Answer
Verified
477.9k+ views
Hint: Let us plot the given equations in a coordinate plane to get the required value of x and y.
Complete step-by-step answer:
As we are asked to solve the given equations graphically.
So, we had to plot the equation on the graph.
For plotting a linear equation, we need at least 2 points through which the equation passes.
So, for our convenience we find those two points which lie on x-axis and y-axis and also satisfy the equation.
So, as we know that any point lying on the x-axis has y-coordinate equal to zero.
And, any point lying on the y-axis has x-coordinate equal to zero.
Given equations are,
2x – 3y = 1 (1)
4x – 3y + 1 = 0 (2)
So, let equation 1 passes through two points A (a, 0) and B (0, b).
And equation 2 passes through two points C (c, 0) and D (0, d).
Now finding the value of points A, B, C and D to plot them on the graph.
As point A (a, 0) lies on the equation 1. So, it must satisfy equation 1.
2a – 0 = 1
a = \[\dfrac{1}{2}\]
As point B (0, b) lies on the equation 1. So, it must satisfy equation 1.
0 – 3b = 1
b = \[\dfrac{{ - 1}}{3}\]
As point C (c, 0) lies on equation 2. So, it must satisfy equation 2.
4c – 0 + 1 = 0
c = \[\dfrac{{ - 1}}{4}\]
As point D (0, d) lies on equation 2. So, it must satisfy equation 2.
0 – 3d + 1 = 0
d = \[\dfrac{1}{3}\]
Now line joining points A\[\left( {\dfrac{1}{2},{\text{ 0}}} \right)\] and B\[\left( {0,{\text{ }}\dfrac{{ - 1}}{3}} \right)\] will be the equation 2x – 3y = 1.
And the line joining points C\[\left( {\dfrac{{ - 1}}{4},{\text{ 0}}} \right)\] and D\[\left( {0,{\text{ }}\dfrac{1}{3}} \right)\] will be the equation 4x – 3y + 1 = 0.
Now we can see from the above graph that both the given lines intersect each other at P (-1, -1).
So, the solution of the given equation will be x = -1, and y = -1.
Hence, the correct answer will be B.
Note: Whenever we come up with this type of problem then to plot any line on a coordinate plane first, we should find points where the equation of line intersects x and y axis. And after plotting those points in the graph we will clearly see the intersection points of all the equations. And this will be the required solution of the problem.
Complete step-by-step answer:
As we are asked to solve the given equations graphically.
So, we had to plot the equation on the graph.
For plotting a linear equation, we need at least 2 points through which the equation passes.
So, for our convenience we find those two points which lie on x-axis and y-axis and also satisfy the equation.
So, as we know that any point lying on the x-axis has y-coordinate equal to zero.
And, any point lying on the y-axis has x-coordinate equal to zero.
Given equations are,
2x – 3y = 1 (1)
4x – 3y + 1 = 0 (2)
So, let equation 1 passes through two points A (a, 0) and B (0, b).
And equation 2 passes through two points C (c, 0) and D (0, d).
Now finding the value of points A, B, C and D to plot them on the graph.
As point A (a, 0) lies on the equation 1. So, it must satisfy equation 1.
2a – 0 = 1
a = \[\dfrac{1}{2}\]
As point B (0, b) lies on the equation 1. So, it must satisfy equation 1.
0 – 3b = 1
b = \[\dfrac{{ - 1}}{3}\]
As point C (c, 0) lies on equation 2. So, it must satisfy equation 2.
4c – 0 + 1 = 0
c = \[\dfrac{{ - 1}}{4}\]
As point D (0, d) lies on equation 2. So, it must satisfy equation 2.
0 – 3d + 1 = 0
d = \[\dfrac{1}{3}\]
Now line joining points A\[\left( {\dfrac{1}{2},{\text{ 0}}} \right)\] and B\[\left( {0,{\text{ }}\dfrac{{ - 1}}{3}} \right)\] will be the equation 2x – 3y = 1.
And the line joining points C\[\left( {\dfrac{{ - 1}}{4},{\text{ 0}}} \right)\] and D\[\left( {0,{\text{ }}\dfrac{1}{3}} \right)\] will be the equation 4x – 3y + 1 = 0.
Now we can see from the above graph that both the given lines intersect each other at P (-1, -1).
So, the solution of the given equation will be x = -1, and y = -1.
Hence, the correct answer will be B.
Note: Whenever we come up with this type of problem then to plot any line on a coordinate plane first, we should find points where the equation of line intersects x and y axis. And after plotting those points in the graph we will clearly see the intersection points of all the equations. And this will be the required solution of the problem.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Change the following sentences into negative and interrogative class 10 english CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you graph the function fx 4x class 9 maths CBSE
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Application to your principal for the character ce class 8 english CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE