Solve using​ substitution, elimination, or graphing.
yequals4xplus9
5plusnothingyequals2​(yminus​x)

We are given the system of two equations:

1. \( y = 4x + 9 \)
2. \( 5 + 0y = 2(y - x) \)

Let's solve this system step-by-step. I'll use **substitution** as the first method.

### Step 1: Simplify the second equation
The second equation is:
\[
5 + 0y = 2(y - x)
\]
Since \( 0y \) means that \( y \) disappears, this simplifies to:
\[
5 = 2(y - x)
\]

Now distribute the 2 on the right-hand side:
\[
5 = 2y - 2x
\]

### Step 2: Rearrange the second equation
We want to isolate one of the variables. Rearranging the above equation:
\[
2y - 2x = 5
\]

Divide the entire equation by 2 to simplify:
\[
y - x = \frac{5}{2}
\]

### Step 3: Use substitution
We already know from the first equation that:
\[
y = 4x + 9
\]

Substitute this into the equation \( y - x = \frac{5}{2} \):
\[
(4x + 9) - x = \frac{5}{2}
\]

### Step 4: Simplify and solve for \( x \)
Simplify the left-hand side:
\[
4x - x + 9 = \frac{5}{2}
\]
\[
3x + 9 = \frac{5}{2}
\]

Subtract 9 from both sides:
\[
3x = \frac{5}{2} - 9
\]

Convert 9 into a fraction with a denominator of 2:
\[
3x = \frac{5}{2} - \frac{18}{2}
\]
\[
3x = \frac{-13}{2}
\]

Now divide both sides by 3:
\[
x = \frac{-13}{6}
\]

### Step 5: Solve for \( y \)
Now substitute \( x = \frac{-13}{6} \) back into the first equation \( y = 4x + 9 \):
\[
y = 4\left(\frac{-13}{6}\right) + 9
\]
Simplify the multiplication:
\[
y = \frac{-52}{6} + 9
\]
Convert 9 to a fraction with a denominator of 6:
\[
y = \frac{-52}{6} + \frac{54}{6}
\]
\[
y = \frac{2}{6} = \frac{1}{3}
\]

### Solution:
The solution to the system is:
\[
x = \frac{-13}{6}, \quad y = \frac{1}{3}
\]

---

Would you like more details on any of these steps, or do you have any questions?

### Related Questions:
1. How do you graphically solve a system of equations?
2. What are the key differences between substitution and elimination methods?
3. How can you check if a solution to a system of equations is correct?
4. How do you solve a system of equations using matrices?
5. What are the advantages of each method (substitution, elimination, graphing)?

### Tip:
Always double-check your solutions by plugging them back into the original equations to ensure they satisfy both.

Solve using substitution, elimination, or graphing. y equals 4x + 9, 5 + 0y equals 2(y - x)

Learn how to solve the system of linear equations y = 4x + 9 and 5 + 0y = 2(y - x) using the substitution method. This step-by-step solution covers key algebraic concepts and properties of equality. Suitable for students in Grades 8-10.

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