Introduction for solving linear equations using substitution:
Linear equations substitution is nothing but a process of exchanging a variable in a linear expression with its actual value. In linear algebra the substitution process plays a major role to solve the system of linear equation by substituting the value for the given variable, linear substitution is mainly used to identify a variable in an expression and to find linear relationships between the equations. Here we use the linear substitution method to solve the different types of linear equations.
Having problem with Solving Systems of Linear Inequalities keep reading my upcoming posts, i will try to help you.
Solved Examples on Linear Equation Substitution
Ex 1:
Solve the linear equation by substitution method.
1 / (y - 1) 2 - 4 / (y - 1) + 4 = 0
Solution:
Let s = `1 / (y - 1)` and substitute this term in the given equation.
s 2 – 4s + 4 = 0
By solving the above quadratic equation, we get:
s = 2 and s=2.
Now substitute s by` 1 / (y - 1)` and solve for y
`1 / (y - 1)` = 2
1 = 2(y - 1)
1=2y - 2
3= 2y
Y = `3/2` is the solution for above equation.
Ex 2:
Solve the linear equation by substitution method.
z - 5 `sqrt (z)` = - 6
Solution:
Let s = `sqrt (z)` so that s 2 = z. Substitute z by s and `sqrt (z)` by s 2 respectively to obtain an equation in s.
s 2 - 5 s = - 6
The above equation looks like quadratic form, so rewrite the above term
s 2 - 5 s + 6 = 0
By solving the above equation we get
s = 2 or s = 3
We now substitute s by `sqrt (z)` and solve for z
`sqrt (z)` = 2 or `sqrt ( z )` = 3
z = 4 or z = 9 Is the solution for above equation.
My forthcoming post is on example of a algebraic expression, how to write an algebraic expression in words will give you more understanding about Algebra.
Practice Problems on Linear Equations Substitution for Solving:
1) Solve the linear equation by substitution method.
1 - 2 / (a - c) = 8 at c=1.
Answer: a = 5/7.
2) Solve the linear equation by substitution method.
1 - 1 / (x - z) = -8 / (x 2 - z 2) at z=4
Answer: x = -3.
Linear equations substitution is nothing but a process of exchanging a variable in a linear expression with its actual value. In linear algebra the substitution process plays a major role to solve the system of linear equation by substituting the value for the given variable, linear substitution is mainly used to identify a variable in an expression and to find linear relationships between the equations. Here we use the linear substitution method to solve the different types of linear equations.
Having problem with Solving Systems of Linear Inequalities keep reading my upcoming posts, i will try to help you.
Solved Examples on Linear Equation Substitution
Ex 1:
Solve the linear equation by substitution method.
1 / (y - 1) 2 - 4 / (y - 1) + 4 = 0
Solution:
Let s = `1 / (y - 1)` and substitute this term in the given equation.
s 2 – 4s + 4 = 0
By solving the above quadratic equation, we get:
s = 2 and s=2.
Now substitute s by` 1 / (y - 1)` and solve for y
`1 / (y - 1)` = 2
1 = 2(y - 1)
1=2y - 2
3= 2y
Y = `3/2` is the solution for above equation.
Ex 2:
Solve the linear equation by substitution method.
z - 5 `sqrt (z)` = - 6
Solution:
Let s = `sqrt (z)` so that s 2 = z. Substitute z by s and `sqrt (z)` by s 2 respectively to obtain an equation in s.
s 2 - 5 s = - 6
The above equation looks like quadratic form, so rewrite the above term
s 2 - 5 s + 6 = 0
By solving the above equation we get
s = 2 or s = 3
We now substitute s by `sqrt (z)` and solve for z
`sqrt (z)` = 2 or `sqrt ( z )` = 3
z = 4 or z = 9 Is the solution for above equation.
My forthcoming post is on example of a algebraic expression, how to write an algebraic expression in words will give you more understanding about Algebra.
Practice Problems on Linear Equations Substitution for Solving:
1) Solve the linear equation by substitution method.
1 - 2 / (a - c) = 8 at c=1.
Answer: a = 5/7.
2) Solve the linear equation by substitution method.
1 - 1 / (x - z) = -8 / (x 2 - z 2) at z=4
Answer: x = -3.
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