1. The linear equation 2x – 5y = 7 has (A) A unique solution (B) Two solutions (C) Infinitely many solutions (D) No solution
Explanation:
Assuming 2x - 5y = 7,
By switching out 5y = 2x - 7
The result is y = (2x - 7)/ 5.
Here, we shall obtain various y values for various x values.
As a result, there are innumerable solutions to the linear equation.
2. The equation 2x + 5y = 7 has a unique solution, if x, y are :
(A) Natural numbers (B) Positive real numbers
(C) Real numbers (D) Rational numbers
Explanation:
The formula is 2x + 5y = 7.
Let's think about the value (1, 1).
a. (1, 1) is the answer to 2x + 5y = 7.
b. If positive real numbers are taken into account, the equation 2x + 5y = 7 will have a lot of solutions.
c. If real numbers are taken into account, 2x + 5y = 7 will have an unlimited number of answers.
If rational numbers are taken into account, there are multiple solutions to the equation 2x + 5y = 7.
We are aware that a two-variable linear equation has an infinite number of solutions.
There is a collection of infinite solution points for the linear equations system where an equation's L.H.S changes to an R.H.S.
Therefore, if x and y are natural numbers, the equation has a singular solution.
3. If (2, 0) is a solution of the linear equation 2x + 3y = k, then the value of k is
(A) 4 (B) 6 (C) 5 (D) 2
Explanation:
2x + 3y = k is the given linear equation.
Here, the point is given as (x, y) = (2, 0).
Let's change the values in equation 2 (2) + 3(0) = k that are given.
Calculating further, 4 + 0 = k
So, we obtain k = 4.
Consequently, k has a value of 4.
4. Any solution of the linear equation 2x + 0y + 9 = 0 in two variables is of the form
(A) (– 9/2 ,m) (B) (n,-9/2) © (0,-9/2) (D) (-9,0)
Explanation:
2x + 0y + 9 = 0 is the linear equation that is given.
It can be expressed as 2x + 9 = 0.
2x = -9
x = -9/2
Since the y coefficient is 0, any value can be used with no effect on the solution.
As a result, the two-variable linear equation has the form (-9/2, m).
5. The graph of the linear equation 2x + 3y = 6 cuts the y-axis at the point (A) (2, 0) (B) (0, 3) (C) (3, 0) (D) (0, 2)
Explanation:
2x + 3y = 6 is the given linear equation.
Given that the equation crosses the y-axis, the x-coordinate is zero.
In the linear equation 2 (0) + 3y = 6,
x = 0 is substituted.
0 + 3y = 6
3y = 6
3 y divided by both sides yields 2
therefore (0, 2) are the coordinates.
As a result, the point (0, 2) on the graph of the linear equation is where the y-axis is broken.
6. The equation x = 7, in two variables, can be written as
(A) 1 . x + 1 . y = 7 (B) 1. x + 0. y = 7
(C) 0 . x + 1 . y = 7 (D) 0 . x + 0 . y = 7
Explanation:
The response is B.
Here, the coefficient of 'y' in the formula x = 7 is equal to zero.
As a result, the formula is 1. x + 0. y = 7.
The necessary equation is therefore 1. x + 0. y = 7.
7. Any point on the x-axis is of form (A) (x, y) (B) (0, y) (C) (x, 0) (D) (x, x)
Explanation:
The response is C.
The y-coordinate of any point on the X-axis is equal to zero, or y = 0.
The generic form of each point on the X-axis is hence (x, 0).
8. Any point on the line y = x is of form (A) (a, a) (B) (0, a) (C) (a, 0) (D) (a, – a)
Explanation:
Any point on the line y = x is known to have the same x and y coordinates.
Therefore, any point on the line y = x will be (a, a). Any point on the line y = x will therefore be (a, a).
9. The equation of the x-axis is of form (A) x = 0 (B) y = 0 (C) x + y = 0 (D) x = y
Explanation:
The response is B.
All the X-axis points have the same y-coordinate, which is 0.
As a result, the X-axis equation has the form y = 0.
10. The graph of y = 6 is a line (A) parallel to the x-axis at a distance of 6 units from the origin
(B) parallel to the y-axis at a distance of 6 units from the origin (C) making an intercept 6 on the x-axis. (D) making an intercept 6 on both axes.
Explanation:
y = 6 is the given equation.
Y = 0 can be used to express it.x + 6
Here, y equals 6 for every value of x.
The points are 0, 6, 1, 6, 3, and 6...
These points can be plotted to produce a straight line parallel to the x-axis at a distance of 6 units from the x-axis.
The line is therefore 6 units from the origin and parallel to the x-axis.
