SOLVING SYSTEM OF EQUATION USING SUBSTITUTION
Solving System of Equations Using Substitution
Here we are going to see, how to solve system of equations using substitution method.
(a) Use one of the equations in the system of equations to solve for one of the variables in terms of the other variables.
(b) Substitute the expression obtained in the previous step into the other equations, resulting in a new system of equations with one less variable and one less equation.
(c) Repeat the first two steps until you can solve the remaining system.
(d) Then substitute the values you have found into the previously obtained equations to get the complete solutions.
Solving System of Equations Using Substitution - Examples
Question 1 :
Find all solutions to the given system of equations.
x2 − 2y2 = 3
x + 2y = 1
Solution :
x2 − 2y2 = 3 -------(1)
x + 2y = 1 -------(2)
x = 1 - 2y
Substitute x = 1 - 2y in (1)
(1 - 2y)2 - 2y2 = 3
1 + 4y2 - 4y - 4y2 = 3
-4y = 3 - 1
-4y = 2
y = 2/(-4) = -1/2
x = 1 - 2(-1/2)
= 1 + 1
x = 2
Hence the solution is (2, -1/2).
Question 2 :
Find all solutions to the given system of equations.
(1/x) − (1/y) = 2
4x + y = 3
Solution :
(1/x) − (1/y) = 2 ----(1)
4x + y = 3 ----(2)
y = 3 - 4x
Substitute y = 3 - 4x in (1)
(1/x) - (1/(3 - 4x)) = 2
(3 - 4x - x)/(x(3-4x)) = 2
(3 - 5x)/(3x - 4x2) = 2
3 - 5x = 2(3x - 4x2)
3 - 5x = 6x - 8x2
8x2 - 6x - 5x + 3 = 0
8x2 - 11x + 3 = 0
(8x - 3) (x - 1) = 0
x = 3/8 and x = 1
|
If x = 3/8 y = 3 - 4(3/8) y = 3 - (3/2) y = (6 - 3)/2 y = 3/2 |
If x = 1 y = 3 - 4(1) y = 3 - 4 y = -1 |
Hence the solutions are (3/8, 3/2) and (1, -1).
Question 3 :
Find all solutions to the given system of equations.
x2 + 3y2 = 5
x − 3y = 2
Solution :
x2 + 3y2 = 5 ----(1)
x − 3y = 2 ----(2)
x = 2 + 3y
Substitute x = 2 + 3y in (1)
(2 + 3y)2 + 3y2 = 5
4 + 9y2 + 2(2)(3y) + 3y2 = 5
12y2 + 12y + 4 - 5 = 0
12y2 + 12y - 1 = 0
a = 12, b = 12 and c = -1
y = -b ± √(b2 - 4ac) / 2a
y = -12 ± √(122 - 4(12) (-1) / 2(12)
y = -12 ± √(144 + 48) / 24
y = (-12 ± √192) / 24
y = (-12 ± 8√3) / 24
y = 4(-3 ± 2√3) / 24
y = (-3 ± 2√3) / 6
y = (-3 + 2√3)/6 (or) y = (-3 - 2√3)/6
|
y = (-3 + 2√3)/6 x = 2 + 3y x = 2 + 3(-3 + 2√3)/6 x = 2 + (-3 + 2√3)/2 = (4 - 3 + 2√3)/2 = (1 + 2√3)/2 |
y = (-3 - 2√3)/6 x = 2 + 3y x = 2 + 3(-3 - 2√3)/6 x = 2 + (-3 - 2√3)/2 = (4 - 3 - 2√3)/2 = (1 - 2√3)/2 |
Hence the solutions are ((1 + 2√3)/2, (-3 + 2√3)/6) and ((1 - 2√3)/2, (-3 - 2√3)/6).
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