PAIR OF LINEAR EQUATIONS IN TWO VARIABLES-Notes
Maths - Notes
Linear Equation in Two Variables
A linear equation in two variables is an equation that can be written in the form
\[ax+by+c=0\]
where \(a,\ b,\text{ and }c\) are real numbers, and \(a\text{ and }b\) are not both zero.
Example:
\[2x+3y-5=0\]
This equation has infinitely many solutions because each solution is a pair of values \((x,\ y)\)
that
satisfy the equation.
Pair of Linear Equations in Two Variables
Two equations involving the same variables \(x\) and \(y\) form a pair of linear equations: \[\begin{aligned}a_1x+b_1y+c_1=0\\ a_2x+b_2y+c_2=0\end{aligned}\] Example: \[\begin{aligned}x+3y=6\\ 2x-3y=12\end{aligned}\]
Graphical Representation
Each linear equation represents a straight line on the coordinate plane.
So, a pair of linear equations represents two lines.
The nature of solutions depends on how these lines behave:
- Intersecting Lines \(\Rightarrow\) Unique
Solution:
If the two lines intersect at one point, the pair has one unique solution. - Parallel Lines \(\Rightarrow\) No Solution:
If the two lines never intersect, the pair is inconsistent and has no solution. - Coincident Lines \(\Rightarrow\) Infinitely Many
Solutions:
If both lines lie on top of each other, the pair is dependent, with infinitely many solutions.
Algebraic Conditions for the Nature of Solutions
For given equations:
\[\begin{aligned}a_1x+b_1y+c_1=0\\ a_2x+b_2y+c_2=0\end{aligned}\]| Condition on ratios | Type of lines | Number of solutions |
|---|---|---|
| \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\) | Intersecting | One solution |
| \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\) | Parallel | No solution |
| \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\) | Coincident | Infinitely many solutions |
Example
Hence the lines represented by Equations (1) and (2) are coincident. Therefore, Equations (1) and (2) have infinitely many solutions
Example
Solution
Let number of pant that Champa purchease=\(x\) and skirts be \(y) \[y=2x-2\tag{1}\] \[ \begin{array}{|c|c|c|} \hline x&2&0\\\hline y&2&-2\\\hline \end{array} \] and Also \[y=4x-4\tag{2}\] \[ \begin{array}{|c|c|c|} \hline x&0&1&2\\\hline y&-4&0&4\\\hline \end{array} \]
The two lines intersect at the point (1, 0). So, x = 1, y = 0 is the required solution of the pair of linear equations, i.e., the number of pants she purchased is 1 and she did not buy any skirt.
Algebraic Methods of Solving a Pair of Linear Equations
- Substitution Method:
Solve one equation for one variable and substitute into the other. - Elimination Method:
Steps in the elimination method- Step 1: First multiply both the equations by some suitable non-zero constants to make the coefficients of one variable (either x or y) numerically equal.
- Step 2: Then add or subtract one equation from the other so that one variable
gets
eliminated. If you get an equation in one variable, go to Step 3.
- If in Step 2, we obtain a true statement involving no variable, then the original pair of equations has infinitely many solutions.
- If in Step 2, we obtain a false statement involving no variable, then the original pair of equations has no solution, i.e., it is inconsistent.
- Step 3: If in Step 2, we obtain a false statement involving no variable, then the original pair of equations has no solution, i.e., it is inconsistent.
- Step 4: Substitute this value of x (or y) in either of the original equations to get the value of the other variable.
- Cross Multiplication Method:
For equations: \[ \begin{aligned} a_1x+b_1y+c_1=0\\ a_2x+b_2y+c_2=0 \end{aligned} \] \[ \scriptsize \begin{aligned} \frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2=c_2a_1}=\frac{1}{a_1b_2-a_2b_1} \end{aligned} \](Not in scope of NCERT class IX)
Example
Solve the following pair of equations by substitution method: \[ \begin{aligned} 7x-15y&=2\\ x+2y&=3 \end{aligned} \]
Solution
\[ \begin{align} 7x-15y&=2\tag{1}\\ x+2y&=3\tag{2} \end{align} \] Consider Equation $$x+2y=3$$ $$\implies y=\dfrac{3-x}{2}$$Substituting value of \(y\) in Eqn(1)
$$\begin{aligned}7x-\dfrac{15}{2}\left( 3-x\right) -2&=0\\ 14x-15\left( 3-x\right) -4&=0\\ 14x+15x-45-4&=0\\ 29x-49&=0\\ 29x&=49\\ x&=\dfrac{49}{29}\end{aligned}$$Putting value of \(x\) in equation (2)
$$\begin{aligned}x+2y&=3\\ \dfrac{49}{29}+2y&=3\\ 2y&=3-\dfrac{49}{29}\\ &=\dfrac{87-49}{29}\\ &=\dfrac{38}{29}\\ &=\dfrac{38}{29}\\ y&=\dfrac{38}{2\times 29}\\ &=\dfrac{19}{29}\end{aligned}$$Example
Solve the following question—Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically by the method of substitution.