11. x = 5, y = 2 is a solution of the linear equation (A) x + 2 y = 7 (B) 5x + 2y = 7 (C) x + y = 7 (D) 5 x + y = 7
Explanation:
In (a) x+2y=7, substituting the values x = 5 and y = 2 yields LHS=5+22=5+4=9#7=RHSi.e.LHSRHS in
(b) 5x+2y=7, LHS=5+2=7=RHSi.e.LHS=RHS in
(c), and LHS=5+2=27=RHSi.e.LHSRHS in
Therefore, (c) is the appropriate choice.
12. If a linear equation has solutions (–2, 2), (0, 0) and (2, – 2), then it is of form (A) y – x = 0 (B) x + y = 0 (C) –2x + y = 0 (D) –x + 2y = 0
Explanation:
The right answer is A x + y = 0.
For (-2, 2): -2 + 2 = 0
For (0, 0): 0 + 0 = 0
For (2, -2): 2 + (-2) = 0
Thus, the equation for which the provided points represent the answer is x + y = 0.
13. The positive solutions of the equation ax + by + c = 0 always lie in the
(A) 1st quadrant (B) 2nd quadrant (C) 3rd quadrant (D) 4th quadrant
Explanation:
Think of the favorable answers as (p, q).
Since the abscissa, p is positive, it will be located along the x-axis' positive direction.
Given that the ordinate, q, is positive, it will lie toward the y-axis's positive direction.
Consequently, (p, q) will be in the first quadrant.
As a result, the first quadrant will contain the positive solutions.
14. The graph of the linear equation 2x + 3y = 6 is a line that meets the x-axis at the point(A) (0, 2) (B) (2, 0) (C) (3, 0) (D) (0, 3)
Explanation:
A linear equation is one that has a degree of 1 as its maximum value.
In a linear equation, this means that no variable's exponent can be greater than 1.
2x + 3y = 6 is the given linear equation.
It intersects the x-axis, indicating that y = 0.
Let's replace it in the formula 2x + 3(0) = 6
2x = 6
2 x divided by both sides yields 3
As a result, the linear equation's graph is (3, 0).
15. The graph of the linear equation y = x passes through the point (A) 3/2 3/2)(B) (0,3/2) (C) (1, 1) (D) ½- ½
Explanation:
As far as we are aware, a linear equation is one in which the highest power of the variable is consistently 1. A linear equation is one that has a degree of 1 as its maximum value.
The x and y coordinates in the linear equation y = x are identical.
The graph comes from the suggested choice (1, 1).
Consequently, (1, 1) is where the linear equation's graph intersects.
16. If we multiply or divide both sides of a linear equation with a non-zero number, then the solution of the linear equation : (A) Changes (B) Remains the same (C) Changes in case of multiplication only (D) Changes in case of division only
Explanation:
Take the equation x + 2y = 5 as an example. (1)
In the equation 1 + 2y = 5 2y = 5 - 1 2y = 4, replace x = 1
two sides divided by two equals two.
The answer to the problem is (1, 2)
Multiplying equation (1) by three yields 3x + 6y = 15.
Replace with x = 1 and y = 2
LHS = 3(1) + 6(2) = 3+12=15, which equals RHS.
By a non-zero value multiplied by an equation, the answer remains the same.
Equation (1) is now divided by 2 by x/2 + y = 5/2.
Change equation (1) to read x = 1 and y = 2.
LHS = 1/2 + 2
LCM = (1 + 4)/2 = 5/2 = RHS after taking
By using a non-zero number to divide an equation, the result remains the same.
Consequently, the answer is still the same.
17. How many linear equations in x and y can be satisfied by x = 1 and y = 2?
(A) Only one (B) Two (C) Infinitely many (D) Three
Explanation:
The response is C.
Let axe by and c equal zero in the linear equation.
When we change x to 1 and y to 2, we get the following equation: => a + 2b + c = 0, where a, b, and c are real numbers.
Here, a+ 2b + c = 0 is satisfied for various values of a, b, and c.
Consequently, x = 1 and y = 2 can satisfy an endless number of linear equations in x and y.
18. The point of the form (a, a) always lies on (A) x-axis (B) y-axis (C) On the line y = x (D) On the line x + y = 0
Explanation:
The provided point has the form (a, a).
Here, the values of the x and y coordinates are identical.
It must therefore be on the line y = x.
The point is therefore on the line y = x.
19. The point of the form (a, – a) always lies on the line (A) x = a (B) y = – a (C) y = x (D) x + y = 0
Explanation:
D x + y = 0 x + y = a + (- a) = 0 is the right answer because the provided point has the form (a, -a).
