Question

this graph shows how the number of salad plates that jon has left is related to the number of tables that he has set.
how many tables does jon need to set in order to have just 7 salad plates left to distribute?
\boxed{} tables

Explanation:

Step1: Find initial plates and rate

When tables set $x=0$, plates left $y=10$. When $x=3$, $y=0$. Rate: $\frac{10-0}{0-3}=-\frac{10}{3}$ plates per table. Equation: $y = -\frac{10}{3}x + 10$

Step2: Solve for x when y=7

Substitute $y=7$ into the equation:
$7 = -\frac{10}{3}x + 10$
Subtract 10 from both sides:
$7-10 = -\frac{10}{3}x$
$-3 = -\frac{10}{3}x$
Multiply both sides by $-\frac{3}{10}$:
$x = (-3) \times (-\frac{3}{10}) = \frac{9}{10}$? No, correct rate: $\frac{10}{3}$ plates per table, so per table uses $\frac{10}{3}$? No, better: from (0,10) to (3,0), each table uses $\frac{10}{3}$? No, wait, when 3 tables set, 10 plates used, so per table uses $\frac{10}{3}$? No, the graph is linear, so when plates left is 7, plates used is $10-7=3$. Plates per table: $\frac{10}{3}$ plates per table. So tables needed: $\frac{3}{\frac{10}{3}}=0.9$? No, wrong. Wait, re-express:
Equation: slope $m=\frac{0-10}{3-0}=-\frac{10}{3}$. So $y=-\frac{10}{3}x +10$. Set $y=7$:
$7 = -\frac{10}{3}x +10$
$-\frac{10}{3}x = 7-10=-3$
$x = (-3) \times (-\frac{3}{10}) = \frac{9}{10}=0.9$? No, that can't be. Wait, no, the graph: when x=1, y=10 - 10/3 ≈6.66? No, the graph shows at x=1, y≈7? Wait, the graph's line at x=1 is at y=7. Oh! I misread the graph. At x=1, the line is at y=7.

Step3: Read graph directly

Look for y=7 on the vertical axis, find the corresponding x value on the horizontal axis. The line passes through (1,7).

Answer:

1 tables