A Bunch of Questions on Linear Functions (and direct variation)

A Bunch of Questions on Linear Functions (and direct variation)

  

1.	Which of the ordered pairs             (6, 1), (8, 0), (4, –2), (–4, 6) are solutions for the equation x + 2y = 8?

Solution:

To be solutions the pair of numbers (x,y) must satisfy the equation when you substitute the values into it.

(6,1):    LHS$=6+2\times 1=8=$RHS    YES

(8,0):    LHS$=8+2\times 0= 8 =$RHS    YES

(4,-2):    LHS$=4+2\times -2=4-4=0\ne$RHS    NO

(-4,6):    LHS$=-4+2\times 6=8=$RHS    YES

  

3.	Complete the ordered pairs for the equation 2x + y = 10.             (5,    ), (    , 10), (    , –2), (0,    )

 Solution

When $x=5$:  $2\times 5+y=10\implies y=5$, so $(5,5)$.

 When $y=10$:  $2x+10=10\implies x=0$, so $(0,10)$.

 When $y=-2$:  $2x-2=10\implies x=6$, so $(6,-2)$.

 When $x=0$:  $2\times 0+y=10\implies y=10$, so $(0,10)$.

4.	Find four solutions for the equation 3x + 5y = 15.

Solution: Just use 4 values of $x$ and find the corresponding $y$ values.

When $x=0$, $3(0)+5y=15\implies y=3$ so $(0,3)$.

When $x=1$, $3(1)+5y=15\implies y=12/5$ so $(1,12/5)$.

When $x=2$, $3(2)+5y=15\implies y=9/5$ so $(2,9/5)$.

When $x=3$, $3(3)+5y=15\implies y=6/5$ so $(3,6/5)$.

5.	Graph 2x – y = 4.

 Solution

6.	Graph using the intercept method: $x + 3y = 6$.

  Solution

When $y=0, x=6$

When $x=0, 3y=6\implies y=2$

So the $x$ and $y$ intercepts are $x=6$ and $y=2$.

7.	Graph by first solving for y.             $4x – 3y = 6$

  Solution:

$4x – 3y = 6$

$– 3y = 6-4x$

$3y = -6+4x$

$3y =4x-6$

$y ={4\over 3}x-2$

So $y$ intercept is $2$ and gradient is $m={4\over 3}$.

8.	Graph using the intercept method: 2x + y = 4.

  Solution

When $y=0, 2x=4\implies x=2$

When $x=0, y=4$

So the $x$ and $y$ intercepts are $x=2$ and $y=4$.

9.	Find the slope of the line passing through the points (9, 12) and (8, 4).

 Solution

Slope $=\displaystyle { y_2-y_1 \over x_2-x_1}= { 12-4 \over 9-8}={8\over 1}=8$.

10.	Find the slope of the line passing through the points (–10, 10) and (0, 0).

 Solution

Slope $=\displaystyle { y_2-y_1 \over x_2-x_1}= { 10-0 \over -10-0}={10\over -10}=-1$.

11.	Find the slope of the graphed line.

  Solution: $x=5$

12.	Find the slope of the graphed line.

  Solution:

Slope $=-{rise\over run}={3\over 9}={1\over 3}$

 [It is negative since it slopes downwards]

13.	If y varies directly with x, and y = 110 when x = 100, find the constant of variation k.

 Solution:

 $y=kx$.

$110=k\times 100$

$\therefore k={110\over 100}=1.1$

Update the equation: $\therefore y=1.1 x$ 

15.	The line graph below shows the number of stray dogs in a certain city for the years listed.  (a)  Which year had the least amount of stray dogs Solution: 1999 (b)  Between which two years did the greatest decrease in stray dogs occur?

           Solution: 2002-2003

16.	A student earns $0.65 for each mistake she finds in a text.  Sketch the equation of direct variation. $E=0.65m$

 

17.	Find the slope and y-intercept.             8x + 5y = –50

Solution

Make $y$ the subject and then read the gradient m, and $y$-intercept $b$ from the equation $y=mx+b$.

$8x + 5y = –50$
$5y = –50-8x$
$y = –50/5-8x/5$
$y=-1-{8\over 5}x$
$y=-{8\over 5}x-1$
$\therefore$ gradient $m=-{8\over 5}$, $y$ intercept $=-1$.

18.	Write the equation of the line with slope 4 and y-intercept (0, –5).  Then graph the line.

 $y=mx+b$

$y=4x-5$

19.	Write the equation of the line with slope $-{1\over 2}$ and y-intercept (0, 3).  Then graph the line.

 $y=mx+b$

$y=-0.5x+3$

20.	A line passing through (10, 4) and (–3, y) is perpendicular to a line with slope $-{13\over 14}$.  Find the value of y. The perpendicular line has gradient which is begative reciprocal of $-{13\over 14}$, and therefore equals ${14\over 13}$. ${y-4\over -3-10}={14\over 13}$ $y-4=-14$ $y=-10$

21.	An airplane covered 15 miles of its route while decreasing its altitude by 31,000 feet.  Find the slope of the airplane's line of descent.  Round to the nearest hundredth.  [Hint: 1 mi = 5280 feet.]

22.	Determine which two equations represent parallel lines. (a) $y = {5\over 4}x + 3$ (b) $y ={4\over 5}x + 7$ (c) $y = {4\over 5}x – 7$ (d) $y = –{5\over 4}x + 7$ Solution

(b) and (c) since the number in front of $x$ is the same (in gradient intercept form)

23.

Determine which two equations represent perpendicular lines. (a) $y=9x-9$ (b) $y={1\over 9}x+9$ (c) $y=-{1\over 9}x+9$ (d) $y={1\over 9}x-9$ solution The theory says that if gradients are perpendicular then their product is $-1$. (a) and (c) since the product of the gradients equals $-1$, that is $$9\times -{1\over 9}=-1$$

24.	Write the equation of the line that passes through point (0, –8) with a slope of a ${4\over 5}$.

$y={4\over 5}x-8$
$5y=4x-40$
$4x-5y-40=0$

25.	Write the equation of the line passing through (6, 37) and (1, 12). gradient is $m={37-12\over 6-1}={25\over 5}=5$ $y-y_1=m(x-x_1)$ $y-12=5(x-1)$ $y-12=5x-5$ $y=5x+7$  or $5x-y+7=0$.