Since x + y = 0, the point (a, -a) is always on that line.
(C) Short Answer Questions with Reasoning
20. The point (0, 3) lies on the graph of the linear equation 3x + 4y = 12.
Explanation:
Given is the linear equation 3x + 4y = 12.
The provided point is (0, 3).
In equation 3 (0) + 4 (3) = 12 12 = 12, let's substitute it.
The assertion is accurate as a result.
21. The graph of the linear equation x + 2y = 7 passes through the point (0, 7).
Explanation:
Given is the linear equation x + 2y = 7.
The indicated point is (0, 7).
Let's replace it in the formula 0 + 2 (7) = 7 14 7
22. The graph given below represents the linear equation x + y = 0.
Explanation:
Given is the linear equation x + y = 0.
It can be written as
x = -y
The graph's points are (-3, 3) and (-1, 1)
Taking into account (-3, 3) x = -3 and y = 3 -3 = 3
fulfills the equation
In light of (-1, 1)
x = -1 and y = 1 -1 = 1
fulfills the equation
The assertion is accurate as a result.
23. The graph given below represents the linear equation x = 3 (see Fig. 4.2).
Explanation:
based on the graph
At a distance of 3 units to the right of the origin, a line is perpendicular to the y-axis.
Consequently, x = 3 will be the linear equation.
The assertion is accurate as a result.
24. The coordinates of points in the table:
x 0 1 2 3 4
y 234 –5 6
represent some of the solutions of the equation x – y + 2 = 0.
Explanation:
The given equation, xy+2=0, can be satisfied by the points (0,2), (1,3), (2,4), and (4,6).
The equation xy+2o is solved at each of these positions.
However, if we substitute (3,-5), i.e. 3(-5)+2=0, i.e. 3+5=0, or 10=0, it does not meet the equation and is therefore untrue.
Because point (3, and 5) does not meet the provided equation, the stated statement is untrue.
25. Every point on the graph of a linear equation in two variables does not represent a solution of the linear equation.
Explanation:
The dots that are the answers to the given problem are connected to form a graph of a linear equation in two variables.
A solution to the linear equation must therefore exist at every point on the graph.
As a result, the assertion is untrue.
26. The graph of every linear equation in two variables need not be a line.
Explanation:
Every two-variable linear equation has the form y=mx+c.
In this case, y has a distinct value for each value of x.
Additionally, x and y are directly related.
As a result, xy will never change.
As a result, the graph will continue to have a distinct slope, which causes it to be a straight line. The assertion is untrue
27. Draw the graphs of linear equations y = x and y = – x on the same cartesian plane. What do you observe?
Explanation:
Given, the equations in the linear form are y = x and y = -x.
The equations' graphs must be drawn on the same cartesian plane.
To plot a few points on the graph using the equation y = x, we require a few data points.
For x = 1, y = 1
For x = 4, y = 4
The points of the equation are (1, 1) and (4, 4), respectively.
The graph of y = x is created by connecting the points (1, 1) and (4, 4) on the graph to form a straight line.
We need a few points to plot on the graph when y = -x is taken into consideration.
For x = 3, y = -3
For x = -4, y = 4
As a result, the equation's points are (3, -3) and (-4,4) The graph of y = -x is obtained by combining the points (3, -3) and (-4, 4), which are plotted on the graph as the points (3, -3) and (-4, 4) respectively.
28. Determine the point on the graph of the linear equation 2x + 5y = 19, whose ordinate is 1 ½ times its abscissa.
Explanation:
Given equation is 2x + 5y = 19, which is (1)
Considering that the ordinate is 1.5 times the abscissa
Therefore, y = 3x /2
Hence equation (1) becomes 2x + 5(3x /2) = 19 19x/2 = 19
x thus equals 2
y = 3x/2 = (3 × 2)/2 = 3
The graph's point is therefore (2, 3).
29. Draw the graph of the equation represented by a straight line that is parallel to the x-axis and at a distance 3 units below it.
Explanation:
The equation y = - k, where k is the distance of the line from the x-axis, can be used to calculate any straight line that is perpendicular to the X-axis and below the X-axis.
Here, k = 3
The line's equation will therefore be y=-3.
The needed graph is this one.
30. Draw the graph of the linear equation whose solutions are represented by the points having the sum of the coordinates as 10 units.
Explanation:
The graph's coordinates add up to 10 units, as is obvious.
The linear equation, whose solutions are represented by points with coordinates that add up to 10, must be graphed.
Let the coordinates be x and y.
Assuming x + y = 10,
When x=0
0+y=10,
y=10
Y = 10 - 3
y = 7 when x = 3 and
3 + y = 10.
