A linear system of two equations with two variables is any system that can be written in the form.
a_{1}x + b_{1}y + c_{1} = 0
a_{2}x + b_{2}y + c_{2} = 0
A solution to a system of equations is a value of x and and value of y, when substituted into the equations, satisfies both equations at the same time.
For the example,
3x - y = 7 ----(1)
2x + 3y = 1 ----(2)
3(1) + (2)
9x - 3y + 2x + 3y = 21 + 2
11x = 22
x = 2
By applying the value of x in (1), we get
3(2) - y = 7
6 - y = 7
y = -1
x = 2 and y = −1 is a solution to the system.
Checking :
3x - y = 7 ----(1)
2x + 3y = 1 ----(2)
x = 2 and y = -1 3x - y = 7 3(2) - (-1) = 7 6 + 1 = 7 7 = 7 |
x = 2 and y = -1 2x + 3y = 1 2(2) + 3(-1) = 1 4 - 3 = 1 1 = 1 |
The following method are used to solve system of linear equations
(i) Graphing
(ii) Substitution
(iii) Elimination
(iv) Cross multiplication
Example 1 :
x + y = 7
x - y = 3
Solution :
x + y = 7 ----(1)
x - y = 3 ----(2)
From (1),
y = 7 - x
Substitute some random values for x and solve for y.
The points on the first line are
(-1, 8), (0, 7), (1, 6)
From (2),
y = x - 3
Substitute some random values for x and solve for y.
The points on the second line are
(-1, -4), (0, -3), (1, -2)
Plot the points from the above tables and sketch the graph
The point of intersection of the two lines is the solution.
So, the solution is (x, y) = (5, 2).
Example 2 :
Solve the following system of equations by substitution method.
2x - 3y = -1
y = x - 1
Solution :
2x - 3y = -1 ----(1)
y = x - 1 ----(2)
Step 1 :
From (2), substitute (x - 1) for y into (1).
(1)----> 2x - 3(x - 1) = -1
Simplify.
2x - 3x + 3 = -1
-x - 3 = -1
Add 3 to each side.
-x = 2
Multiply each side (-1).
x = -2
Step 2 :
Substitute -2 for x into (2).
(2)----> y = 2 - 1
y = 1
Therefore, the solution is
(x, y) = (-2, 1)
Example 3 :
Solve the following system of linear equations by elimination method.
x + 2y = 7
x – 2y = 1
Solution :
x + 2y = 7 ----(1)
x – 2y = 1 ----(2)
The coefficients of x and y are equal in both the equations.
(1) + (2) :
2x = 8
x = 8/2
x = 4
By substituting x = 4 in (1), we get
4 + 2y = 7
2y = 7 – 4
2y = 3
y = 3/2
Hence the solution is (4, 3/2).
Example 4 :
Solve the following system of linear equations by cross multiplication method.
3x + 4y = 24
20x - 11y = 47
Solution :
3x + 4y - 24 = 0 ----(1)
20x - 11y - 47 = 0 ----(2)
x/(-188 - 264) = y/(-480 -(-141)) = 1/(-33 - 80)
x/(-452) = y/(-480 + 141)) = 1/(-33 - 80)
x/(-452) = y/(-339) = 1/(-113)
x/(-452) = 1/(-113) x = 4 |
x/(-339) = 1/(-113) x = 3 |
So, the solution is (4, 3).
Example 5 :
If the point (4,-2) lies on the line 5x + 2y + k = 0.
Find the value of k.
Solution :
Here the given point (4,-2) lies on the given line.
To solve this problem we have to apply the given points in the equation as the value of x and y.
Here x = 4 and y = -2
The given equation is 5x + 2y + k = 0.
By applying the values of x and y, we get
5(4) + 2(-2) + k = 0
20 - 4 + k = 0
16 + k = 0
k = -16
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