To solve the system of equations:
1. [tex]\(\frac{1}{3} y = x + \frac{4}{3}\)[/tex]
2. [tex]\(y = 4x + 5\)[/tex],
we need to find the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] that satisfy both equations simultaneously.
### Step 1: Rewrite the First Equation
Multiply both sides of the first equation by 3 to eliminate the fraction:
[tex]\[
\frac{1}{3} y = x + \frac{4}{3} \implies y = 3x + 4
\][/tex]
Now we have the system:
1. [tex]\(y = 3x + 4\)[/tex]
2. [tex]\(y = 4x + 5\)[/tex]
### Step 2: Set the Equations Equal to Each Other
Since both equations are equal to [tex]\(y\)[/tex], set the right-hand sides equal to each other:
[tex]\[
3x + 4 = 4x + 5
\][/tex]
### Step 3: Solve for [tex]\(x\)[/tex]
Rearrange the equation to isolate [tex]\(x\)[/tex]:
[tex]\[
3x + 4 - 4x = 5 \implies -x + 4 = 5 \implies -x = 1 \implies x = -1
\][/tex]
### Step 4: Solve for [tex]\(y\)[/tex]
Now substitute [tex]\(x = -1\)[/tex] back into either original equation to find [tex]\(y\)[/tex]. Let's use the second equation:
[tex]\[
y = 4(-1) + 5 \implies y = -4 + 5 \implies y = 1
\][/tex]
### Step 5: Verify the Solution
Substitute [tex]\(x = -1\)[/tex] and [tex]\(y = 1\)[/tex] back into the first equation to ensure it holds true:
[tex]\[
\frac{1}{3} y = x + \frac{4}{3} \implies \frac{1}{3}(1) = -1 + \frac{4}{3} \implies \frac{1}{3} = \frac{1}{3}
\][/tex]
The solution satisfies both equations, so the values are correct.
### Final Answer
The solution to the system of equations is:
[tex]\[
x = -1 \quad \text{and} \quad y = 1
\][/tex]
