Question

Estimate the solution to the system of equations.
You can use the interactive graph below to find the solution.

\[\begin{cases}
7x-y=7
\\\\
x+2y=6
\end{cases}\]
Choose 1 answer:
Choose 1 answer:
(Choice A)


\[x= 1\dfrac{1}3, y=2\dfrac{1}3\]
A


\[x= 1\dfrac{1}3, y=2\dfrac{1}3\]
(Choice B)


\[x=2\dfrac{1}3, y=2\dfrac{1}3\]
B


\[x=2\dfrac{1}3, y=2\dfrac{1}3\]
(Choice C)


\[x= 1\dfrac{1}3, y=1\dfrac{1}3\]
C


\[x= 1\dfrac{1}3, y=1\dfrac{1}3\]
(Choice D)


\[x=2\dfrac{1}3, y=1\dfrac{1}3\]
D


\[x=2\dfrac{1}3, y=1\dfrac{1}3\]

Answer 1

To solve the system of equations

\[
\begin{cases}
7x - y = 7 \quad (1) \\
x + 2y = 6 \quad (2)
\end{cases}
\]

we can use substitution or elimination. Here, we'll use substitution.

From equation (1), we can express \(y\) in terms of \(x\):

\[
y = 7x - 7 \quad (3)
\]

Now, substitute equation (3) into equation (2):

\[
x + 2(7x - 7) = 6
\]

This simplifies to:

\[
x + 14x - 14 = 6
\]

Combining the terms gives:

\[
15x - 14 = 6
\]

Now, add 14 to both sides:

\[
15x = 20
\]

Divide both sides by 15:

\[
x = \frac{20}{15} = \frac{4}{3} = 1 \frac{1}{3}
\]

Now substitute \(x = 1 \frac{1}{3}\) back into equation (3) to find \(y\):

\[
y = 7(1 \frac{1}{3}) - 7
\]

Convert \(1 \frac{1}{3}\) to an improper fraction:

\[
1 \frac{1}{3} = \frac{4}{3}
\]

So \(y\) becomes:

\[
y = 7 \cdot \frac{4}{3} - 7 = \frac{28}{3} - 7 = \frac{28}{3} - \frac{21}{3} = \frac{7}{3} = 2 \frac{1}{3}
\]

Thus, the solution to the system is:

\[
x = 1 \frac{1}{3}, \quad y = 2 \frac{1}{3}
\]

Looking at the choices provided, the correct answer is:

**Choice A:** \( x = 1 \frac{1}{3}, y = 2 \frac{1}{3} \)