When y = 10 - 5 and x = 5, the result is y = 5.
When x = 10 and 10 + 10 + 10 - 10 y = 0, y = 0.
The points are as follows: (0, 10), (3, 7), (5, 5) and (10, 0).
The graph of the equation x + y = 10 is obtained by plotting the points on the graph.
31. Write the linear equation such that each point on its graph has an ordinate 3 times its abscissa.
Explanation:
Given that the ordinate is three times the abscissa
The graph's linear equation must be written down.
Times the x-coordinate equals the y-coordinate.
So, y = 3x
If x = 1, then (1) y = 3
Y = 3(2) y = 6 when x = 2.
The points are therefore (1, 3), and (2, 6).
Plotting the graph's points (1, 3), and (2, 6),
We obtain a straight line AB by connecting the points, and its equation is y = 3x.
Consequently, y = 3x is the linear equation.
32. If the point (3, 4) lies on the graph of 3y = ax + 7, then find the value of a.
Explanation:
Given: The equation 3y = ax + 7 is linear.
By changing the values of x and y in the given equation, we may determine the value of 'a'.
We get 3(4) = a(3) + 7 by changing the values of x = 3 and y = 4 in the equation 3y = axe + 7.
12 = 3a + 7
3a = 5 a = 5/3
Consequently, a = 5/3.
33. How many solution(s) of equation 2x + 1 = x – 3 are there on the : (i) Number line (ii) Cartesian plane
Explanation:
The preceding equation can be represented as 1.x + 0.y = - 4 and is 2x + 1 = x - 3 2x - x = -3 - 1 x = -4....(ii)
(i) On the x-axis, a number line represents each real value of x. As a result, the number line's point at x = -4 is exactly one on the line.
(ii) In contrast, x + 4 = 0 depicts a straight line that is perpendicular to the y-axis. On a line in the Cartesian plane, there are an infinite number of points.
34. Find the solution of the linear equation x + 2y = 8 which represents a point on
(i) x-axis (ii) y-axis
Explanation:
The formula is x + 2y = 8.(i)
(i) Enter y = 0 in Eq. when the point is on the x-axis. (i). We discover that x + 2 0 = 8 x = 8 and that the necessary point is at (8, 0).
(ii) Enter x = 0 in Eq(i) when the point is on the y-axis. We discover that
0 + 2y = 8
y=82=4
The necessary point is therefore (0, 4).
35. For what value of c, the linear equation 2x + cy = 8 has equal values of x and y for its solution.
Explanation:
Given that, 2x + cy = 8 is a linear equation.
The solution to the equation has equal values for x and y.
The value of c must be determined.
Assuming x = y
In the preceding equation, substitute y = x, 2x + c(x) = 8,
cx = 8 - 2x,
and c = (8 - 2x)/x.
where x is greater than zero.
Consequently, the formula for c is (8 - 2x)/x.
36. Let y vary directly as x. If y = 12 when x = 4, then write a linear equation. What is the value of y when x = 5?
Explanation:
As x changes, y does also.
Y equals 12 when x equals 4.
In order to determine the value of y for x = 5, we must write a linear equation.
Since y is directly related to X, Y ∝ X
Consequently, y = kx is the linear equation.
where k is a randomly chosen constant.
From the starting point, 12 = k(4),
k = 12/4,
k = 3.
Consequently, y = 3x is the linear equation.
In the equation, enter x = 5,
y = 3(5),
y = 15.
Consequently, y has a value of 15.
37. Show that the points A (1, 2), B (– 1, – 16) and C (0, – 7) lie on the graph of the linear equation y = 9x – 7.
Explanation:
First, create the equation's graph.
y = 9x - 7
When x = 2 and y = 9, the linear equation at this point is: 2 - 7 = 18 - 7 = 11.
Y = 9 -2 - 7 when x = -2.
= -18 - 7 = -25
x2−2y11−25
Two points are located here: D(2, 11) and E(-2, -25).
In order to create the graph, plot the points and connect D and E.
On the graph paper, we now plot the provided points A(1, 2), B(-1, -16), and C(0, -7). We can see that every point is on the DE line.
38. The following observed values of x and y are thought to satisfy a linear equation. Write the linear equation :
x 6, – 6
y –2, 6
Explanation:
Points (6, -2) and (-6, 6) are given.
The positions (6, -2) and (-6, 6) should satisfy the linear equation y = mx + c.
At the location (6, -2), -2 = 6m + c.(i)
At point (-6, 6), 6 = -6m + c, as well.(ii) We obtain, after solving equations (i) and (ii),
replacing c = 2 in equation (i)
6m+c=−2
−6m+c= 6
________________
0+2c= 4
________________
⇒[c=2]
⇒6m+2=−2
⇒6m=−2−2
⇒6m=−4
⇒m=−46
⇒[m=−23]
In the linear equation y=mx+c, we get m=23 and c=2.
