3.1-Graphical Method

3.1-Graphical Method Important Formulae

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Grade 10 → Math → Pair of Equations in Two Variables → 3.1-Graphical Method

After successful completion of this topic, you should be able to:

  • Interpret the concepts of linear equations in order to represent any given situation algebraically and graphically.
  • Demonstrate given two linear equations in order to comment on the nature/behaviour of the lines representing the linear equations.

The graphical method is a visual approach used to solve a pair of linear equations in two variables. This method involves plotting the equations on a coordinate plane and identifying the point(s) where the lines intersect. The intersection point represents the solution of the equations.

1. Steps to Solve Using the Graphical Method

To solve a pair of equations graphically, follow these steps:

  • Step 1: Write down the given pair of equations. For example:
  • $$\begin{align*} (1) \quad 2x + y &= 10 \\ (2) \quad x - y &= 1 \end{align*}$$

  • Step 2: Convert each equation into slope-intercept form ($y = mx + b$) if necessary. For equation (1):
  • $$y = 10 - 2x$$

  • For equation (2):
  • $$y = x - 1$$

  • Step 3: Create a table of values for each equation. Choose values for $x$ and calculate corresponding $y$ values.
  • For equation (1):

    • If $x = 0$, then $y = 10$
    • If $x = 5$, then $y = 0$

    For equation (2):

    • If $x = 0$, then $y = -1$
    • If $x = 2$, then $y = 1$
  • Step 4: Plot the points on a coordinate plane:
    • For equation (1): Points (0, 10) and (5, 0)
    • For equation (2): Points (0, -1) and (2, 1)
  • Step 5: Draw the lines for each equation using the plotted points.
  • Step 6: Identify the intersection point of the two lines. This point represents the solution of the system of equations.
2. Example Problem

Consider the following pair of equations:

$$\begin{align*} (1) \quad x + 2y &= 8 \\ (2) \quad 3x - y &= 1 \end{align*}$$

Using the graphical method, we can solve these equations:

  • For equation (1): Rearranging gives:
  • $$2y = 8 - x \implies y = 4 - \frac{1}{2}x$$

  • For equation (2): Rearranging gives:
  • $$-y = 1 - 3x \implies y = 3x - 1$$

  • Now create a table of values for both equations:
  • For equation (1):

    • If $x = 0$, then $y = 4$
    • If $x = 8$, then $y = 0$

    For equation (2):

    • If $x = 0$, then $y = -1$
    • If $x = 1$, then $y = 2
  • Plot these points on a graph and draw the lines.
  • Identify the intersection point, which represents the solution.
3. Key Points to Remember
  • The graphical method is useful for visualizing the relationship between equations.
  • The solution corresponds to the intersection of the lines, which can indicate one solution, no solution, or infinitely many solutions depending on the nature of the equations.
  • This method is best for simple equations or for understanding the concepts of linear relationships.
4. Types of Solutions

In the graphical method, three scenarios may occur:

  • One Solution: The lines intersect at one point, indicating a unique solution.
  • No Solution: The lines are parallel and never intersect, indicating that the equations are inconsistent.
  • Infinitely Many Solutions: The lines coincide, meaning the equations represent the same line and have an infinite number of solutions.
5. Practice Problems

To strengthen your understanding of the graphical method, try solving the following pairs of equations:

  • 1. $x + y = 5$ and $2x - y = 1$
  • 2. $2x + 3y = 12$ and $x - y = 2$
  • 3. $3x + y = 9$ and $x + 2y = 6$

By practicing these problems, you will gain confidence in using the graphical method to solve linear equations.

Form the pair of linear equations in the following problems, and find their solutions graphically.
(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
(ii) 5 pencils and 7 pens together cost Rs. 50, whereas 7 pencils and 5 pens together cost Rs. 46. Find the cost of one pencil and that of one pen.

Solution:

Forming and Solving Linear Equations

(i) Let $b$ be the number of boys and $g$ be the number of girls.
Equations: $b + g = 10$
$g = b + 4$
(ii) Let $p$ be the cost of one pencil and $q$ be the cost of one pen.
Equations: $5p + 7q = 50$
$7p + 5q = 46$

Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Draw the graphs of the equations x- y+ 1 = 0 and 3x+ 2y- 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

Solution:
\[x- y+ 1 = 0\] \[x+ 1 = y\] Get cordinates of two points to draw the line:
When x = 0, y = 1 + 0 = 1 ....$\rightarrow$ (0,1)
When x = 1, y = 1 + 1 = 2 ....$\rightarrow$ (1,2)

\[3x+ 2y- 12 = 0\] \[3x+ 2y = 12\] \[2y = 12 - 3x\] \[y = 6 - 1.5x\] Get cordinates of two points to draw the line:
When x = 0, y = 6 - 1.5(0) = 6 ....$\rightarrow$ (0,6)
When x = 2, y = 6 - 1.5(2) = 3 ....$\rightarrow$ (2,3)

Vertices of the shaded triangle are: (2,3), (-1,0) and (4,0)