private teacher classes will be taken only for physics and chemistry
With over 15 years of dedicated teaching experience, I am an accomplished and qualified educator specializing in Spoken English, Math, Science, Social...
Do you need help in finding the best teacher matching your requirements?
Post your requirement nowI am a teacher of mathematics. I am giving online tuitions and tuitions at my home in Jaipur(Rajasthan). I have done M.Sc. , B.Ed. in mathematics....
I have been teaching class 10 students for the past 8 years. I teach students by making fundamentals clear. I prepare them for their board exam by...
Hello dear students, I am PRANJAL MISHRA, I teach Mathematics (upto A level) . I can improve your understanding in Mathematics using my innovative...
I am an experienced tutor with over 5 years of experience in teaching English, Tamil, Maths, Science, Social Science, History, Political Science across...
Great
This is Debabrato Chatterjee online Maths and Science tutor having 13+years of experience. I have already and presently teaching students of IB,...
I am R.Vinotha ., M.Sc., M.Phil., B.Ed in mathematics.I am the trainer/tutor of this course .I was allotted 12 regular classes for a month.I am using...
Sushma attended Class 10 Tuition
"He teaches well and explains till the concept is understood well"
Mary attended Class 10 Tuition
"not started yet"
Susan attended Class 10 Tuition
"Constructive teaching"
Ananya attended Class 10 Tuition
"Very efficient, friendly & caring."
Anna attended Class 10 Tuition
"He is prompt. He knows the subject and teaches well."
Swaraj attended Class 10 Tuition
"Satnam Sir is a very good teacher, he single handedly taught me my whole of Java..."
Binit attended Class 10 Tuition
"narayan sir has guided me very well in the whole academic year. He is very dedicated..."
Anurag attended Class 10 Tuition
"I was introduced to Narayan Sir by one of my classmates who has been taking tuition..."
Linear equations in 2 variables Lessons
Ask a Question
Post a LessonAnswered on 18 Apr Learn CBSE/Class 9/Mathematics/Algebra/Linear equations in 2 variables
Nazia Khanum
Let's denote:
The total fare can be calculated as the sum of the initial fare and the fare for the subsequent distance.
So, the equation can be expressed as:
y=10+6(x−1)y=10+6(x−1)
Where:
Now, let's substitute x=15x=15 into the equation to find the total fare for a 15 km journey.
y=10+6(15−1)y=10+6(15−1) y=10+6(14)y=10+6(14) y=10+84y=10+84 y=94y=94
The total fare for a 15 km journey would be Rs. 94.
Answered on 18 Apr Learn CBSE/Class 9/Mathematics/Algebra/Linear equations in 2 variables
Nazia Khanum
Problem Analysis:
Given the equation 2x−y=p2x−y=p and a solution point (1,−2)(1,−2), we need to find the value of pp.
Solution:
Step 1: Substitute the Given Solution into the Equation
Substitute the coordinates of the given solution point (1,−2)(1,−2) into the equation:
2(1)−(−2)=p2(1)−(−2)=p
Step 2: Solve for pp
2+2=p2+2=p 4=p4=p
Step 3: Final Result
p=4p=4
Conclusion:
The value of pp for the equation 2x−y=p2x−y=p when the point (1,−2)(1,−2) is a solution is 44.
Answered on 18 Apr Learn CBSE/Class 9/Mathematics/Algebra/Linear equations in 2 variables
Nazia Khanum
Graph of the Equation x - y = 4
Graphing the Equation:
To draw the graph of the equation x−y=4x−y=4, we'll first rewrite it in slope-intercept form, which is y=mx+by=mx+b, where mm is the slope and bb is the y-intercept.
Given equation: x−y=4x−y=4
Rewriting in slope-intercept form:
y=x−4y=x−4
Now, let's plot the graph using this equation.
Plotting the Graph:
Find y-intercept:
Set x=0x=0 in the equation y=x−4y=x−4
y=0−4y=0−4
y=−4y=−4
So, the y-intercept is at the point (0,−4)(0,−4).
Find x-intercept:
To find the x-intercept, set y=0y=0 in the equation y=x−4y=x−4.
0=x−40=x−4
x=4x=4
So, the x-intercept is at the point (4,0)(4,0).
Drawing the Graph:
Now, plot the points (0,−4)(0,−4) and (4,0)(4,0) on the Cartesian plane and draw a straight line passing through these points. This line represents the graph of the equation x−y=4x−y=4.
Intersecting with the x-axis:
To find where the graph line meets the x-axis, we need to find the point where y=0y=0.
Substitute y=0y=0 into the equation x−y=4x−y=4:
x−0=4x−0=4
x=4x=4
So, when the graph line meets the x-axis, the coordinates of the point are (4,0)(4,0).
Answered on 18 Apr Learn CBSE/Class 9/Mathematics/Algebra/Linear equations in 2 variables
Nazia Khanum
Graphing the Equation x + 2y = 6
To graph the equation x+2y=6x+2y=6, we'll first rewrite it in slope-intercept form (y=mx+by=mx+b):
x+2y=6x+2y=6 2y=−x+62y=−x+6 y=−12x+3y=−21x+3
Plotting the Graph
To plot the graph, we'll identify two points and draw a line through them:
Intercept Method:
Slope Method: From the slope-intercept form y=−12x+3y=−21x+3, the slope is -1/2, meaning the line decreases by 1 unit in the y-direction for every 2 units in the x-direction.
Plotting the Points and Drawing the Line
Using the intercepts and the slope, we plot the points (0, 3) and (-6, 0), then draw a line through them.
Finding the Value of x when y = -3
Given y=−3y=−3, we substitute this value into the equation y=−12x+3y=−21x+3 and solve for x:
−3=−12x+3−3=−21x+3 −12x=−3−3−21x=−3−3 −12x=−6−21x=−6 x=−6×(−2)x=−6×(−2) x=12x=12
Conclusion
Answered on 18 Apr Learn CBSE/Class 9/Mathematics/Algebra/Linear equations in 2 variables
Nazia Khanum
Understanding Linear Equations: Linear equations are fundamental in mathematics, representing straight lines on a coordinate plane. They're expressed in the form of ax+b=0ax+b=0, where aa and bb are constants.
Identifying Axis: In the context of linear equations, the term "axis" typically refers to either the x-axis or the y-axis on a Cartesian plane.
Analyzing the Equation: The linear equation provided is x−2=0x−2=0.
Finding the Axis: To determine which axis the given linear equation is parallel to, let's analyze the equation:
Equation Form:
Solving for x:
Interpretation:
Conclusion: The linear equation x−2=0x−2=0 is parallel to the y-axis.
Ask a Question