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Algebra

Algebra relates to CBSE/Class 9/Mathematics

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Answered on 18 Apr Learn CBSE/Class 9/Mathematics/Algebra/Linear equations in 2 variables

Nazia Khanum

Introduction: In this problem, we are tasked with verifying whether the values x=2x=2 and y=1y=1 satisfy the linear equation 2x+3y=72x+3y=7. Verification: We'll substitute the given values of xx and yy into the equation and check if it holds true. Given Equation: 2x+3y=72x+3y=7 Substituting Given... read more

Introduction: In this problem, we are tasked with verifying whether the values x=2x=2 and y=1y=1 satisfy the linear equation 2x+3y=72x+3y=7.

Verification: We'll substitute the given values of xx and yy into the equation and check if it holds true.

Given Equation: 2x+3y=72x+3y=7

Substituting Given Values:

  • Substitute x=2x=2 and y=1y=1 into the equation. 2(2)+3(1)=72(2)+3(1)=7

Solving the Equation: 4+3=74+3=7 7=77=7

Conclusion:

  • Since the equation simplifies to 7=77=7, it confirms that x=2x=2 and y=1y=1 satisfy the linear equation 2x+3y=72x+3y=7.

Therefore, the given values x=2x=2 and y=1y=1 indeed satisfy the linear equation 2x+3y=72x+3y=7.

 
 
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Answered on 18 Apr Learn CBSE/Class 9/Mathematics/Algebra/Linear equations in 2 variables

Nazia Khanum

Solutions for 2x + 3y = 8 Introduction: In this problem, we're tasked with finding solutions to the equation 2x + 3y = 8. There are multiple solutions that satisfy this equation. Let's explore four of them: Solution 1: Using Integer Values Choose a set of integer values for x and solve for y. Let's... read more

Solutions for 2x + 3y = 8

Introduction: In this problem, we're tasked with finding solutions to the equation 2x + 3y = 8. There are multiple solutions that satisfy this equation. Let's explore four of them:

Solution 1: Using Integer Values

  • Choose a set of integer values for x and solve for y.
  • Let's say x = 2.
  • Substitute x = 2 into the equation: 2(2) + 3y = 8.
  • Solve for y: 4 + 3y = 8.
  • 3y = 8 - 4.
  • 3y = 4.
  • y = 4/3.
  • So, one solution is (2, 4/3).

Solution 2: Using Fractional Values

  • Choose fractional values for x and solve for y.
  • Let's say x = 1/2.
  • Substitute x = 1/2 into the equation: 2(1/2) + 3y = 8.
  • Solve for y: 1 + 3y = 8.
  • 3y = 8 - 1.
  • 3y = 7.
  • y = 7/3.
  • Another solution is (1/2, 7/3).

Solution 3: Using a Variable for y

  • Express y in terms of x and a constant.
  • Rearrange the equation to isolate y: 3y = 8 - 2x.
  • Divide both sides by 3: y = (8 - 2x)/3.
  • So, a solution can be represented as (x, (8 - 2x)/3).

Solution 4: Using Graphical Method

  • Graph the equation on a coordinate plane.
  • Plot the points where the line intersects the x-axis and the y-axis.
  • Determine the coordinates of these points as solutions.
  • By plotting, we find that two points of intersection are (4, 0) and (0, 8/3).
  • Thus, solutions are (4, 0) and (0, 8/3).

Conclusion: The equation 2x + 3y = 8 has multiple solutions, including both integer and fractional values of x and y. Additionally, solutions can also be represented using variables. Graphically, the solutions are the points where the line intersects the axes.

 
 
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Answered on 18 Apr Learn CBSE/Class 9/Mathematics/Algebra/Linear equations in 2 variables

Nazia Khanum

Graph of the Equation 2x – 3y = 12 To draw the graph of the equation 2x−3y=122x−3y=12, let's 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. Rewrite Equation in Slope-Intercept Form 2x−3y=122x−3y=12−3y=−2x+12−3y=−2x+12y=23x−4y=32x−4 Plotting... read more

Graph of the Equation 2x – 3y = 12

To draw the graph of the equation 2x−3y=122x−3y=12, let's 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.

Rewrite Equation in Slope-Intercept Form

2x−3y=122x−3y=12
−3y=−2x+12−3y=−2x+12
y=23x−4y=32x−4

Plotting the y-intercept and Slope

  1. Y-intercept: When x=0x=0,
    y=23(0)−4y=32(0)−4
    y=−4y=−4
    So, the y-intercept is (0, -4).

  2. Slope: The coefficient of xx is 2332, which represents the slope.
    For every increase of 1 in xx, yy increases by 2332.
    For every decrease of 1 in xx, yy decreases by 2332.

Plotting Points and Drawing the Graph

Now, let's plot some points to draw the graph:

  • x = 3: y=23(3)−4=2−4=−2y=32(3)−4=2−4=−2
    Point: (3, -2)

  • x = 6: y=23(6)−4=4−4=0y=32(6)−4=4−4=0
    Point: (6, 0)

  • x = -3: y=23(−3)−4=−2−4=−6y=32(−3)−4=−2−4=−6
    Point: (-3, -6)

Plotting the Graph

With these points, we can draw a straight line passing through them.

