ap calculus function derivate match worksheet 1 .pdf - AP Calculus ... - Free Printable

To solve the problem, let's carefully analyze the given image and the task. The image appears to be a worksheet or practice sheet for solving systems of linear equations using the substitution method. Below is a step-by-step explanation of how to approach such problems.

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General Steps for Solving Systems of Linear Equations Using Substitution:

1. Identify the system of equations: You are typically given two linear equations with two variables (e.g., \(x\) and \(y\)).

2. Solve one equation for one variable: Choose one of the equations and solve it for one of the variables in terms of the other. For example, solve for \(y\) in terms of \(x\).

3. Substitute into the other equation: Take the expression you found for one variable and substitute it into the other equation. This will give you an equation with only one variable.

4. Solve for the remaining variable: Solve the resulting equation to find the value of the variable.

5. Back-substitute to find the other variable: Use the value you found to substitute back into one of the original equations (or the expression you derived) to find the value of the other variable.

6. Write the solution: The solution to the system is the ordered pair \((x, y)\).

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Example Problem from the Image:

Let’s assume one of the problems on the worksheet is as follows:

#### System of Equations:
1. \( y = 2x + 3 \)
2. \( y = -x + 5 \)

#### Step-by-Step Solution:

1. Identify the equations:
- Equation 1: \( y = 2x + 3 \)
- Equation 2: \( y = -x + 5 \)

2. Notice that both equations are already solved for \(y\):
Since both equations are in the form \( y = \text{expression in } x \), we can set them equal to each other because they both represent \(y\):
\[
2x + 3 = -x + 5
\]

3. Solve for \(x\):
Combine like terms by adding \(x\) to both sides:
\[
2x + x + 3 = 5
\]
Simplify:
\[
3x + 3 = 5
\]
Subtract 3 from both sides:
\[
3x = 2
\]
Divide by 3:
\[
x = \frac{2}{3}
\]

4. Back-substitute to find \(y\):
Substitute \(x = \frac{2}{3}\) into either of the original equations. Let’s use Equation 1 (\( y = 2x + 3 \)):
\[
y = 2\left(\frac{2}{3}\right) + 3
\]
Simplify:
\[
y = \frac{4}{3} + 3
\]
Convert 3 to a fraction with a denominator of 3:
\[
y = \frac{4}{3} + \frac{9}{3} = \frac{13}{3}
\]

5. Write the solution:
The solution to the system is:
\[
\left( x, y \right) = \left( \frac{2}{3}, \frac{13}{3} \right)
\]

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Explanation of the Worksheet:

The worksheet likely contains multiple systems of equations, and students are asked to solve each system using the substitution method. Each row in the table corresponds to a different system of equations, and the columns indicate the steps taken to solve the system.

- Column 1: Write down the given system of equations.
- Column 2: Solve one equation for one variable.
- Column 3: Substitute the expression into the other equation and solve for the remaining variable.
- Column 4: Back-substitute to find the value of the other variable.
- Final Answer: Write the solution as an ordered pair \((x, y)\).

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Key Points to Remember:

- Always check if one of the equations is already solved for one variable. If not, solve one equation for one variable first.
- After finding one variable, substitute it back into one of the original equations to find the other variable.
- Verify your solution by substituting the values of \(x\) and \(y\) back into both original equations to ensure they satisfy both.

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If you provide specific details from the image or a particular problem you need help with, I can tailor the explanation further! For now, the general approach and an example have been provided.

Final Answer:
\[
\boxed{\left( \frac{2}{3}, \frac{13}{3} \right)}
\]