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A linear equation in two variables is an equation which can be put in the form of ax + by + c = 0, where a, b and c are real numbers and both a and b are non zero (a ≠ 0, b ≠ 0). In the equation, a and b are the coefficients and c is the constant.
Read More: Pair of Linear Equations in Two Variables Formula
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Key Terms: Equations, variables, degree, straight lines, graphs, elimination method
Linear Equations
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Equation: An equation refers to a statement where any two mathematical expressions containing at least one variable are equal.
Linear Equation: In simple terms, an equation between two variables that can be plotted in the form of a straight line on a graph is called a linear equation. The degree of a linear equation will always be one.
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What are Linear Equations in Two Variables?
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If two linear equations have two same variables, then they are called the pair of linear equations in two variables. The most general form of a pair of linear equations is –
a1x + b1y + c1 = 0
and
a2x + b2y + c2 = 0
where a1,b1,a2 and b2 ≠ 0.
The graph of a linear equation is a straight line. Each solution (x,y) for any linear equation corresponds to a point on the line.
Read More: Pair of Linear Equations in Two Variables Important Questions
Explanation
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A pair of linear equations in two variables can be represented and solved using 2 methods –
- Graphical Method
- Algebraic Method
Graphical Method
Take values of x and y variables for the two given equations. Then, draw two lines using the (x,y) coordinates in a graph. Two lines will be drawn as we are given two equations. Now, there are three possible outcomes:
- The two lines can intersect each other at one particular point. If this happens, then the particular point will be the only solution for the pair of linear equations. These kinds of equations are called consistent pairs of equations.
- The two lines can coincide with each other. If this happens, then there will be infinite solutions for the pair of linear equations; as all the points on the line will be the solutions for the pair of linear equations. These kinds of equations are called consistent or dependent pairs of equations.
- The two lines can be parallel to each other. If this happens, then there won't be any solutions for the pair of linear equations, as the lines are not gonna intersect at any point. These kinds of equations are called inconsistent pairs of equations.
Algebraic Method
Step A- Find the value of one variable, say x in terms of the other variable, i.e., y from either of the given two equations.
Step B- Substitute the found value of x in the other equation, and reduce it to an equation in one variable, i.e. in terms of y, which can be solved.
Step C- Now, substitute the value of y (found in the previous step) in the equation used in the first step to get the value of x.
- Elimination Method
Step A- First, multiply both the equations by such a number, so that the coefficient of any one variable, say x becomes equal.
Step B- Now, add or subtract the equations so that the one variable, x gets eliminated as the coefficients of x were the same.
Step C- Solve the equation in that leftover variable, y to find its value.
Step D- Now, substitute the calculated value of y in any one of given equations to find the value of the other variable, x.
- Cross Multiplication Method
Step A- First, write the given equations in the form of
a1x + b1y + c1 = 0
and
a2x + b2y + c2 = 0
Step B- Make a diagram as shown below:
Cross Multiplication Method
The arrows indicate the pairs which are to be multiplied. We have to subtract the product of the upward arrow pairs from the product of the downward arrow pairs.
Step C- Now, write the equations using the diagram shown above, like this -
\(\frac{x}{b_1c_2 - b_2c_1}\)= \(\frac{y}{c_1a_2 - c_2a_1}\)= \(\frac{1}{a_1b_2 - a_2b_1}\)
From the above equation we can find values of x and y, provided that a1b2 - b2a1 ≠ 0.
Read More: Difference between linear and non-linear equations
Sample Questions
Ques. Solve the following pairs of linear equations in two variables using graphical method :-
4x - y = 2, x + 3y = 7 (2 marks)
Solution: Here,
For 4x - y = 2
x = (2 + y) / 4
Now, for y = 0, x = 0
y = 1, x = 1/2
y = 2, x = 1
Again,
x + 3y = 7
x = 7 - 3y
Now, for y = 0, x = 7
y = 1, x = 4
y = 2, x = 1
Now, if we take (0,0), (1,½), (2,1) as (x,y) coordinates to draw a line representing 4x - y = 2 equation and (0,7), (1,4), (2,1) as (x,y) coordinates to draw a line in the same rectangular coordinate system, then we will find that the lines will intersect at only one point (1,2).
Hence, the solutions are x = 1 and y = 2.
