Question

unit 3 lesson 13 - practice problems

1. for each equation, find ( y ) when ( x = -3 ). then find ( x ) when ( y = 2 )

a. ( y = 6x + 8 )
b. ( y = \frac{2}{3}x )
c. ( y = -x + 5 )
d. ( y = \frac{3}{4}x - 2\frac{1}{2} )
e. ( y = 1.5x + 11 )

1. match each graph of a linear relationship to a situation that most reasonably reflects its context.
2. ( y ) is the weight of a kitten ( x ) days after birth.
3. ( y ) is the distance left to go in a car ride after ( x ) hours of driving at a constant rate toward its destination.
4. ( y ) is the temperature, in degrees c, of a gas being warmed in a laboratory experiment.
5. ( y ) is the amount of calories consumed eating ( x ) crackers.
6. true or false: the points ( (6, 13) ), ( (21, 33) ), and ( (99, 137) ) all lie on the same line. the equation of the line is ( y = \frac{4}{3}x + 5 ). explain or show your reasoning.

Explanation:

Problem 1

Part a: $y=6x+8$

Step1: Find $y$ at $x=-3$

Substitute $x=-3$ into equation:
$y = 6(-3) + 8 = -18 + 8$

Step2: Calculate $y$ value

$y = -10$

Step3: Find $x$ at $y=2$

Substitute $y=2$ into equation:
$2 = 6x + 8$

Step4: Solve for $x$

$6x = 2 - 8 = -6$
$x = \frac{-6}{6} = -1$

Part b: $y=\frac{2}{3}x$

Step1: Find $y$ at $x=-3$

Substitute $x=-3$ into equation:
$y = \frac{2}{3}(-3)$

Step2: Calculate $y$ value

$y = -2$

Step3: Find $x$ at $y=2$

Substitute $y=2$ into equation:
$2 = \frac{2}{3}x$

Step4: Solve for $x$

$x = 2 \times \frac{3}{2} = 3$

Part c: $y=-x+5$

Step1: Find $y$ at $x=-3$

Substitute $x=-3$ into equation:
$y = -(-3) + 5 = 3 + 5$

Step2: Calculate $y$ value

$y = 8$

Step3: Find $x$ at $y=2$

Substitute $y=2$ into equation:
$2 = -x + 5$

Step4: Solve for $x$

$-x = 2 - 5 = -3$
$x = 3$

Part d: $y=\frac{3}{4}x - 2\frac{1}{2}$

Rewrite $2\frac{1}{2}$ as $\frac{5}{2}$

Step1: Find $y$ at $x=-3$

Substitute $x=-3$ into equation:
$y = \frac{3}{4}(-3) - \frac{5}{2} = -\frac{9}{4} - \frac{10}{4}$

Step2: Calculate $y$ value

$y = -\frac{19}{4} = -4\frac{3}{4}$

Step3: Find $x$ at $y=2$

Substitute $y=2$ into equation:
$2 = \frac{3}{4}x - \frac{5}{2}$

Step4: Isolate $x$ term

$\frac{3}{4}x = 2 + \frac{5}{2} = \frac{4}{2} + \frac{5}{2} = \frac{9}{2}$

Step5: Solve for $x$

$x = \frac{9}{2} \times \frac{4}{3} = 6$

Part e: $y=1.5x+11$

Rewrite $1.5$ as $\frac{3}{2}$

Step1: Find $y$ at $x=-3$

Substitute $x=-3$ into equation:
$y = 1.5(-3) + 11 = -4.5 + 11$

Step2: Calculate $y$ value

$y = 6.5$

Step3: Find $x$ at $y=2$

Substitute $y=2$ into equation:
$2 = 1.5x + 11$

Step4: Isolate $x$ term

$1.5x = 2 - 11 = -9$

Step5: Solve for $x$

$x = \frac{-9}{1.5} = -6$

1. A kitten's weight starts at 0 at birth and increases over time, matching Graph A (starts at origin, positive slope).
2. Distance left to a destination starts at a positive value and decreases over time, matching Graph D (positive y-intercept, negative slope).
3. A gas being warmed starts below zero (initial cold temperature) and increases, matching Graph C (negative y-intercept, positive slope).
4. Calories from crackers start at a positive minimum (even 0 crackers has 0, but the graph has a positive intercept representing base calories, or a slow positive increase), matching Graph B (positive y-intercept, small positive slope).

Step1: Test point $(6,13)$

Substitute $x=6$ into $y=\frac{4}{3}x+5$:
$y = \frac{4}{3}(6) + 5 = 8 + 5 = 13$

Step2: Test point $(21,33)$

Substitute $x=21$ into $y=\frac{4}{3}x+5$:
$y = \frac{4}{3}(21) + 5 = 28 + 5 = 33$

Step3: Test point $(99,137)$

Substitute $x=99$ into $y=\frac{4}{3}x+5$:
$y = \frac{4}{3}(99) + 5 = 132 + 5 = 137$

Answer:

a. $y=-10$ when $x=-3$; $x=-1$ when $y=2$
b. $y=-2$ when $x=-3$; $x=3$ when $y=2$
c. $y=8$ when $x=-3$; $x=3$ when $y=2$
d. $y=-\frac{19}{4}$ when $x=-3$; $x=6$ when $y=2$
e. $y=6.5$ when $x=-3$; $x=-6$ when $y=2$

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Problem 2