Points where the Graph Intersects the Axes

X-axis

To find where the graph intersects the x-axis, we set y=0y=0 and solve for xx:

0=23x−40=32x−4
23x=432x=4
x=4×32x=24×3
x=6x=6

So, the graph intersects the x-axis at x=6x=6, which corresponds to the point (6, 0).

Y-axis

To find where the graph intersects the y-axis, we set x=0x=0 and solve for yy:

y=23(0)−4y=32(0)−4
y=−4y=−4

So, the graph intersects the y-axis at y=−4y=−4, which corresponds to the point (0, -4).

Summary

  • X-axis intersection: (6, 0)
  • Y-axis intersection: (0, -4)

This information helps us visualize and understand the behavior of the equation 2x−3y=122x−3y=12 on the coordinate plane.

 
 
 
 
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Answered on 18 Apr Learn CBSE/Class 9/Mathematics/Algebra/Linear equations in 2 variables

Nazia Khanum

Graph of 9x – 5y + 160 = 0 To graph the equation 9x – 5y + 160 = 0, we'll first rewrite it in slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept. Step 1: Rewrite the equation in slope-intercept form 9x – 5y + 160 = 0 Subtract 9x from both sides: -5y =... read more

Graph of 9x – 5y + 160 = 0

To graph the equation 9x – 5y + 160 = 0, we'll first rewrite it in slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept.

Step 1: Rewrite the equation in slope-intercept form

9x – 5y + 160 = 0

Subtract 9x from both sides:

-5y = -9x - 160

Divide both sides by -5 to isolate y:

y = (9/5)x + 32

Now we have the equation in slope-intercept form.

Step 2: Identify the slope and y-intercept

The slope (m) is 9/5 and the y-intercept (b) is 32.

Step 3: Plot the y-intercept and use the slope to find additional points

Now, let's plot the y-intercept at (0, 32). From there, we'll use the slope to find another point. The slope of 9/5 means that for every 5 units we move to the right along the x-axis, we move 9 units upwards along the y-axis.

So, starting from (0, 32), if we move 5 units to the right, we move 9 units up to get the next point.

Step 4: Plot the points and draw the line

Plot the y-intercept at (0, 32) and the next point at (5, 41). Then, draw a line through these points to represent the graph of the equation.

Finding the value of y when x = 5

To find the value of y when x = 5, we'll substitute x = 5 into the equation and solve for y.

9x – 5y + 160 = 0

9(5) – 5y + 160 = 0

45 – 5y + 160 = 0

Combine like terms:

-5y + 205 = 0

Subtract 205 from both sides:

-5y = -205

Divide both sides by -5 to solve for y:

y = 41

So, when x = 5, y = 41.

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Answered on 18 Apr Learn CBSE/Class 9/Mathematics/Algebra/Linear equations in 2 variables

Nazia Khanum

Finding Solutions of Line AB Equation Given Information: Line AB is represented by the equation. A graph depicting Line AB is provided. Procedure: Identify Points on Line AB: Locate four points on the graph that lie on Line AB. Determine Coordinates: Extract the coordinates of these points. Substitute... read more

Finding Solutions of Line AB Equation

Given Information:

  • Line AB is represented by the equation.
  • A graph depicting Line AB is provided.

Procedure:

  1. Identify Points on Line AB: Locate four points on the graph that lie on Line AB.
  2. Determine Coordinates: Extract the coordinates of these points.
  3. Substitute Coordinates: Substitute the coordinates into the equation of Line AB.
  4. Verify Solutions: Confirm that the substituted coordinates satisfy the equation of Line AB.

1. Identify Points on Line AB:

  • Locate four distinct points where the line intersects the axes or stands out on the graph.

2. Determine Coordinates:

  • Note down the coordinates (x, y) of each identified point.

3. Substitute Coordinates:

  • Use the coordinates obtained to substitute into the equation of Line AB.
  • The equation of a line is typically in the form y = mx + b, where m is the slope and b is the y-intercept.

4. Verify Solutions:

  • Confirm that the substituted coordinates satisfy the equation of Line AB.
  • The substituted values should make the equation true when solved.

Example:

  • Suppose the equation representing Line AB is y = 2x + 3.
  • Points on the graph are (0, 3), (1, 5), (2, 7), and (-1, 1).
  • Substituting these coordinates into the equation:
    • For (0, 3): 3 = 2(0) + 3 (True)
    • For (1, 5): 5 = 2(1) + 3 (True)
    • For (2, 7): 7 = 2(2) + 3 (True)
    • For (-1, 1): 1 = 2(-1) + 3 (True)
  • All points satisfy the equation, confirming they lie on Line AB.

Conclusion:

  • By following these steps, you can find solutions of the equation representing Line AB from the provided graph.
 
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