Question

Which ordered pair is a solution to the system of linear equations One-half x minus three-fourths y = StartFraction 11 Over 60 EndFraction and Two-fifths x + one-sixth y = StartFraction 3 Over 10 EndFraction?
(one-fifth, negative two-thirds)
(one-fifth, two-thirds)
(two-thirds, negative one-fifth)
(two-thirds, one-fifth)

Answer 1

To determine which ordered pair is a solution to the given system of linear equations, we can substitute each pair into both equations to see if they satisfy both.

The equations we have are:

1. \(\frac{1}{2} x - \frac{3}{4} y = \frac{11}{60}\)
2. \(\frac{2}{5} x + \frac{1}{6} y = \frac{3}{10}\)

Now let's test each ordered pair:

### 1. \((\frac{1}{5}, -\frac{2}{3})\)

**For the first equation:**
\[
\frac{1}{2} \cdot \frac{1}{5} - \frac{3}{4} \cdot \left(-\frac{2}{3}\right) = \frac{1}{10} + \frac{1}{2} = \frac{1}{10} + \frac{5}{10} = \frac{6}{10} = \frac{3}{5}
\]
**Check:**
\(\frac{3}{5} \neq \frac{11}{60}\)

**For the second equation:**
\[
\frac{2}{5} \cdot \frac{1}{5} + \frac{1}{6} \cdot \left(-\frac{2}{3}\right) = \frac{2}{25} - \frac{1}{9}
\]
**Finding a common denominator (225):**
\(\frac{2}{25} = \frac{18}{225}\), \(-\frac{1}{9} = -\frac{25}{225}\)
So:
\[
\frac{18}{225} - \frac{25}{225} = -\frac{7}{225} \neq \frac{3}{10}
\]

### 2. \((\frac{1}{5}, \frac{2}{3})\)

**For the first equation:**
\[
\frac{1}{2} \cdot \frac{1}{5} - \frac{3}{4} \cdot \frac{2}{3} = \frac{1}{10} - \frac{1}{2} = \frac{1}{10} - \frac{5}{10} = -\frac{4}{10} = -\frac{2}{5}
\]
**Check:**
\(-\frac{2}{5} \neq \frac{11}{60}\)

**For the second equation:**
\[
\frac{2}{5} \cdot \frac{1}{5} + \frac{1}{6} \cdot \frac{2}{3} = \frac{2}{25} + \frac{1}{9}
\]
**Finding a common denominator (225):**
\(\frac{2}{25} = \frac{18}{225}, \frac{1}{9} = \frac{25}{225}\)
So:
\[
\frac{18}{225} + \frac{25}{225} = \frac{43}{225} \neq \frac{3}{10}
\]

### 3. \((\frac{2}{3}, -\frac{1}{5})\)

**For the first equation:**
\[
\frac{1}{2} \cdot \frac{2}{3} - \frac{3}{4} \cdot \left(-\frac{1}{5}\right) = \frac{1}{3} + \frac{3}{20}
\]
**Finding a common denominator (60):**
\(\frac{1}{3} = \frac{20}{60}, \frac{3}{20} = \frac{9}{60}\)
So:
\[
\frac{20}{60} + \frac{9}{60} = \frac{29}{60} \neq \frac{11}{60}
\]

**For the second equation:**
\[
\frac{2}{5} \cdot \frac{2}{3} + \frac{1}{6} \cdot \left(-\frac{1}{5}\right) = \frac{4}{15} - \frac{1}{30}
\]
**Finding a common denominator (30):**
\(\frac{4}{15} = \frac{8}{30}\)
So:
\[
\frac{8}{30} - \frac{1}{30} = \frac{7}{30} \neq \frac{3}{10}
\]

### 4. \((\frac{2}{3}, \frac{1}{5})\)

**For the first equation:**
\[
\frac{1}{2} \cdot \frac{2}{3} - \frac{3}{4} \cdot \frac{1}{5} = \frac{1}{3} - \frac{3}{20}
\]
**Finding a common denominator (60):**
\(\frac{1}{3} = \frac{20}{60}\), \(\frac{3}{20} = \frac{9}{60}\)
So:
\[
\frac{20}{60} - \frac{9}{60} = \frac{11}{60}
\]
**Check: This works!**

**For the second equation:**
\[
\frac{2}{5} \cdot \frac{2}{3} + \frac{1}{6} \cdot \frac{1}{5} = \frac{4}{15} + \frac{1}{30}
\]
**Finding a common denominator (30):**
\(\frac{4}{15} = \frac{8}{30}\)
So:
\[
\frac{8}{30} + \frac{1}{30} = \frac{9}{30} = \frac{3}{10}
\]
**Check: This works too!**

So the ordered pair that satisfies both equations is \((\frac{2}{3}, \frac{1}{5})\).