⇒y=−23x+2⇒y=−2x+63⇒3y=−2x+6
When the linear equation's graph,
Cuts the x-axis (i)
Put y = 0 in equation 2x + 3y = 6
to get 2x = 6 and x = 3.
When the linear equation's graph,
Cuts the y-axis (ii)
Add x = 0 to the equation 2x + 3y = 6 to obtain 2.0+3y=6 3y=6 y=2
As a result, the x-axis and y-axis of the graph of the linear equation are split at the points (3, 0) and (0, 2), respectively.
39. Draw the graph of the linear equation 3x + 4y = 6. At what points, the graph cuts the x-axis and the y-axis?
Explanation:
Given that 3x + 4y = 6 is a linear equation.
The x-axis and the y-axis must be intersected at specific points on the equation's graph, which must be drawn.
You may rewrite the equation as 4y = 6 - 3x
y = (6 - 3x)/4.
If x = 0, then
y = (6 - 3(0)) / 4
and y = 6/4
and y = 3/2.
If y = 0, then 0 = (6 - 3x)/4
6 - 3x = 0
and 3x = 6
x = 6/3
x = 2
The points are therefore (2, 0) and (0, 3/2)
When the points are plotted on the graph,
We create a straight line by connecting the points, which is the graph of equation 3x + 4y = 6.
The graph's x-axis is broken at the point (2, 0), as can be seen.
The graph's y-axis is broken at (0, 3/2)
40. The linear equation that converts Fahrenheit (F) to Celsius (C) is given by the relation 5F 160 C /9
(i) If the temperature is 86°F, what is the temperature in Celsius?
(ii) If the temperature is 35°C, what is the temperature in Fahrenheit?
(iii) If the temperature is 0°C what is the temperature in Fahrenheit and if the
temperature is 0°F, what is the temperature in Celsius?
(iv) What is the numerical value of the temperature which is the same in both scales?
Explanation:
Given that C = (5F - 160)/9 may be used to convert between Fahrenheit (F) and Celsius (C),
C = F given
We must determine the temperature's numerical value.
You can rewrite the relationship as 9C = 5F - 160.
In the provided relation, 9F = 5F - 160, replace C = F.
9F - 5F = -160
4F = -160
F = -160/4
F = -40
As a result, the temperature is represented by the number 40.
41. If the temperature of a liquid can be measured in Kelvin units as x°K or in Fahrenheit units as y°F, the relation between the two systems of measurement of temperature is given by the linear equation y = 9 /5 (x – 273) + 32
(i) Find the temperature of the liquid in Fahrenheit if the temperature of the liquid is 313°K. (ii) If the temperature is 158° F, then find the temperature in Kelvin.
Explanation:
Given that a liquid's temperature is x°K in Kelvin units.
A liquid's temperature is expressed as y°F in Fahrenheit units.
The equation y = (9/5)(x - 273) + 32 describes the relationship between the two temperature measuring systems.
If the temperature is 158° F, we must determine the temperature in Kelvin.
Given that y=158° F
When the value of y is substituted in the given relationship, 158 = (9/5)(x - 273) + 32
158 - 32 = (9/5)(x - 273)
14(5) x - 273 = 70
x = 70 + 273
x = 343°K where 126(5/9) = x - 273 x - 273
The temperature is therefore 343°K.
42. The force exerted to pull a cart is directly proportional to the acceleration produced in the body. Express the statement as a linear equation of two variables and draw the graph of the same by taking the constant mass equal to 6 kg. Read from the graph, the force required when the acceleration produced is (i) 5 m/sec2 (ii) 6 m/sec2
Explanation:
Given that the acceleration caused by pulling a cart is directly proportional to the force applied, this makes sense.
By using the constant mass of 6 kg, we must build a linear equation and graph its solution.
The force needed when the acceleration is 5 m/sec2 must be deduced from the graph.
A given, F
So, F = ma
Where m, or constant mass, is an arbitrary constant.
Assuming m = 6 kg
F = 6a (i) now. When the acceleration is 5 m/sec2, the force needed is F = 6(5).
F = 30 N
The force is 30 N (i) as a result. When the acceleration is 6 m/sec2, the force required is F = 6(6) F = 36 N.
The force is therefore 36 N.
The rating is (5,
The graph of the equation F = 6a is a straight line that we are able to produce.
CHAPTER-4, LINEAR EQUATIONS IN TWO VARIABLES