Ques. Solve the following pairs of linear equations in two variables using substitution method:
a) x - y = 6, 2x - 3y = 9
b) 3x - y = 7, 2x + 3y = 12 (4 marks)
Solution: Given,
x - y = 6 ----- (i)
2x - 3y = 9 ---- (ii)
Again,
x - y = 12
=> x = 6 + y ----- (iii)
Now, using (iii) in (ii),
=> 2(6 + y) - 6y = 9
=> 12 + 2y - 6y = 9
=> 12 - 4y = 9
=> 4y = 3
=> y = ¾ ---- (iv)
Again, using (iv) in (iii),
=> x = 6 + ¾
=> x = 27/4
Hence, the solutions are, x = 27/4 and y = 3/4
We can verify the answer using the method given below -
Let's put the found values of x and y in equation (i)
x - y = 6
=> 27/4 - ¾ = 6
=> 24/4 = 6
=> 6 = 6
=> L.H.S = R.H.S
Hence, the obtained solution is correct.
b) 3x - y = 7, 2x + 3y = 12
Solution -
Given,
3x - y = 7----- (i)
2x + 3y = 12 ---- (ii)
Again,
3x - y = 7
=> x = (7 + y) / 3----- (iii)
Now, using (iii) in (ii),
=> 2 [(7 + y) /3] + 3y = 12
=> (14 + 2y)/3 + 3y = 12
=> 14 + 2y + 9y = 36
=> 14 + 11y = 36
=> y = 22/11
=> y = 2 ---- (iv)
Again, using (iv) in (iii),
=> x = (7 + 2)/3
=> x = 9/3
=> x = 3
Hence, the solutions are, x = 3 and y = 2.
We can verify the answer using the method given below -
Let's put the found values of x and y in equation (i)
3x - y = 7
=> (3 X 3)- 2 = 7
=> 9 - 2 = 7
=> L.H.S = R.H.S
Hence, the obtained solution is correct.
Ques. Solve the following pairs of linear equations in two variables using elimination method :-
a) 2x - y = 12, 3x - 4y = 18
b) x + y = 4, 2x + y = 6 (3 marks)
Solution: Given,
2x - y = 12 ----- (i)
3x - 4y = 18 ----- (ii)
Now, multiplying (i) by 4,
8x - 4y = 48 ----- (iii)
Now,
(ii) - (iii) => 3x - 4y - 8x + 4y = 18 - 48
=> -5x = -30
=> x = 6 ----- (iv)
Again, using (iv) in (i),
12 - y = 12
=> y = 0
Hence, the solutions are, x = 6, y = 0.
Alternatively,
Given,
2x - y = 12 ----- (i)
3x - 4y = 18 ----- (ii)
Now, multiplying (i) by 3,
6x - 3y = 36 ----- (iii)
and, multiplying (ii) by 2,
6x - 8y = 36 ------ (iv)
Now,
(iii) - (iv) => 6x - 3y - 6x + 8y = 0
=> y = 0
=> y = 0 ----- (v)
Again, using (v) in (i),
2x - 0 = 12
=> x = 6.
Hence, the solutions are, x = 6, y = 0.
b) x + y = 4, 2x + y = 6
Solution:
Given,
x + y = 4 ----- (i)
2x + y = 6 ----- (ii)
Now, multiplying (i) by 2,
2x + 2y = 8 ----- (iii)
Now,
(ii) - (iii) => 2x + y - 2x - 2y = 6 - 8
=> -y = -2
=> y = 2 ----- (iv)
Again, using (iv) in (i),
x + 2 = 4
=> x = 2
Hence, the solutions are, x = 2, y = 2.
Alternatively,
Given,
x + y = 4 ----- (i)
2x + y = 6 ----- (ii)
Now,
(ii) - (i) => 2x + y - x - y = 6 - 4
=> x = 2 ------ (iii)
Again, using (iii) in (i),
2 + y = 4
=> y = 2
Hence, the solutions are, x = 2, y =2 .
Ques: Solve the following pairs of linear equations in two variables using cross multiplication method:
a) x + y = 4, 3x - 2y = 7
b) x + 2y = 7, 2x - 4y = -2 (2 marks)
Solution: Given,
x + y - 4 = 0
3x -2y - 7 = 0
Now, by method of cross multiplication,
x/ (b1c2 - b2c1) = y/ (c1a2 - c2a1) =
1/ (a1b2 - a2b1)
Now substituting the values we get,
x/ -7 - 8 = y/ -12 + 7 = 1/ -2 -3
=> x/ -15 = y/ -5 = 1/ -5
Now,
x/ -15 = 1/ -5
=> 5x = 15
=> x = 3
And,
y/ -5 = 1/ -5
=> 5y = 5
=> y = 1
Hence, the solutions are, x = 3, y = 1.
b) x + 2y = 7, 2x - 4y = -2
Solution
Given,
x + 2y - 7 = 0
2x - 4y + 2 = 0
Now, by method of cross multiplication,
x/ (b1c2 - b2c1) = y/ (c1a2 - c2a1) =
1/ (a1b2 - a2b1)
Now substituting the values we get,
x/ 4 - 28 = y/ -14 - 2 = 1/ -4 - 4
=> x/ -24 = y/ -16 = 1/ -8
Now,
x/ -24 = 1/ -8
=> 8x = 24
=> x = 3
And,
y/ -16 = 1/ -8
=> 8y = 16
=> y = 2
Hence, the solutions are, x = 3, y = 2.
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