4.2-Solutions of Linear Equations

4.2-Solutions of Linear Equations Important Formulae

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Grade 9 → Math → Linear Equations → 4.2-Solutions of Linear Equations

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

  • Solve a system of linear equations using method of substitution, method of elimination and by graph.

A linear equation is an equation of the first degree, which means it can be expressed in the form:

$$ax + by + c = 0$$

where \(a\), \(b\), and \(c\) are constants, and \(x\) and \(y\) are variables. The solution of a linear equation is the value of \(x\) and \(y\) that satisfies the equation.

In the context of two variables, a linear equation can be represented graphically as a straight line. The points on this line represent the solutions of the equation.

Types of Solutions

There are three types of solutions for linear equations:

  • Unique Solution: This occurs when the two lines intersect at a single point. For example, if we have the equations:

    • $$2x + 3y = 6$$

    $$x - y = 1$$

    The unique solution can be found using methods like substitution or elimination.

  • No Solution: This occurs when the two lines are parallel and never intersect. For instance:

    • $$2x + 3y = 6$$

    $$4x + 6y = 15$$

    These lines will never meet, hence there is no solution.

  • Infinitely Many Solutions: This occurs when the two equations represent the same line. For example:

    • $$2x + 3y = 6$$

    $$4x + 6y = 12$$

    Here, any point on the line is a solution, hence infinitely many solutions exist.

Methods to Solve Linear Equations

There are various methods to find the solution of linear equations:

  • Graphical Method: Plotting both equations on a graph and identifying the point of intersection.
  • Substitution Method: Solving one equation for one variable and substituting that value into the other equation.

    • For example, if we solve the first equation for \(y\):

    $$y = \frac{6 - 2x}{3}$$

    We can substitute this value into the second equation.

  • Elimination Method: Adding or subtracting the equations to eliminate one of the variables. For instance, if we have:

    • $$2x + 3y = 6$$

    $$4x + 6y = 12$$

    We can multiply the first equation by 2 and subtract:

    $$4x + 6y - (4x + 6y) = 12 - 12$$

    This will yield a consistent equation, leading to infinite solutions.

Interpretation of Solutions

The nature of the solution can be interpreted in terms of the graphical representation:

  • If the lines intersect, the solution is the point of intersection.
  • If the lines are parallel, there is no solution.
  • If the lines coincide, then every point on the line is a solution.

Understanding the concept of solutions in linear equations is fundamental in algebra, as it lays the groundwork for more advanced topics in mathematics.


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Which one of the following options is true, and why?
y = 3x + 5 has:

(i) A unique solution,

(ii) Only two solutions, 

(iii) Infinitely many solutions

Solution:

There are two variables and only one equation, which means we can get infinite pairs of (x,y) which can satisfy the given equation. Hence, it has infinitely many solutions.

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Write four solutions for each of the following equations:
(i) 2x+y=7
(ii) $\pi$x+y=9
(iii) x=4y

Solution:

Solutions to Equations

(i) For $2x + y = 7$:
1. Let $x = 0 \Rightarrow y = 7$
2. Let $x = 1 \Rightarrow y = 5$
3. Let $x = 2 \Rightarrow y = 3$
4. Let $x = 3 \Rightarrow y = 1$

(ii) For $\pi x + y = 9$:
1. Let $x = 0 \Rightarrow y = 9$
2. Let $x = 1 \Rightarrow y = 9 - \pi$
3. Let $x = 2 \Rightarrow y = 9 - 2\pi$
4. Let $x = 3 \Rightarrow y = 9 - 3\pi$

(iii) For $x = 4y$:
1. Let $y = 0 \Rightarrow x = 0$
2. Let $y = 1 \Rightarrow x = 4$
3. Let $y = 2 \Rightarrow x = 8$
4. Let $y = 3 \Rightarrow x = 12$

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Check which of the following are solutions of the equation x – 2y = 4 and which are not:
(i) (0,2)
(ii) (2,0)
(iii) (4,0)
(iv) ( $\sqrt{2}$,4$\sqrt{2}$)
(v) (1,1)

Solution:

Checking Solutions of the Equation x – 2y = 4

For each point, substitute $x$ and $y$ into the equation:

(i) (0,2): $0 - 2(2) = -4 \neq 4$ (not a solution)
(ii) (2,0): $2 - 2(0) = 2 \neq 4$ (not a solution)
(iii) (4,0): $4 - 2(0) = 4$ (solution)
(iv) ($\sqrt{2}$,4$\sqrt{2}$): $\sqrt{2} - 2(4\sqrt{2}) = -7\sqrt{2} \neq 4$ (not a solution)
(v) (1,1): $1 - 2(1) = -1 \neq 4$ (not a solution)

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Find the value of k, if x=2, y=1 is a solution of the equation 2x + 3y = k.

Solution:

Finding the Value of k

Substituting $x = 2$ and $y = 1$ into the equation $2x + 3y = k$:

$2(2) + 3(1) = k$

$4 + 3 = k$

$k = 7$

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