into the expression for \( x \) from Step 1: \[ x=-3.5+12(0.5) \Rightarrow x=-3.5+6 \Rightarrow x=2.5. \]

Step 5: State the solution for \( x \)

The value of \( x \) in the solution of the system of equations is \( x=2.5 \).

Key Concepts

Substitution Method

Understanding the substitution method in solving systems of linear equations is equivalent to learning a fundamental technique in algebra. This method involves replacing one variable with another equivalent expression to reduce the system to a single variable equation. Here’s how it's done:

• First, isolate one variable in one of the equations.
• Next, express that variable in terms of the other variable.
• Substitute this expression into the other equation.
• Solve for the single variable that's left.
• Finally, back substitute to find the value of the other variable.

When following these steps, make sure each algebraic manipulation is clear and correct to avoid any errors. It's vital to check your solution by substituting the values back into the original equations. If they satisfy both equations, you have the correct solution.

System of Linear Equations

A system of linear equations consists of two or more equations made up of two or more variables, where each equation is linear. This means that the variables are not multiplied by each other or raised to any power other than 1. Solving a system of linear equations like the one from our exercise,

• Look for equations that can be easily manipulated; you may find one is already set up nicely for the substitution method.
• If the system is well-suited for substitution, one equation will often solve neatly for one variable.
• Remember, the goal is to find a common solution for all equations: a set of values for the variables that make all of the equations true simultaneously.

In real-life terms, think of it like finding the perfect balance point between different forces or conditions, where everything aligns just right. Each equation in the system can be seen as a straight line, and the solution is the point or points where the lines intersect.

Algebraic Expressions

Algebraic expressions are combinations of variables, numbers, and at least one arithmetic operation. In the context of the system of equations we're working with, algebraic expressions are used to represent one variable in terms of another. This is done so we can substitute this expression into the other equation and solve for one variable at a time. Elements of an algebraic expression include:

• Variables (e.g., x and y)
• Coefficients (numerical factors of the variables like 5 in 5x)
• Constants (numbers on their own, e.g., 3.5)
• Operational signs (plus, minus, multiply, divide)

Algebraic thinking involves recognizing patterns, understanding and constructing functions, and analyzing and interpreting various forms of data. It's important to be comfortable manipulating these expressions, as this skill is frequently used in more complex areas of mathematics and applied fields.