Question

For the equation y=3/5x+7, find the missing values in the following ordered pairs.
​A(___,10​)
​B(−10​,___)
C (1/3,___)

Answer 1

To find the missing values in the ordered pairs based on the equation \( y = \frac{3}{5}x + 7 \), we can substitute the values of \( x \) or \( y \) as required.

### For Point \( A(___, 10) \):
We need to find \( x \) such that \( y = 10 \):
\[
10 = \frac{3}{5}x + 7
\]
Subtract 7 from both sides:
\[
10 - 7 = \frac{3}{5}x
\]
\[
3 = \frac{3}{5}x
\]
Now, multiply both sides by \( \frac{5}{3} \):
\[
x = 3 \cdot \frac{5}{3} = 5
\]
So, Point \( A \) is \( A(5, 10) \).

### For Point \( B(-10, ___) \):
We need to find \( y \) when \( x = -10 \):
\[
y = \frac{3}{5}(-10) + 7
\]
Calculating:
\[
y = -6 + 7 = 1
\]
So, Point \( B \) is \( B(-10, 1) \).

### For Point \( C\left(\frac{1}{3}, ___\right) \):
We need to find \( y \) when \( x = \frac{1}{3} \):
\[
y = \frac{3}{5}\left(\frac{1}{3}\right) + 7
\]
Calculating:
\[
y = \frac{3}{15} + 7 = \frac{1}{5} + 7 = \frac{1}{5} + \frac{35}{5} = \frac{36}{5}
\]
So, Point \( C \) is \( C\left(\frac{1}{3}, \frac{36}{5}\right) \).

### Summary of Points:
- \( A(5, 10) \)
- \( B(-10, 1) \)
- \( C\left(\frac{1}{3}, \frac{36}{5}\right) \)