Grade 9 - Linear-Equations-in-Two-Variables

(Sample Printed worksheet)

Answer The Following

1)At what point does line represented by the equation 4x + 5y = 7 intersects a line which is parallel to the x-axis, and at a distance 1 units from the origin and in the negative direction of y-axis.
2)An internet service provider charges $ 1.4 for the first minute and $ 0.9 per minute for subsequent minutes of a call. If duration of connection is represented as d, and amount charged is represented as c, find the linear equation for this relationship.
3)A line passe through points (3, 2) and (-4, -5). Find the y-intercept of the line.
4)The positive solutions of the equation ax + by + c = 0 always lie in which quadrant?
5)If solutions of a linear equation are (-7, -7), (0, 0) and (7, 7), find the equation.

Choose correct answer(s) from given choice

6)The Taxi fare in Philadelphia is $ 20 for the first kilometer and $ 11 per kilometer for subsequent distance covered. If distance is represented as d, and fare is represented as f, find the linear equation for this relationship.
a. f = 20d + 11 b. f = 11d + 9
c. f = 11d + 20 d. f = 20d + 9
7)The equation of x-axis is

a. x = 0 b. y = 0
c. x + y = 0 d. x = y
8)The equation of straight line which is parallel to y-axis, and is at a distance of p from y-axis is
a. y - x = p b. x = p
c. y + x = p d. y = p
9)Find the linear equation represented in the graph below.

a. y = -x + 2 b. y = x + 2
c. y = 2 d. y = x
10)Find the point where linear equation 5x + 6y = 48 intersects with y-axis.
a. (5, 0) b. (0, 8)
c. (0, 6) d. (8, 0)

Answers

1) (3, -1)
Let's consider the second line first.
The line which is parallel to the x-axis and is at a distance 1 units from the origin in the negative direction of the y-axis is defined by the following equation
  y=-1 So, now we know that at the point of intersection, the value of y = -1
The equation of the first line is
 4x + 5y = 7
Subtituting for y with the value -1 in this equation, we get
 x = 3
So the answer is that the intersection is at the point (3, -1)

2) c = 0.9d + 0.5
Step 1
We are given the following facts
- The charge for the first minute is is $ 1.4
  - The charge per minute after that is $ 0.9
Step 2
We can see that the charge will be dependent on the time spent in minutes
So we set c on the left hand side
c = Some linear function of d
Step 3
We know that after the first minute, the rate is $ 0.9 per minute.
So if the connection lasts for d minutes, there will be a charge $ 1.4 for the first minute, and $ 0.9 for d - 1 minutes.
Step 4
This means the equation is c = 1.4 + ((d - 1) x 0.9)
Step 5
Simplifying, we get c = 0.9d + 0.5

3) -1
Step 1
Equation of line y = m x + c
Step 2
Substitute first point in the equation
2 = 3 m + c
m = (2 - c)/3 ________________(1)
Step 3
Substitute second point in the equation
-5 = -4 m + c
m = (-5 - c)/-4 ________________(2)
Step 4
On equating value of m from both equations,
(2 - c)/3 = (-5 - c)/-4
-8 + 4c = -15 - 3c
7 c = -7
c = -1

4) First quadrant

5) x = y
The points (a,b) that solve a linear equation would satisfy ax+by=c, where a,b,c are constants.
Substituting (0,0) we see that c = 0.
Substituting the other two points (-7, -7) and (7, 7), we get the following equations
-7x + (-7y)=0 and 7x + 7y = 0
From these two equations, we can therefore eliminate the variables and see that the answer is x = y

6) b. f = 11d + 9
Step 1
We are given the following facts
- The fare for the first kilometer is $ 20
  - The fare per kilometer after that is $ 11
Step 2
We can see that the fare will be dependent on the distance
So we set f on the left hand side
f = Some linear function of d
Step 3
We know that after the first kilometer, the rate is $ 11 per kilometer.
So if we travel d kilometers, we will get charged $ 20 for the first kilometer, and $ 11 for d - 1 kilometers.
Step 4
This means the equation is f = 20 + ((d - 1) x 11)
Step 5
Simplifying, we get f = 11d + 9

7) b. y = 0
Take a look at a graph



You can see that for the x-axis, the value of y is always 0. So the equation is y=0

8) b. x = p
If a line is parallel to the y-axis, then x value of it is constant for all values of y.
Take a look at the image to see this case

Further, if the line is distance p away from the y-axis, it also means that this constant value of x is p.
So the equation for that line is x=p

9) b. y = x + 2
The general equation of a line is y=mx+c
So we have to find m and c
To find c, note from the equation that c is the value of y when x=0 (i.e. the equation becomes y=m*0 + c, or y=c).
Look at the graph to see if this is a vertical line. If it is not (we'll see the case where it is later in this tip), then what the value of y is when the equation crosses the vertical axis
We see that the value of y at this point is 2. So c=2
The next part is finding m
The best way to consider m is to think of it as the slope of the line.
Think of it as the change in y for a given change in x.
Consider the two equations,
   y1 = mx1 + c, and
   y2 = mx2 + c
Now we subtract the first equation from the second
We get y1 - y2 = mx1 + c - (mx2 + c)
Simplifying,
  (y1 - y2) = m(x1 - x2)
or   m = (y1 - y2)/(x1 - x2)
Now, substitute the two points seen in the graph.
  m = (3 - (1))/(1 - (-1))
Also, note that this is the reason why we don't apply this when the line is vertical, because the denominator would be 0, and the equation is meaningless
This is solved to get the value of m, and get the answer m=1

Now, if the line is a vertical one, then you can solve it by inspection.

So the answer is y=x + 2.

10) b. (0, 8)
We are told to find the point where the equation intersects with the y axis.
Now, at that point the value of x will be zero.
So we need to substitute x=0 into the equation.
From there, we can then solve to find the value of y to be 0. So the point is (0,8)

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