Problem 13 - Algebra Form and Function

Question

Solve the systems of equations. $$ \left\\{\begin{array}{l} y=20-4 x \\ y=30-5 x \end{array}\right. $$

Step-by-Step Solution

Answer

Answer: The values of x and y that satisfy the given system of equations are x = 10 and y = -20.

Step 1: Write the given equations

The given system of equations is: $$ \left\{\begin{array}{l} y=20-4x \\ y=30-5x \end{array}\right. $$

Step 2: Use the substitution method

Since both equations are already solved for y, we can set them equal to each other: $$ 20-4x = 30-5x $$ Now we can solve for x.

Step 3: Solve for x

Add 5x to both sides of the equation: $$ 20-4x+5x = 30-5x+5x $$ Simplify by combining like terms: $$ 20+x = 30 $$ Subtract 20 from both sides to isolate x: $$ x = 30-20 $$ So, the value of x is: $$ x=10 $$

Step 4: Substitute the value of x into one of the original equations

We can use either equation, but for this example, let's use the first equation: $$ y = 20 - 4x $$ Substitute the value of x found in Step 3: $$ y = 20 - 4(10) $$

Step 5: Solve for y

Now, we can solve for y: $$ y = 20 - 40 $$ So, the value of y is: $$ y = -20 $$

Step 6: Write the solution as an ordered pair

The solution to the system of equations is the ordered pair (x, y): $$ (10, -20) $$ Thus, the values of x and y that satisfy the given system of equations are x = 10 and y = -20.

Key Concepts

Substitution Method

The substitution method is a powerful technique for solving systems of equations. It involves solving one of the equations for one variable and then substituting that expression into the other equation. This simplifies the system by reducing it to a single equation with one variable. In our example, both equations were already solved for \( y \). Thus, setting them equal is an efficient first step:

• Equation 1: \( y = 20 - 4x \)
• Equation 2: \( y = 30 - 5x \)

By equating both expressions for \( y \), we eliminate \( y \) from the system. This results in \( 20 - 4x = 30 - 5x \), enabling us to solve directly for \( x \). Substitution is particularly useful when one of the equations is already solved for a variable, as it simplifies the process considerably.

Solving Linear Equations

Solving linear equations involves finding the variable's value that makes an equation true. In practice, linear equations of one variable like \( 20 - 4x = 30 - 5x \) are solved through straightforward algebraic operations. These operations maintain the equality while isolating the variable.

1. First, add \( 5x \) to both sides: \( 20 - 4x + 5x = 30 - 5x + 5x \).
2. This simplifies to \( 20 + x = 30 \).
3. Next, subtract \( 20 \) from both sides to isolate \( x \): \( x = 10 \).

After performing these steps, we've solved for \( x \) using basic algebra. This results in \( x = 10 \), which we then use in substitution to find \( y \), completing the solution.

Algebraic Solutions

Algebraic solutions involve using algebraic manipulations to find the values of unknowns in equations. This concept embraces various techniques: simplification, rearrangement, and substitution. With our system of equations, the initial step solves for a single variable, \( x \), then uses substitution back into an equation to find \( y \). Once \( x = 10 \) was found, we substituted it into one of the original equations:- Substituting into \( y = 20 - 4x \), where \( x = 10 \), gives \( y = 20 - 40 \).The calculation shows that \( y = -20 \), allowing us to write the solution as the ordered pair \( (10, -20) \). This process of solving ensures the solution satisfies both equations, demonstrating the consistency and interconnected nature of algebraic solutions.