Solution
Graphical Representation
Coordinates from Equation(1) \[ \begin{array}{|c|c|c|} \hline x&0&7 &14\\\hline y&6&7&8\\\hline \end{array} \] Coordinates from Equation(2) \[ \begin{array}{|c|c|c|} \hline x&6&9 &30\\\hline y&0&1&8\\\hline \end{array} \]
In a shop the cost of 2 pencils and 3 erasers is ₹9 and the cost of 4 pencils and 6 erasers is ₹18. Find the cost of each pencil and each eraser.
Solution
let cost of pencil is a cost of eraser is \(y\) $$2x+3y=9\tag{1}$$ and $$4x+6y=18\tag{2}$$ Consider equation (1) $$\begin{aligned}2x+3y&=9\\ 2x&=9-3y\\ x&=\dfrac{9-3y}{2}\end{aligned}$$ Substituting value of \(x\) in Equation-(2) $$\begin{aligned}4x+6y&=1\\ 4\left( \dfrac{9-3y}{2}\right) +6y&=18\\ \Rightarrow 18-6y+6y&=18\\ \Rightarrow 18&=18\end{aligned}$$This statement is true for all values of \(y\). However, we do not get a specific value of \(y\) as a solution. Therefore, we cannot obtain a specific value of \(x\). This situation has arisen because both the given equations are the same. Therefore, Equations (1) and (2) have infinitely many solutions. We cannot find a unique cost of a pencil and an eraser, because there are many common solutions, to the given situation.
Example
The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each of them manages to save ₹ 2000 per month, find their monthly incomes.
Solution
Let income be \(x\) and expenditure be \(y\)
Income of both person is in ratio of 9: 7
therefore, Income becomes \(9x\) & \(7x\)
Expenditure of both the person is in ratio of 4: 3
therefore Expenditure \(4y\) & \(3y\)
Their respective savings = 2000
Multiply equation-(1) by 3 and Equation-(2) by 4
$$\begin{align} (9x-4y &=2000) \times 3\\ (7x-3y &=2000) \times 4\\ 27x-12y &=6000\tag{3} \end{align}$$ $$28x-12y=8000\tag{4}$$ Substrating Eqn-(3) from Eqn-(4) $$\begin{aligned}28x-12y-\left( 27x-12y\right) &=8000-6000\\ 28x-27x-12y+12y&=2000\\ x&=2000\end{aligned}$$ Hence, Incomes of both persons $$\begin{aligned}9x&=9\times 200\\ &=\text{₹}18000\\\\ 7x&=7\times 2000\\ &=\text{₹}14000\end{aligned}$$Example
The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there?
Solution
Let unit digit of first number be \(x\)
and ten's digit of the first number be \(y\)
then Number can be written as
On Reversing the digits number will become
$$10x+y\tag{2}$$Sum of these number = 66, therefore
$$\begin{align}10y+x+10x+y&=66\\ \Rightarrow 11y+11x&=66\\ \Rightarrow x+y&=6\end{align}\tag{3}$$Difference of digits = 2, therefore
$$x-y=2\tag{4}$$Adding equations (3) and (4)
$$\begin{aligned}x+y+x-y&=6+2\\ 2x&=8\\ x&=\dfrac{8}{2}\\ &=4\end{aligned}$$Putting value of \(x=4\) in equation (4)
$$\begin{aligned}x-y&=2\\ 4-y&=2\\ -y&=2-4\\ y&=2\end{aligned}$$So numbers = 42 and 24
Important Points
- A pair of linear equations in two variables can be represented, and solved, by
the:
- graphical method
- algebraic method
- Graphical Method :
The graph of a pair of linear equations in two variables is represented by two lines.- If the lines intersect at a point, then that point gives the unique solution of the two equations. In this case, the pair of equations is consistent.
- If the lines coincide, then there are infinitely many solutions — each point on the line being a solution. In this case, the pair of equations is dependent (consistent) .
- If the lines are parallel, then the pair of equations has no solution. In this case, the pair of equations is inconsistent.
- Algebraic Methods : We have discussed the following methods for
finding the
solution(s) of a pair of linear equations :
- Substitution Method
- Elimination Method
- If a pair of linear equations is given by \(a_1x + b_1y + c_1= 0\) and
\(a_2x + b_2y + c_2= 0\), then the following situations can arise :
- \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\):
In this case, the pair of linear equations is consistent. - \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\):
In this case, the pair of linear equations is inconsistent. - \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\):
In this case, the pair of linear equations is dependent and consistent.
- \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\):
- There are several situations which can be mathematically represented by two equations that are not linear to start with. But we alter them so that they are reduced to a pair of linear equations.
