Question

Two linear equations are shown.

A coordinate grid with 2 lines. The first line is labeled y equals StartFraction one-third EndFraction x plus 2 and passes through (negative 6, 0) and (0, 2). The second line is labeled y equals StartFraction 4 over 3 EndFraction minus 5.
What is the solution to the system of equations?

(7, 4)
(7, StartFraction 13 over 3 EndFraction)
(8, StartFraction 14 over 3 EndFraction)
(9, 7)

Answer 1

To solve the system of equations, we need to find the point where the two lines intersect. The first equation is:

1. \( y = \frac{1}{3}x + 2 \)

The second equation appears to be incorrectly written. It likely should be of the form \( y = \frac{4}{3}x - 5 \) if we assume you meant it should be a linear equation.

Let's assume the second equation is:

2. \( y = \frac{4}{3}x - 5 \)

To find the solution, we will set these two equations equal to each other:

\[
\frac{1}{3}x + 2 = \frac{4}{3}x - 5
\]

Now we can solve for \( x \).

1. Multiply the entire equation by 3 to eliminate fractions:

\[
x + 6 = 4x - 15
\]

2. Rearranging gives:

\[
6 + 15 = 4x - x
\]

\[
21 = 3x
\]

3. Divide by 3:

\[
x = 7
\]

Now that we have \( x = 7 \), we will substitute this value back into either equation to find \( y \). Using the first equation:

\[
y = \frac{1}{3}(7) + 2 = \frac{7}{3} + 2 = \frac{7}{3} + \frac{6}{3} = \frac{13}{3}
\]

Therefore, the solution to the system of equations is:

\[
(x, y) = (7, \frac{13}{3})
\]

Thus, the answer is:

**(7, \( \frac{13}{3} \))**