Question

Verify if (Five-thirds, negative two-thirds) is a solution to the system of equations.

y = one-half x minus three-halves

y = –x + 1

Is the ordered pair a solution to the system?
Yes, because there are infinite solutions to the system of equations.
No, because substituting these values into the first equation creates a false statement.
Yes, because substituting these values into both equations forms two true statements.
No, because substituting these values into the second equation forms a false statement.

Answer 1

To verify if the ordered pair \((\frac{5}{3}, -\frac{2}{3})\) is a solution to the system of equations, we will substitute \(x = \frac{5}{3}\) and \(y = -\frac{2}{3}\) into both equations.

1. Check the first equation:
\[
y = \frac{1}{2}x - \frac{3}{2}
\]
Substituting the values:
\[
-\frac{2}{3} = \frac{1}{2} \cdot \frac{5}{3} - \frac{3}{2}
\]

Calculate the right side:
\[
\frac{1}{2} \cdot \frac{5}{3} = \frac{5}{6}
\]

Now, substituting:
\[
-\frac{2}{3} = \frac{5}{6} - \frac{3}{2}
\]
To subtract, we need a common denominator. The common denominator between 6 and 2 is 6. Rewrite \(-\frac{3}{2}\) as \(-\frac{9}{6}\):
\[
-\frac{2}{3} = \frac{5}{6} - \frac{9}{6}
\]
Simplifying the right side:
\[
-\frac{2}{3} = \frac{5 - 9}{6} = -\frac{4}{6} = -\frac{2}{3}
\]

The first equation holds true.

2. Check the second equation:
\[
y = -x + 1
\]
Substituting the values:
\[
-\frac{2}{3} = -\left(\frac{5}{3}\right) + 1
\]

Calculate the right side:
\[
-\left(\frac{5}{3}\right) + 1 = -\frac{5}{3} + 1
\]
Rewrite 1 as \(\frac{3}{3}\):
\[
-\frac{5}{3} + \frac{3}{3} = -\frac{5 - 3}{3} = -\frac{2}{3}
\]

The second equation holds true.

Since substituting the values into both equations yields true statements, the correct answer is:

**Yes, because substituting these values into both equations forms two true statements.**