24 questions · timed · auto-graded
Natural numbers.
Solution:
The equation 2x + 5y = 7 has a unique solution, if x and y are natural numbers.
If we take x = 1 and y = 1, the given equation is satisfied.
Infinitely many solutions.
Solution:
The linear equation 3x - 5y = 15 has infinitely many solutions since any every point on this line will be a solution of this equation.
For different values of x, we will get get the corresponding different values of y.
Since x can take infinitely many values, y will also have infinite values.
Hence, the line will have infinitely many solutions.
(x, 0)
Solution:
Any point on x-axis is of the form (x, 0), where $\text{x}\neq0,$
Since its y-coordinate will be 0 always.
The line x + y = 0
Solution:
A point which lies on the x-axis has its y-coordinate = 0
While a point which lies on the y-axis has its x-coordinate = 0.
So, the points of the form (a, -a) will not lie on either axes.
Also, it does not satisfy the equation on of the line y = x.
The point of the form (a, -a) lies on the line x + y = 0 since it satisifes the equation of the given line.
(0, 2)
Solution:
When a graph meets the y-axis, the x coordinate is zero.
Thus, substituting x = 0 in the given equation,
We get:
2(0) + 3y = 6
⇒ 3y = 6
⇒ y = 2
Hence, the required point is (0, 2).
Parallel to the x-axis at a distance of 2 units below the x-axis.
Solution:
The graph of y + 2 = 0 is a line parallel to the x-axis at a distance of 2 units below the x-axis.
y = 0
Solution:
The equation of the x-axis is y = 0.
(5, 0)
Solution:
When a graph meets the x-axis, the y coordinate is zero.
Thus, substituting y = 0 in the given equation,
We get:
2x + 5(0) = 10
⇒ 2x = 10
⇒ x = 5
Hence, the required point is (5, 0).
Parallel to the y-axis at a distance of 4 units from the origin.
Solution:
The graph of x = 4 is a line parallel to the y-axis at a distance of 4 units from the origin.
4
Solution:
Since, (2, 0) is a solution of the linear equation 2x + 3y = k, substituting x = 2 and y = 0 in the given equation,
We have:
2(2) + 3(0) = k
⇒ 4 + 0 = k
⇒ k = 4
y = 0
Solution:
The equation of the y-axis is x = 0.
Parallel to the x-axis at a distance of 5 units from the origin.
Solution:
The graph of y = 5 is a line parallel to the x-axis at a distance of 5 units from the origin.
x + y = 7
Solution:
Substituting x = 5 and y = 2 in L.H.S. of equation x + y = 7,
We get:
LHS
= 5 + 2
7 = RHS
Hence, x = 5 and y = 2 is a solution of the linear equation x + y = 7.
Parallel to the y-axis at a distance of 3 units to the left of y-axis.
Solution:
The graph of x + 3 = 0 is a line parallel to the y-axis at a distance of 3 units to the left of y-axis.
(2, 3)
Solution:
Since, the y coordinate is 3, the graph of the line y = 3 passes through the point (2, 3).
Infinitely many.
Solution:
Infinitely many linear equations can be satisfied by x = 2 and y = 3.
(0, y)
Solution:
Any point on y-axis is of the form (0, y), where $\text{y}\neq0,$
Since its x-coordinate will always be 0.
(3, 2)
Solution:
The line x = 3 passes through the point (3, 2).
$\text{a}\neq0,\ \text{b}\neq0$
$\text{a}\neq0,\ \text{b}=0$
$\text{a}=0,\ \text{b}\neq0$
$\text{a}=0,\ \text{c}=0$
$\text{a}\neq0,\ \text{b}\neq0$
Solution:
A linear equation in tow variables x and y is of the form ax + by + c = 0, where $\text{a}\neq0$ and $\text{b}\neq0,$ since if either a or be is 0, the degree of the equation would be but it would not be a linear equation in tow variables.
If both a and b are 0, then the equation is not linear.
$\frac{2}{5}$
$\frac{5}{3}$
$\frac{3}{5}$
$\frac{2}{7}$
$\frac{5}{3}$
Solution:
Since the point (3, 4) lies on the graph of 3y = ax + 7,
substituting x = 3 and y = 4 in the given equation,
We get:
3(4) = a(3) + 7
⇒ 12 = 3a + 7
⇒ 3a = 5
$\Rightarrow\text{a}=\frac{5}{3}$
$\Big(\frac{-1}{2},\frac{1}{2}\Big)$
$\Big(\frac{3}{2},\frac{-3}{2}\Big)$
$(0,-1)$
$(1, 1)$
$(1, 1)$
Solution:
The given linear equation is x = y = 0.
We have to check which of the point satisfy the given equation.
consider option (a):
Substituting $\text{x}=-\frac{1}{2}$ and $\text{y}=\frac{1}{2}$ in the LHS if the given linear equation
$\therefore\ \text{x}-\text{y}=-\frac{1}{2}-\frac{1}{2}=-1\neq\text{RHS}$
$\therefore\ \text{x}=-\frac{1}{2}$ and $\text{y}=\frac{1}{2}$ does not satisfy the given linear equation.
Consider option (b):
Substituting $\text{a}=\frac{3}{2}$ and $\text{y}=-\frac{3}{2}$ in the LHS if the given linear equation on
$\therefore\ \text{x}-\text{y}=\frac{3}{2}+\frac{3}{2}=3\neq\text{RHS}$
$\therefore\ \text{x}=-\frac{3}{2}$ and $\text{y}=-\frac{3}{2}$ does not satisfy the given linear eqation on.
Consider option (d):
Substitution x = 1 and y = 1 in the LHS if the given linear equation
$\therefore\ $x - y = 1 - 1 = 0 = RHS
$\therefore\ $x = 1 and y = 1 satisfies the given linear equation.
The line y = x
Solution:
Given, a point of the form (a, a), where $\text{a}\neq0$
When a = 1, the point is (1, 1)
When a = 2, the point is (2, 2) ... and so on.
Plot the points (1, 1) and (2, 2) ... and so on.
Join the points and extend them in both the direction.
You will get equation of the line y = x.
(-3, 2)
Solution:
The line y = -3 does not pass through the point (-3, 2) since $\text{y}\neq2.$
x + y = 0
Solution:
Since given that each of the three points is a solution of the linear equation, all three points have to satisfy the linear equation.
We need to check for each of the four given equations.
Substituting x = -2 and y = 2 in option (b),
We get:
LHS
= x + y
= -2 + 2
0 = RHS
$\therefore\ $x = -2 and y = 2
Satisfy the given linear equation.
Substituting x = 0 and y = 0 in option (b),
We get:
LHS
= x + y
= 0 + 0
0 = RHS
$\therefore\ $x = 0 and y = 0
Satisfy the given linear equation.
Substituting x = -2 and y = 2 in option (b),
We get:
LHS
= x + y
= 2 - 2
0 = RHS
$\therefore\ $x = 2 and y = -2
Satisfy the given linear equation.
So, clearly all the three points satisfy the equation
x + y = 0.