Your cart is empty.
Practice each question, then open the answer or worked solution.
Draw the graphs of linear equations y = x and y = – x on the same Cartesian plane. What do you observe?
According to the question,
y = x ——- eq (i)
Values of x and y satisfying the equation=
| x | – 1 | 0 | 1 |
| y | – 1 | 0 | 1 |
y = -x ——- (ii)
Values of x and y satisfying the equation=
| x | – 1 | 0 | 1 |
| y | 1 | 0 | – 1 |
Plotting the graph:
From the above graph,
We observe that the two lines y = x and y = – x intersect each other at O (0, 0).
Determine the point on the graph of the linear equation 2x + 5y = 19 whose ordinate is 1½ times its abscissa.
From the question, we have,
2x + 5y = 19 …(i)
According to the question,
Ordinate is 1½ times its abscissa
⇒ y = 1½ x = (3/2) x
Substituting y = (3/2)x in eq. (i)
We get,
2x + 5 (3/2) x = 19
(19/2)x = 19
x = 2
Substituting x = 2 in eq. (i)
We get
2x + 5y = 19
2(2) + 5y = 19
y = (19 – 4)/5 = 3
Hence, we get x =2 and y = 3
Thus, point (2, 3) is the required solution.
Draw the graph of the equation represented by a straight line which is parallel to the x-axis and at 3 units below.
According to the question,
We get the linear equation,
y = – 3
Values of x and y satisfying the equation=
| x | – 1 | 0 | 1 |
| y | – 3 | – 3 | – 3 |
Plotting the graph:
Draw the graph of the linear equation whose solutions are represented by the points having the sum of the coordinates as 10 units.
According to the question,
We get the linear equation,
x + y = 10
We get,
x = 10 – y
Values of x and y satisfying the equation=
| x | 10 | 5 | 0 |
| y | 0 | 5 | 10 |
Plotting the graph:
Write the linear equation such that each point on its graph has an ordinate 3 times its abscissa.
According to the question,
A linear equation such that each point on its graph has an ordinate(y) which is 3 times its abscissa(x).
So we get,
⇒ y = 3x.
Hence, y = 3x is the required linear equation.