Question

1. use your knowledge of systems of equations and linear functions to answer the question below. show your work 2. \\(\

$$\begin{cases} 3x + y = 25 \\\\ y = x - 3 \\end{cases}$$

\\) substitute \\(\underline{\quad\quad}\\) for \\(y\\) in the first equation \\(3x + (\underline{\quad\quad}) = 25\\) \\(\underline{\quad\quad} - 3 = 25\\) \\(\underline{\quad\quad} + 3 \quad \underline{\quad\quad} + 3\\) \\(\underline{\quad\quad} = 28\\) \\(\underline{\quad\quad} = \underline{\quad\quad}\\) since \\(x = \underline{\quad\quad}\\), \\(x\\) in one of the equations to find the value of \\(y\\) \\(y = x - 3\\) 4. answer each problem below. show your work. solve each system by substitution. check your answer. 3. \\(\

$$\begin{cases} y = 4x \\\\ y = 2x + 6 \\end{cases}$$

\\)

Explanation:

Problem 2: Solve the system \(

$$\begin{cases} 3x + y = 25 \\ y = x - 3 \end{cases}$$

\) using substitution.

Step 1: Substitute \(y\) in the first equation

We know \(y = x - 3\), so substitute \(y\) into \(3x + y = 25\).
\(3x + (x - 3) = 25\)

Step 2: Simplify and solve for \(x\)

Combine like terms: \(3x + x - 3 = 25\)
\(4x - 3 = 25\)
Add 3 to both sides: \(4x - 3 + 3 = 25 + 3\)
\(4x = 28\)
Divide by 4: \(\frac{4x}{4} = \frac{28}{4}\)
\(x = 7\)

Step 3: Find \(y\) using \(y = x - 3\)

Substitute \(x = 7\) into \(y = x - 3\):
\(y = 7 - 3\)
\(y = 4\)

Problem 3: Solve the system \(

$$\begin{cases} y = 4x \\ y = 2x + 6 \end{cases}$$

\) using substitution.

Step 1: Set the two expressions for \(y\) equal

Since \(y = 4x\) and \(y = 2x + 6\), we can set \(4x = 2x + 6\).

Step 2: Solve for \(x\)

Subtract \(2x\) from both sides: \(4x - 2x = 2x + 6 - 2x\)
\(2x = 6\)
Divide by 2: \(\frac{2x}{2} = \frac{6}{2}\)
\(x = 3\)

Step 3: Find \(y\) using \(y = 4x\)

Substitute \(x = 3\) into \(y = 4x\):
\(y = 4(3)\)
\(y = 12\)

Answer:

s:

• For Problem 2: \(x = 7\), \(y = 4\) (Solution: \((7, 4)\))
• For Problem 3: \(x = 3\), \(y = 12\) (Solution: \((3, 12)\))